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K’th Smallest/Largest Element in Unsorted Array | Set 1

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  • Difficulty Level : Medium
  • Last Updated : 27 Jun, 2022

Given an array and a number k where k is smaller than the size of the array, we need to find the k’th smallest element in the given array. It is given that all array elements are distinct.

Examples:  

Input: arr[] = {7, 10, 4, 3, 20, 15}, k = 3 
Output: 7

Input: arr[] = {7, 10, 4, 3, 20, 15}, k = 4 
Output: 10 

We have discussed a similar problem to print k largest elements

Method 1 (Simple Solution) 
A simple solution is to sort the given array using an O(N log N) sorting algorithm like Merge Sort, Heap Sort, etc, and return the element at index k-1 in the sorted array. 
The Time Complexity of this solution is O(N log N) 

 

C++




// Simple C++ program to find k'th smallest element
#include <bits/stdc++.h>
using namespace std;
 
// Function to return k'th smallest element in a given array
int kthSmallest(int arr[], int n, int k)
{
    // Sort the given array
    sort(arr, arr + n);
 
    // Return k'th element in the sorted array
    return arr[k - 1];
}
 
// Driver program to test above methods
int main()
{
    int arr[] = { 12, 3, 5, 7, 19 };
    int n = sizeof(arr) / sizeof(arr[0]), k = 2;
    cout << "K'th smallest element is "
         << kthSmallest(arr, n, k);
    return 0;
}
 
// This code is contributed by Sania Kumari Gupta (kriSania804)


C




// Simple C program to find k'th smallest element
#include <stdio.h>
#include <stdlib.h>
 
// Compare function for qsort
int cmpfunc(const void* a, const void* b)
{
    return (*(int*)a - *(int*)b);
}
 
// Function to return k'th smallest element in a given array
int kthSmallest(int arr[], int n, int k)
{
    // Sort the given array
    qsort(arr, n, sizeof(int), cmpfunc);
 
    // Return k'th element in the sorted array
    return arr[k - 1];
}
 
// Driver program to test above methods
int main()
{
    int arr[] = { 12, 3, 5, 7, 19 };
    int n = sizeof(arr) / sizeof(arr[0]), k = 2;
    printf("K'th smallest element is %d",
           kthSmallest(arr, n, k));
    return 0;
}
 
// This code is contributed by Sania Kumari Gupta (kriSania804)


Java




// Java code for kth smallest element
// in an array
import java.util.Arrays;
import java.util.Collections;
 
class GFG {
    // Function to return k'th smallest
    // element in a given array
    public static int kthSmallest(Integer[] arr,
                                  int k)
    {
        // Sort the given array
        Arrays.sort(arr);
 
        // Return k'th element in
        // the sorted array
        return arr[k - 1];
    }
 
    // driver program
    public static void main(String[] args)
    {
        Integer arr[] = new Integer[] { 12, 3, 5, 7, 19 };
        int k = 2;
        System.out.print("K'th smallest element is " + kthSmallest(arr, k));
    }
}
 
// This code is contributed by Chhavi


Python3




# Python3 program to find k'th smallest
# element
 
# Function to return k'th smallest
# element in a given array
def kthSmallest(arr, n, k):
 
    # Sort the given array
    arr.sort()
 
    # Return k'th element in the
    # sorted array
    return arr[k-1]
 
# Driver code
if __name__=='__main__':
    arr = [12, 3, 5, 7, 19]
    n = len(arr)
    k = 2
    print("K'th smallest element is",
          kthSmallest(arr, n, k))
 
# This code is contributed by
# Shrikant13


C#




// C# code for kth smallest element
// in an array
using System;
 
class GFG {
 
    // Function to return k'th smallest
    // element in a given array
    public static int kthSmallest(int[] arr,
                                  int k)
    {
 
        // Sort the given array
        Array.Sort(arr);
 
        // Return k'th element in
        // the sorted array
        return arr[k - 1];
    }
 
    // driver program
    public static void Main()
    {
        int[] arr = new int[] { 12, 3, 5,
                                7, 19 };
        int k = 2;
        Console.Write("K'th smallest element"
                      + " is " + kthSmallest(arr, k));
    }
}
 
// This code is contributed by nitin mittal.


PHP




<?php
// Simple PHP program to find
// k'th smallest element
 
// Function to return k'th smallest
// element in a given array
function kthSmallest($arr, $n, $k)
{
     
    // Sort the given array
    sort($arr);
 
    // Return k'th element
    // in the sorted array
    return $arr[$k - 1];
}
 
    // Driver Code
    $arr = array(12, 3, 5, 7, 19);
    $n =count($arr);
    $k = 2;
    echo "K'th smallest element is ", kthSmallest($arr, $n, $k);
 
// This code is contributed by anuj_67.
?>


Javascript




<script>
 
// Simple Javascript program to find k'th smallest element
 
// Function to return k'th smallest element in a given array
function kthSmallest(arr, n, k)
{
    // Sort the given array
    arr.sort((a,b) => a-b);
 
    // Return k'th element in the sorted array
    return arr[k - 1];
}
 
// Driver program to test above methods
    let arr = [ 12, 3, 5, 7, 19 ];
    let n = arr.length, k = 2;
    document.write("K'th smallest element is " + kthSmallest(arr, n, k));
 
//This code is contributed by Mayank Tyagi
</script>


Output

K'th smallest element is 5

Method 2 (using set from C++ STL)

we can find the kth smallest element in time complexity better than O(N log N). we know the Set in C++ STL is implemented using Binary Search Tree and we also know that the time complexity of all cases(searching, inserting, deleting ) in BST is log (n) in the average case and O(n) in the worst case. We are using set because it is mentioned in the question that all the elements in an array are distinct.

The following is the C++ implementation of the above method.

C++




/* the following code demonstrates how to find kth smallest
element using set from C++ STL */
 
#include <bits/stdc++.h>
using namespace std;
 
int main()
{
 
    int arr[] = { 12, 3, 5, 7, 19 };
    int n = sizeof(arr) / sizeof(arr[0]);
    int k = 4;
 
    set<int> s(arr, arr + n);
    set<int>::iterator itr
        = s.begin(); // s.begin() returns a pointer to first
                     // element in the set
    advance(itr, k - 1); // itr points to kth element in set
 
    cout << *itr << "\n";
 
    return 0;
}


Output

12

Time Complexity:  O(N*log N) in Average Case and O(N) in Worst Case
Auxiliary Space: O(N)

Method 3 (Using Min Heap – HeapSelect) 
We can find k’th smallest element in time complexity better than O(N Log N). A simple optimization is to create a Min Heap of the given n elements and call extractMin() k times. 

The following is C++ implementation of above method.  

C++




// A C++ program to find k'th smallest element using min heap
#include <climits>
#include <iostream>
using namespace std;
 
// Prototype of a utility function to swap two integers
void swap(int* x, int* y);
 
// A class for Min Heap
class MinHeap {
    int* harr; // pointer to array of elements in heap
    int capacity; // maximum possible size of min heap
    int heap_size; // Current number of elements in min heap
public:
    MinHeap(int a[], int size); // Constructor
    void MinHeapify(int i); // To minheapify subtree rooted with index i
    int parent(int i) { return (i - 1) / 2; }
    int left(int i) { return (2 * i + 1); }
    int right(int i) { return (2 * i + 2); }
 
    int extractMin(); // extracts root (minimum) element
    int getMin() { return harr[0]; } // Returns minimum
};
 
MinHeap::MinHeap(int a[], int size)
{
    heap_size = size;
    harr = a; // store address of array
    int i = (heap_size - 1) / 2;
    while (i >= 0) {
        MinHeapify(i);
        i--;
    }
}
 
// Method to remove minimum element (or root) from min heap
int MinHeap::extractMin()
{
    if (heap_size == 0)
        return INT_MAX;
 
    // Store the minimum value.
    int root = harr[0];
 
    // If there are more than 1 items, move the last item to root
    // and call heapify.
    if (heap_size > 1) {
        harr[0] = harr[heap_size - 1];
        MinHeapify(0);
    }
    heap_size--;
 
    return root;
}
 
// A recursive method to heapify a subtree with root at given index
// This method assumes that the subtrees are already heapified
void MinHeap::MinHeapify(int i)
{
    int l = left(i);
    int r = right(i);
    int smallest = i;
    if (l < heap_size && harr[l] < harr[i])
        smallest = l;
    if (r < heap_size && harr[r] < harr[smallest])
        smallest = r;
    if (smallest != i) {
        swap(&harr[i], &harr[smallest]);
        MinHeapify(smallest);
    }
}
 
// A utility function to swap two elements
void swap(int* x, int* y)
{
    int temp = *x;
    *x = *y;
    *y = temp;
}
 
// Function to return k'th smallest element in a given array
int kthSmallest(int arr[], int n, int k)
{
    // Build a heap of n elements: O(n) time
    MinHeap mh(arr, n);
 
    // Do extract min (k-1) times
    for (int i = 0; i < k - 1; i++)
        mh.extractMin();
 
    // Return root
    return mh.getMin();
}
 
// Driver program to test above methods
int main()
{
    int arr[] = { 12, 3, 5, 7, 19 };
    int n = sizeof(arr) / sizeof(arr[0]), k = 2;
    cout << "K'th smallest element is " << kthSmallest(arr, n, k);
    return 0;
}


Java




// A Java program to find k'th smallest element using min heap
import java.util.*;
class GFG
{
 
  // A class for Max Heap
  class MinHeap
  {
    int[] harr; // pointer to array of elements in heap
    int capacity; // maximum possible size of min heap
    int heap_size; // Current number of elements in min heap
 
    int parent(int i) { return (i - 1) / 2; }
    int left(int i) { return ((2 * i )+ 1); }
    int right(int i) { return ((2 * i) + 2); }
    int getMin() { return harr[0]; } // Returns minimum
 
    // to replace root with new node x and heapify() new root
    void replaceMax(int x)
    {
      this.harr[0] = x;
      minHeapify(0);
    }
    MinHeap(int a[], int size)
    {
      heap_size = size;
      harr = a; // store address of array
      int i = (heap_size - 1) / 2;
      while (i >= 0)
      {
        minHeapify(i);
        i--;
      }
    }
 
    // Method to remove maximum element (or root) from min heap
    int extractMin()
    {
      if (heap_size == 0)
        return Integer.MAX_VALUE;
 
      // Store the maximum value.
      int root = harr[0];
 
      // If there are more than 1 items, move the last item to root
      // and call heapify.
      if (heap_size > 1)
      {
        harr[0] = harr[heap_size - 1];
        minHeapify(0);
      }
      heap_size--;
      return root;
    }
 
    // A recursive method to heapify a subtree with root at given index
    // This method assumes that the subtrees are already heapified
    void minHeapify(int i)
    {
      int l = left(i);
      int r = right(i);
      int smallest = i;
      if (l < heap_size && harr[l] < harr[i])
        smallest = l;
      if (r < heap_size && harr[r] < harr[smallest])
        smallest = r;
      if (smallest != i)
      {
        int t = harr[i];
        harr[i] = harr[smallest];
        harr[smallest] = t;
        minHeapify(smallest);
      }
    }
  };
 
  // Function to return k'th largest element in a given array
  int kthSmallest(int arr[], int n, int k)
  {
 
    // Build a heap of first k elements: O(k) time
    MinHeap mh = new MinHeap(arr, n);
 
    // Process remaining n-k elements. If current element is
    // smaller than root, replace root with current element
    for (int i = 0; i < k - 1; i++)
      mh.extractMin();
 
    // Return root
    return mh.getMin();
  }
 
  // Driver program to test above methods
  public static void main(String[] args)
  {
    int arr[] = { 12, 3, 5, 7, 19 };
    int n = arr.length, k = 2;
    GFG gfg = new GFG();
    System.out.print("K'th smallest element is " +
                     gfg.kthSmallest(arr, n, k));
  }
}
 
// This code is contributed by avanitrachhadiya2155


Python3




# Python3 program to find k'th smallest element
# using min heap
 
# Class for Min Heap
class MinHeap:
   
    # Constructor
    def __init__(self, a, size):
       
        # list of elements in the heap
        self.harr = a
         
        # maximum possible size of min heap
        self.capacity = None
         
        # current number of elements in min heap
        self.heap_size = size
 
        i = int((self.heap_size - 1) / 2)
        while i >= 0:
            self.minHeapify(i)
            i -= 1
 
    def parent(self, i):
        return (i - 1) / 2
 
    def left(self, i):
        return 2 * i + 1
 
    def right(self, i):
        return 2 * i + 2
 
    # Returns minimum
    def getMin(self):
        return self.harr[0]
 
    # Method to remove minimum element (or root)
    # from min heap
    def extractMin(self):
        if self.heap_size == 0:
            return float("inf")
 
        # Store the minimum value
        root = self.harr[0]
 
        # If there are more than 1 items, move the last item
        # to root and call heapify
        if self.heap_size > 1:
            self.harr[0] = self.harr[self.heap_size - 1]
            self.minHeapify(0)
        self.heap_size -= 1
        return root
 
    # A recursive method to heapify a subtree with root at
    # given index. This method assumes that the subtrees
    # are already heapified
    def minHeapify(self, i):
        l = self.left(i)
        r = self.right(i)
        smallest = i
        if ((l < self.heap_size) and
            (self.harr[l] < self.harr[i])):
            smallest = l
        if ((r < self.heap_size) and
            (self.harr[r] < self.harr[smallest])):
            smallest = r
        if smallest != i:
            self.harr[i], self.harr[smallest] = (
                self.harr[smallest], self.harr[i])
            self.minHeapify(smallest)
 
# Function to return k'th smallest element in a given array
def kthSmallest(arr, n, k):
   
    # Build a heap of n elements in O(n) time
    mh = MinHeap(arr, n)
 
    # Do extract min (k-1) times
    for i in range(k - 1):
        mh.extractMin()
 
    # Return root
    return mh.getMin()
 
# Driver code
arr = [12, 3, 5, 7, 19]
n = len(arr)
k = 2
print("K'th smallest element is", kthSmallest(arr, n, k))
 
# This Code is contributed by Kevin Joshi


C#




using System;
public class GFG
{
 
  public class MinHeap
  {
    int[] harr; // pointer to array of elements in heap
    //    int capacity; // maximum possible size of min heap
    int heap_size; // Current number of elements in min heap
 
    int parent(int i) { return (i - 1) / 2; }
    int left(int i) { return ((2 * i )+ 1); }
    int right(int i) { return ((2 * i) + 2); }
    public int getMin() { return harr[0]; } // Returns minimum
 
    // to replace root with new node x and heapify() new root
    public void replaceMax(int x)
    {
      this.harr[0] = x;
      minHeapify(0);
    }
 
    public MinHeap(int[] a, int size)
    {
      heap_size = size;
      harr = a; // store address of array
      int i = (heap_size - 1) / 2;
      while (i >= 0)
      {
        minHeapify(i);
        i--;
      }
    }
 
    // Method to remove maximum element (or root) from min heap
    public int extractMin()
    {
      if (heap_size == 0)
        return Int32.MaxValue;
 
      // Store the maximum value.
      int root = harr[0];
 
      // If there are more than 1 items, move the last item to root
      // and call heapify.
      if (heap_size > 1)
      {
        harr[0] = harr[heap_size - 1];
        minHeapify(0);
      }
      heap_size--;
      return root;
    }
 
    // A recursive method to heapify a subtree with root at given index
    // This method assumes that the subtrees are already heapified
    public void minHeapify(int i)
    {
      int l = left(i);
      int r = right(i);
      int smallest = i;
      if (l < heap_size && harr[l] < harr[i])
        smallest = l;
      if (r < heap_size && harr[r] < harr[smallest])
        smallest = r;
      if (smallest != i)
      {
        int t = harr[i];
        harr[i] = harr[smallest];
        harr[smallest] = t;
        minHeapify(smallest);
      }
    }
  };
 
  // Function to return k'th largest element in a given array
  int kthSmallest(int[] arr, int n, int k)
  {
 
    // Build a heap of first k elements: O(k) time
    MinHeap mh = new MinHeap(arr, n);
 
    // Process remaining n-k elements. If current element is
    // smaller than root, replace root with current element
    for (int i = 0; i < k - 1; i++)
      mh.extractMin();
 
    // Return root
    return mh.getMin();
  }
 
  // Driver program to test above methods
  static public void Main (){
    int[] arr = { 12, 3, 5, 7, 19 };
    int n = arr.Length, k = 2;
    GFG gfg = new GFG();
    Console.Write("K'th smallest element is " +
                  gfg.kthSmallest(arr, n, k));
  }
}
 
// This code is contributed by rag2127


Javascript




<script>
// A Javascript program to find k'th smallest element using min heap
 
// A class for Max Heap
class MinHeap {
 
 
    parent(i) {
        return (i - 1) / 2;
    };
    left(i) {
        return ((2 * i) + 1);
    };
    right(i) {
        return ((2 * i) + 2);
    }
    getMin() {
        return this.harr[0];
    } // Returns minimum
 
    // to replace root with new node x and heapify() new root
    replaceMax(x) {
        this.harr[0] = x;
        minHeapify(0);
    }
    constructor(a, size) {
        this.heap_size = size;
        this.harr = a; // store address of array
        let i = (this.heap_size - 1) / 2;
        while (i >= 0) {
            this.minHeapify(i);
            i--;
        }
    }
 
    // Method to remove maximum element (or root) from min heap
    extractMin() {
        if (this.heap_size == 0)
            return Number.MAX_SAFE_INTEGER;
 
        // Store the maximum value.
        let root = this.harr[0];
 
        // If there are more than 1 items, move the last item to root
        // and call heapify.
        if (this.heap_size > 1) {
            this.harr[0] = this.harr[this.heap_size - 1];
            this.minHeapify(0);
        }
        this.heap_size--;
        return root;
    }
 
    // A recursive method to heapify a subtree with root at given index
    // This method assumes that the subtrees are already heapified
    minHeapify(i) {
        let l = this.left(i);
        let r = this.right(i);
        let smallest = i;
        if (l < this.heap_size && this.harr[l] < this.harr[i])
            smallest = l;
        if (r < this.heap_size && this.harr[r] < this.harr[smallest])
            smallest = r;
        if (smallest != i) {
            let t = this.harr[i];
            this.harr[i] = this.harr[smallest];
            this.harr[smallest] = t;
            this.minHeapify(smallest);
        }
    }
};
 
// Function to return k'th largest element in a given array
function kthSmallest(arr, n, k) {
 
    // Build a heap of first k elements: O(k) time
    let mh = new MinHeap(arr, n);
 
    // Process remaining n-k elements. If current element is
    // smaller than root, replace root with current element
    for (let i = 0; i < k - 1; i++)
        mh.extractMin();
 
    // Return root
    return mh.getMin();
}
 
// Driver program to test above methods
let arr = [12, 3, 5, 7, 19];
let n = arr.length, k = 2;
document.write("K'th smallest element is " + kthSmallest(arr, n, k));
 
// This code is contributed by gfgking.
 
</script>


Output

K'th smallest element is 5

Time complexity: O(n + kLogn).

Space complexity: O(n) for call stack

 

Method 4 (Using Max-Heap) 
We can also use Max Heap for finding the k’th smallest element. Following is an algorithm. 
1) Build a Max-Heap MH of the first k elements (arr[0] to arr[k-1]) of the given array. O(k)
2) For each element, after the k’th element (arr[k] to arr[n-1]), compare it with root of MH. 
……a) If the element is less than the root then make it root and call heapify for MH 
……b) Else ignore it. 
// The step 2 is O((n-k)*logk)
3) Finally, the root of the MH is the kth smallest element.
Time complexity of this solution is O(k + (n-k)*Logk)

The following is C++ implementation of the above algorithm 

C++




// A C++ program to find k'th smallest element using max heap
#include <climits>
#include <iostream>
using namespace std;
 
// Prototype of a utility function to swap two integers
void swap(int* x, int* y);
 
// A class for Max Heap
class MaxHeap {
    int* harr; // pointer to array of elements in heap
    int capacity; // maximum possible size of max heap
    int heap_size; // Current number of elements in max heap
public:
    MaxHeap(int a[], int size); // Constructor
    void maxHeapify(int i); // To maxHeapify subtree rooted with index i
    int parent(int i) { return (i - 1) / 2; }
    int left(int i) { return (2 * i + 1); }
    int right(int i) { return (2 * i + 2); }
 
    int extractMax(); // extracts root (maximum) element
    int getMax() { return harr[0]; } // Returns maximum
 
    // to replace root with new node x and heapify() new root
    void replaceMax(int x)
    {
        harr[0] = x;
        maxHeapify(0);
    }
};
 
MaxHeap::MaxHeap(int a[], int size)
{
    heap_size = size;
    harr = a; // store address of array
    int i = (heap_size - 1) / 2;
    while (i >= 0) {
        maxHeapify(i);
        i--;
    }
}
 
// Method to remove maximum element (or root) from max heap
int MaxHeap::extractMax()
{
    if (heap_size == 0)
        return INT_MAX;
 
    // Store the maximum value.
    int root = harr[0];
 
    // If there are more than 1 items, move the last item to root
    // and call heapify.
    if (heap_size > 1) {
        harr[0] = harr[heap_size - 1];
        maxHeapify(0);
    }
    heap_size--;
 
    return root;
}
 
// A recursive method to heapify a subtree with root at given index
// This method assumes that the subtrees are already heapified
void MaxHeap::maxHeapify(int i)
{
    int l = left(i);
    int r = right(i);
    int largest = i;
    if (l < heap_size && harr[l] > harr[i])
        largest = l;
    if (r < heap_size && harr[r] > harr[largest])
        largest = r;
    if (largest != i) {
        swap(&harr[i], &harr[largest]);
        maxHeapify(largest);
    }
}
 
// A utility function to swap two elements
void swap(int* x, int* y)
{
    int temp = *x;
    *x = *y;
    *y = temp;
}
 
// Function to return k'th largest element in a given array
int kthSmallest(int arr[], int n, int k)
{
    // Build a heap of first k elements: O(k) time
    MaxHeap mh(arr, k);
 
    // Process remaining n-k elements.  If current element is
    // smaller than root, replace root with current element
    for (int i = k; i < n; i++)
        if (arr[i] < mh.getMax())
            mh.replaceMax(arr[i]);
 
    // Return root
    return mh.getMax();
}
 
// Driver program to test above methods
int main()
{
    int arr[] = { 12, 3, 5, 7, 19 };
    int n = sizeof(arr) / sizeof(arr[0]), k = 4;
    cout << "K'th smallest element is " << kthSmallest(arr, n, k);
    return 0;
}


Java




// A Java program to find k'th smallest element using max heap
import java.util.*;
class GFG
{
 
  // A class for Max Heap
  class MaxHeap
  {
    int[] harr; // pointer to array of elements in heap
    int capacity; // maximum possible size of max heap
    int heap_size; // Current number of elements in max heap
 
    int parent(int i) { return (i - 1) / 2; }
    int left(int i) { return (2 * i + 1); }
    int right(int i) { return (2 * i + 2); }
    int getMax() { return harr[0]; } // Returns maximum
 
    // to replace root with new node x and heapify() new root
    void replaceMax(int x)
    {
      this.harr[0] = x;
      maxHeapify(0);
    }
    MaxHeap(int a[], int size)
    {
      heap_size = size;
      harr = a; // store address of array
      int i = (heap_size - 1) / 2;
      while (i >= 0)
      {
        maxHeapify(i);
        i--;
      }
    }
 
    // Method to remove maximum element (or root) from max heap
    int extractMax()
    {
      if (heap_size == 0)
        return Integer.MAX_VALUE;
 
      // Store the maximum value.
      int root = harr[0];
 
      // If there are more than 1 items, move the last item to root
      // and call heapify.
      if (heap_size > 1)
      {
        harr[0] = harr[heap_size - 1];
        maxHeapify(0);
      }
      heap_size--;
      return root;
    }
 
    // A recursive method to heapify a subtree with root at given index
    // This method assumes that the subtrees are already heapified
    void maxHeapify(int i)
    {
      int l = left(i);
      int r = right(i);
      int largest = i;
      if (l < heap_size && harr[l] > harr[i])
        largest = l;
      if (r < heap_size && harr[r] > harr[largest])
        largest = r;
      if (largest != i)
      {
        int t = harr[i];
        harr[i] = harr[largest];
        harr[largest] = t;
        maxHeapify(largest);
      }
    }
  };
 
  // Function to return k'th largest element in a given array
  int kthSmallest(int arr[], int n, int k)
  {
 
    // Build a heap of first k elements: O(k) time
    MaxHeap mh = new MaxHeap(arr, k);
 
    // Process remaining n-k elements.  If current element is
    // smaller than root, replace root with current element
    for (int i = k; i < n; i++)
      if (arr[i] < mh.getMax())
        mh.replaceMax(arr[i]);
 
    // Return root
    return mh.getMax();
  }
 
  // Driver program to test above methods
  public static void main(String[] args)
  {
    int arr[] = { 12, 3, 5, 7, 19 };
    int n = arr.length, k = 4;
    GFG gfg = new GFG();
    System.out.print("K'th smallest element is "
                     gfg.kthSmallest(arr, n, k));
  }
}
 
// This code is contributed by Rajput-Ji


Python3




# Python3 program to find k'th smallest element
# using max heap
 
# Class for Max Heap
class MaxHeap:
    # Constructor
    def __init__(self, a, size):
        # list of elements in the heap
        self.harr = a
        # maximum possible size of max heap
        self.capacity = None
        # current number of elements in max heap
        self.heap_size = size
 
        i = int((self.heap_size - 1) / 2)
        while i >= 0:
            self.maxHeapify(i)
            i -= 1
 
    def parent(self, i):
        return (i - 1) / 2
 
    def left(self, i):
        return 2 * i + 1
 
    def right(self, i):
        return 2 * i + 2
 
    # Returns maximum
    def getMax(self):
        return self.harr[0]
 
    # to replace root with new node x and heapify() new root
    def replaceMax(self, x):
        self.harr[0] = x
        self.maxHeapify(0)
 
    # Method to remove maximum element (or root)
    # from max heap
    def extractMin(self):
        if self.heap_size == 0:
            return float("inf")
 
        # Store the maximum value.
        root = self.harr[0]
 
        # If there are more than 1 items, move the
        # last item to root and call heapify
        if self.heap_size > 1:
            self.harr[0] = self.harr[self.heap_size - 1]
            self.maxHeapify(0)
        self.heap_size -= 1
        return root
 
    # A recursive method to heapify a subtree with root at
    # given index. This method assumes that the subtrees
    # are already heapified
    def maxHeapify(self, i):
        l = self.left(i)
        r = self.right(i)
        largest = i
        if ((l < self.heap_size) and
            (self.harr[l] > self.harr[i])):
            largest = l
        if ((r < self.heap_size) and
            (self.harr[r] > self.harr[largest])):
            largest = r
        if largest != i:
            self.harr[i], self.harr[largest] = (
                self.harr[largest], self.harr[i])
            self.maxHeapify(largest)
 
 
# Function to return k'th smallest element in a given array
def kthSmallest(arr, n, k):
    # Build a heap of first k elements in O(k) time
    mh = MaxHeap(arr, k)
 
    # Process remaining n-k elements. If current element is
    # smaller than root, replace root with current element
    for i in range(k, n):
        if arr[i] < mh.getMax():
            mh.replaceMax(arr[i])
 
    # Return root
    return mh.getMax()
 
 
# Driver code
arr = [12, 3, 5, 7, 19]
n = len(arr)
k = 4
print("K'th smallest element is", kthSmallest(arr, n, k))
 
# Code contributed by Kevin Joshi


C#




// A C# program to find k'th smallest element using max heap
using System;
 
public class GFG {
 
    // A class for Max Heap
    public
 
        class MaxHeap {
        public
 
            int[] harr; // pointer to array of elements in
                        // heap
        public
 
            int capacity; // maximum possible size of max
                          // heap
        public
 
            int heap_size; // Current number of elements in
                           // max heap
 
        public
 
            int
            parent(int i)
        {
            return (i - 1) / 2;
        }
        public
 
            int
            left(int i)
        {
            return (2 * i + 1);
        }
        public
 
            int
            right(int i)
        {
            return (2 * i + 2);
        }
        public
 
            int
            getMax()
        {
            return harr[0];
        } // Returns maximum
 
        // to replace root with new node x and heapify() new
        // root
        public
 
            void
            replaceMax(int x)
        {
            this.harr[0] = x;
            maxHeapify(0);
        }
        public
 
            MaxHeap(int[] a, int size)
        {
            heap_size = size;
            harr = a; // store address of array
            int i = (heap_size - 1) / 2;
            while (i >= 0) {
                maxHeapify(i);
                i--;
            }
        }
 
        // Method to remove maximum element (or root) from
        // max heap
        public
 
            int
            extractMax()
        {
            if (heap_size == 0)
                return int.MaxValue;
 
            // Store the maximum value.
            int root = harr[0];
 
            // If there are more than 1 items, move the last
            // item to root and call heapify.
            if (heap_size > 1) {
                harr[0] = harr[heap_size - 1];
                maxHeapify(0);
            }
            heap_size--;
            return root;
        }
 
        // A recursive method to heapify a subtree with root
        // at given index This method assumes that the
        // subtrees are already heapified
        public
 
            void
            maxHeapify(int i)
        {
            int l = left(i);
            int r = right(i);
            int largest = i;
            if (l < heap_size && harr[l] > harr[i])
                largest = l;
            if (r < heap_size && harr[r] > harr[largest])
                largest = r;
            if (largest != i) {
                int t = harr[i];
                harr[i] = harr[largest];
                harr[largest] = t;
                maxHeapify(largest);
            }
        }
    };
 
    // Function to return k'th largest element in a given
    // array
    int kthSmallest(int[] arr, int n, int k)
    {
 
        // Build a heap of first k elements: O(k) time
        MaxHeap mh = new MaxHeap(arr, k);
 
        // Process remaining n-k elements.  If current
        // element is smaller than root, replace root with
        // current element
        for (int i = k; i < n; i++)
            if (arr[i] < mh.getMax())
                mh.replaceMax(arr[i]);
 
        // Return root
        return mh.getMax();
    }
 
    // Driver code
    public static void Main(String[] args)
    {
        int[] arr = { 12, 3, 5, 7, 19 };
        int n = arr.Length, k = 4;
        GFG gfg = new GFG();
        Console.Write("K'th smallest element is "
                      + gfg.kthSmallest(arr, n, k));
    }
}
 
// This code is contributed by gauravrajput1


Output

K'th smallest element is 12

Method 5 (QuickSelect) 
This is an optimization over method 1 if QuickSort is used as a sorting algorithm in first step. In QuickSort, we pick a pivot element, then move the pivot element to its correct position and partition the surrounding array. The idea is, not to do complete quicksort, but stop at the point where pivot itself is k’th smallest element. Also, not to recur for both left and right sides of pivot, but recur for one of them according to the position of pivot. The worst case time complexity of this method is O(n2), but it works in O(n) on average. 

C++




#include <climits>
#include <iostream>
using namespace std;
 
int partition(int arr[], int l, int r);
 
// This function returns k'th smallest element in arr[l..r] using
// QuickSort based method.  ASSUMPTION: ALL ELEMENTS IN ARR[] ARE DISTINCT
int kthSmallest(int arr[], int l, int r, int k)
{
    // If k is smaller than number of elements in array
    if (k > 0 && k <= r - l + 1) {
        // Partition the array around last element and get
        // position of pivot element in sorted array
        int pos = partition(arr, l, r);
 
        // If position is same as k
        if (pos - l == k - 1)
            return arr[pos];
        if (pos - l > k - 1) // If position is more, recur for left subarray
            return kthSmallest(arr, l, pos - 1, k);
 
        // Else recur for right subarray
        return kthSmallest(arr, pos + 1, r, k - pos + l - 1);
    }
 
    // If k is more than number of elements in array
    return INT_MAX;
}
 
void swap(int* a, int* b)
{
    int temp = *a;
    *a = *b;
    *b = temp;
}
 
// Standard partition process of QuickSort().  It considers the last
// element as pivot and moves all smaller element to left of it
// and greater elements to right
int partition(int arr[], int l, int r)
{
    int x = arr[r], i = l;
    for (int j = l; j <= r - 1; j++) {
        if (arr[j] <= x) {
            swap(&arr[i], &arr[j]);
            i++;
        }
    }
    swap(&arr[i], &arr[r]);
    return i;
}
 
// Driver program to test above methods
int main()
{
    int arr[] = { 12, 3, 5, 7, 4, 19, 26 };
    int n = sizeof(arr) / sizeof(arr[0]), k = 3;
    cout << "K'th smallest element is " << kthSmallest(arr, 0, n - 1, k);
    return 0;
}


Java




// Java code for kth smallest element in an array
import java.util.Arrays;
import java.util.Collections;
 
class GFG {
    // Standard partition process of QuickSort.
    // It considers the last element as pivot
    // and moves all smaller element to left of
    // it and greater elements to right
    public static int partition(Integer[] arr, int l,
                                int r)
    {
        int x = arr[r], i = l;
        for (int j = l; j <= r - 1; j++) {
            if (arr[j] <= x) {
                // Swapping arr[i] and arr[j]
                int temp = arr[i];
                arr[i] = arr[j];
                arr[j] = temp;
 
                i++;
            }
        }
 
        // Swapping arr[i] and arr[r]
        int temp = arr[i];
        arr[i] = arr[r];
        arr[r] = temp;
 
        return i;
    }
 
    // This function returns k'th smallest element
    // in arr[l..r] using QuickSort based method.
    // ASSUMPTION: ALL ELEMENTS IN ARR[] ARE DISTINCT
    public static int kthSmallest(Integer[] arr, int l,
                                  int r, int k)
    {
        // If k is smaller than number of elements
        // in array
        if (k > 0 && k <= r - l + 1) {
            // Partition the array around last
            // element and get position of pivot
            // element in sorted array
            int pos = partition(arr, l, r);
 
            // If position is same as k
            if (pos - l == k - 1)
                return arr[pos];
 
            // If position is more, recur for
            // left subarray
            if (pos - l > k - 1)
                return kthSmallest(arr, l, pos - 1, k);
 
            // Else recur for right subarray
            return kthSmallest(arr, pos + 1, r, k - pos + l - 1);
        }
 
        // If k is more than number of elements
        // in array
        return Integer.MAX_VALUE;
    }
 
    // Driver program to test above methods
    public static void main(String[] args)
    {
        Integer arr[] = new Integer[] { 12, 3, 5, 7, 4, 19, 26 };
        int k = 3;
        System.out.print("K'th smallest element is " + kthSmallest(arr, 0, arr.length - 1, k));
    }
}
 
// This code is contributed by Chhavi


Python3




# This function returns k'th smallest element
# in arr[l..r] using QuickSort based method.
# ASSUMPTION: ALL ELEMENTS IN ARR[] ARE DISTINCT
import sys
 
def kthSmallest(arr, l, r, k):
 
    # If k is smaller than number of
    # elements in array
    if (k > 0 and k <= r - l + 1):
     
        # Partition the array around last
        # element and get position of pivot
        # element in sorted array
        pos = partition(arr, l, r)
 
        # If position is same as k
        if (pos - l == k - 1):
            return arr[pos]
        if (pos - l > k - 1): # If position is more,
                              # recur for left subarray
            return kthSmallest(arr, l, pos - 1, k)
 
        # Else recur for right subarray
        return kthSmallest(arr, pos + 1, r,
                            k - pos + l - 1)
 
    # If k is more than number of
    # elements in array
    return sys.maxsize
 
# Standard partition process of QuickSort().
# It considers the last element as pivot and
# moves all smaller element to left of it
# and greater elements to right
def partition(arr, l, r):
 
    x = arr[r]
    i = l
    for j in range(l, r):
        if (arr[j] <= x):
            arr[i], arr[j] = arr[j], arr[i]
            i += 1
    arr[i], arr[r] = arr[r], arr[i]
    return i
 
# Driver Code
if __name__ == "__main__":
     
    arr = [12, 3, 5, 7, 4, 19, 26]
    n = len(arr)
    k = 3;
    print("K'th smallest element is",
           kthSmallest(arr, 0, n - 1, k))
 
# This code is contributed by ita_c


C#




// C# code for kth smallest element
// in an array
using System;
 
class GFG {
 
    // Standard partition process of QuickSort.
    // It considers the last element as pivot
    // and moves all smaller element to left of
    // it and greater elements to right
    public static int partition(int[] arr,
                                int l, int r)
    {
        int x = arr[r], i = l;
        int temp = 0;
        for (int j = l; j <= r - 1; j++) {
 
            if (arr[j] <= x) {
                // Swapping arr[i] and arr[j]
                temp = arr[i];
                arr[i] = arr[j];
                arr[j] = temp;
 
                i++;
            }
        }
 
        // Swapping arr[i] and arr[r]
        temp = arr[i];
        arr[i] = arr[r];
        arr[r] = temp;
 
        return i;
    }
 
    // This function returns k'th smallest
    // element in arr[l..r] using QuickSort
    // based method. ASSUMPTION: ALL ELEMENTS
    // IN ARR[] ARE DISTINCT
    public static int kthSmallest(int[] arr, int l,
                                  int r, int k)
    {
        // If k is smaller than number
        // of elements in array
        if (k > 0 && k <= r - l + 1) {
            // Partition the array around last
            // element and get position of pivot
            // element in sorted array
            int pos = partition(arr, l, r);
 
            // If position is same as k
            if (pos - l == k - 1)
                return arr[pos];
 
            // If position is more, recur for
            // left subarray
            if (pos - l > k - 1)
                return kthSmallest(arr, l, pos - 1, k);
 
            // Else recur for right subarray
            return kthSmallest(arr, pos + 1, r,
                               k - pos + l - 1);
        }
 
        // If k is more than number
        // of elements in array
        return int.MaxValue;
    }
 
    // Driver Code
    public static void Main()
    {
        int[] arr = { 12, 3, 5, 7, 4, 19, 26 };
        int k = 3;
        Console.Write("K'th smallest element is " + kthSmallest(arr, 0, arr.Length - 1, k));
    }
}
 
// This code is contributed
// by 29AjayKumar


Javascript




<script>
 
// JavaScript code for kth smallest
// element in an array
 
    // Standard partition process of QuickSort.
    // It considers the last element as pivot
    // and moves all smaller element to left of
    // it and greater elements to right
    function partition( arr , l , r)
    {
        var x = arr[r], i = l;
        for (j = l; j <= r - 1; j++) {
            if (arr[j] <= x) {
                // Swapping arr[i] and arr[j]
                var temp = arr[i];
                arr[i] = arr[j];
                arr[j] = temp;
 
                i++;
            }
        }
 
        // Swapping arr[i] and arr[r]
        var temp = arr[i];
        arr[i] = arr[r];
        arr[r] = temp;
 
        return i;
    }
 
    // This function returns k'th smallest element
    // in arr[l..r] using QuickSort based method.
    // ASSUMPTION: ALL ELEMENTS IN ARR ARE DISTINCT
    function kthSmallest( arr , l , r , k) {
        // If k is smaller than number of elements
        // in array
        if (k > 0 && k <= r - l + 1) {
            // Partition the array around last
            // element and get position of pivot
            // element in sorted array
            var pos = partition(arr, l, r);
 
            // If position is same as k
            if (pos - l == k - 1)
                return arr[pos];
 
            // If position is more, recur for
            // left subarray
            if (pos - l > k - 1)
                return kthSmallest(arr, l, pos - 1, k);
 
            // Else recur for right subarray
            return kthSmallest(arr, pos + 1, r,
            k - pos + l - 1);
        }
 
        // If k is more than number of elements
        // in array
        return Number.MAX_VALUE;
    }
 
    // Driver program to test above methods
     
        var arr = [ 12, 3, 5, 7, 4, 19, 26 ];
        var k = 3;
        document.write("K'th smallest element is " +
        kthSmallest(arr, 0, arr.length - 1, k));
 
// This code contributed by Rajput-Ji
 
</script>


Output

K'th smallest element is 5

Method 6 (Map STL) 

A map-based STL approach is although very much similar to the quickselect and counting sort algorithm but much easier to implement. We can use an ordered map and map each element with its frequency. And as we know that an ordered map would store the data in a sorted manner, we keep on adding the frequency of each element till it does not become greater than or equal to k so that we reach the k’th element from the start i.e. the k’th smallest element.

Eg –

Array={7,0,25,6,16,17,0}

k=3

K’th Smallest/Largest Element in Unsorted Array

Now in order to get the k’th largest element, we need to add the frequencies till it becomes greater than or equal to 3. It is clear from the above that the frequency of 0 + frequency of 6 will become equal to 3 so the third smallest number in the array will be 6.

We can achieve the above using an iterator to traverse the map.

C++




#include <bits/stdc++.h>
using namespace std;
int Kth_smallest(map<int, int> m, int k)
{
    int freq = 0;
    for (auto it = m.begin(); it != m.end(); it++) {
        freq += (it->second); // adding the frequencies of
                              // each element
        if (freq >= k) // if at any point frequency becomes
                       // greater than or equal to k then
                       // return that element
        {
            return it->first;
        }
    }
    return -1; // returning -1 if k>size of the array which
               // is an impossible scenario
}
int main()
{
    int n = 5;
    int k = 2;
    vector<int> arr = { 12, 3, 5, 7, 19 };
    map<int, int> m;
    for (int i = 0; i < n; i++) {
        m[arr[i]] += 1; // mapping every element with it's
                        // frequency
    }
    int ans = Kth_smallest(m, k);
  if(k==1){
    cout << "The " << k << "st smallest element is " << ans
         << endl;
  }
  else if(k==2){
    cout << "The " << k << "nd smallest element is " << ans
         << endl;
  }
  else if(k==3){
    cout << "The " << k << "rd smallest element is " << ans
         << endl;
  }
  else{
    cout << "The " << k << "th smallest element is " << ans
         << endl;
  }
    return 0;
}


Java




// Java program for the above approach
import java.util.*;
 
class GFG {
    static int Kth_smallest(TreeMap<Integer, Integer> m,
                            int k)
    {
        int freq = 0;
        for (Map.Entry it : m.entrySet())
        {
           
            // adding the frequencies of each element
            freq += (int)it.getValue();
 
            // if at any point frequency becomes
            // greater than or equal to k then
            // return that element
            if (freq >= k) {
                return (int)it.getKey();
            }
        }
 
        return -1; // returning -1 if k>size of the array
                   // which is an impossible scenario
    }
   
  // Driver code
    public static void main(String[] args)
    {
        int n = 5;
        int k = 2;
        int[] arr = { 12, 3, 5, 7, 19 };
        TreeMap<Integer, Integer> m = new TreeMap<>();
        for (int i = 0; i < n; i++) {
            // mapping every element with
            // it's
            // frequency
            m.put(arr[i], m.getOrDefault(arr[i], 0) + 1);
        }
        int ans = Kth_smallest(m, k);
        if(k==1){
    System.out.println(
            "The " + k + "st smallest element is " + ans);
  }
  else if(k==2){
    System.out.println(
            "The " + k + "nd smallest element is " + ans);
  }
  else if(k==3){
    System.out.println(
            "The " + k + "rd smallest element is " + ans);
  }
  else{
    System.out.println(
            "The " + k + "th smallest element is " + ans);
  }
         
    }
}
 
// This code is contributed by harshit17.


Output

The 2nd smallest element is 5

There are two more solutions that are better than the above-discussed ones: One solution is to do a randomized version of quickSelect() and the other solution is the worst-case linear time algorithm (see the following posts).

Method 7 (Max heap using STL):

We can implement max and min heap using a priority queue.
To find the kth minimum element in an array we will max heapify the array until the size of the heap becomes k. 
After that for each entry we will pop the top element from the heap/Priority Queue. 

Below is the implementation of the above approach:

C++




// C++ code to implement the approach
#include<bits/stdc++.h>
using namespace std;
 
// Function to find the kth smallest array element
int kthSmallest(int arr[], int n, int k) {
   
    // For finding min element we need (Max heap)priority queue
    priority_queue<int> pq;
     
    for(int i = 0; i < k; i++)
    {           
        // First push first K elememts into heap
        pq.push(arr[i]); 
    }
    // Now check from k to last element
    for(int i = k; i < n; i++) 
    {
       
        // If current element is < top that means
        // there are  other k-1 lesser elements
        // are present at bottom thus, pop that element
        // and add kth largest element into the heap till curr
        // at last all the greater element than kth element will get pop off
          // and at the top of heap there will be kth smallest element
        if(arr[i] < pq.top())
        {
            pq.pop();
            // Push curr element
            pq.push(arr[i]);  
        }
    }
   
    // Return top of element
    return pq.top();   
}
 
 
// Driver's code:
int main()
{
    int n = 10;
    int arr[n] = {10, 5, 4 , 3 ,48, 6 , 2 , 33, 53, 10};
    int k = 4;
    cout<< "Kth Smallest Element is: "<<kthSmallest(arr, n, k);
 
}


Java




// Java code to implement the approach
import java.util.*;
 
//Custom comparator class to form the Max heap
class MinHeapComparator implements Comparator<Integer> {
 
    @Override
    public int compare(Integer number1, Integer number2) {
        int value = number1.compareTo(number2);
     
        // Elements are sorted in reverse order
        if (value > 0) {
            return -1;
        }
        else if (value < 0) {
            return 1;
        }
        else {
            return 0;
        }
    }
}
class GFG{
 
// Function to find kth smallest array element
static int kthSmallest(int []v, int N, int K)
{
     //For finding min element we need (Max heap)priority queue
    PriorityQueue<Integer> heap1 = new PriorityQueue<Integer>(new MinHeapComparator());
    
    for (int i = 0; i < N; ++i) {
 
        // Insert elements into
        // the priority queue
        heap1.add(v[i]);
 
        //If current element is less than top, that means there are
        //other k-1 lesser elements are present at bottom
        // thus pop that element and add kth largest element into the heap till curr
        // at last all the greater element than kth element will get pop off
        // and at the top of heap there will be kth smallest element
        if (heap1.size() > K) {
            heap1.remove();
        }
    }
 
    //Return the top of the heap as kth smallest element
    return heap1.peek();
}
 
// Driver code
public static void main(String[] args)
{
    // Given array
    int []vec = {10, 5, 4 , 3 ,48, 15, 6 , 2 , 33, 53, 10};
 
    // Size of array
    int N = vec.length;
 
    // Given K
    int K = 4;
 
    // Function Call
    System.out.println("Kth Smallest Element: " + kthSmallest(vec, N, K)) ;
}
}


Python3




# Python code to implement the approach
import heapq
 
# Function to find the kth smallest array element
def kthSmallest(arr, n, k):
   
    # For finding min element we need (Max heap)priority queue
    pq = []
    for i in range(k):
       
        # First push first K elememts into heap
        heapq.heappush(pq, arr[i])
        heapq._heapify_max(pq)
         
    # Now check from k to last element
    for i in range(k, n):
       
        # If current element is < first that means
        # there are  other k-1 lesser elements
        # are present at bottom thus, pop that element
        # and add kth largest element into the heap till curr
        # at last all the greater element than kth element will get pop off
        # and at the top of heap there will be kth smallest element
        if arr[i] < pq[0]:
            heapq.heappop(pq)
             
            # Push curr element
            heapq.heappush(pq, arr[i])
            heapq._heapify_max(pq)
    # Return first of element
    return pq[0]
 
# Driver's code:
n = 10
arr = [10, 5, 4, 3, 48, 6, 2, 33, 53, 10]
k = 4
print("Kth Smallest Element is:", kthSmallest(arr, n, k))
 
 
# This code is contributed by Tapesh(tapeshdua420)


C#




// C# program to implement
// the above approach
using System;
using System.Collections.Generic;
 
class GFG
{
  // Function to find the kth smallest array element
  static int kthSmallest(int[] arr, int n, int k) {
 
    // For finding min element we need (Max heap)priority queue
    List<int> pq = new List<int>();
 
    for(int i = 0; i < k; i++)
    {          
      // First push first K elememts into heap
      pq.Add(arr[i]);
    }
    // Now check from k to last element
    for(int i = k; i < n; i++)
    {
 
      // If current element is < top that means
      // there are  other k-1 lesser elements
      // are present at bottom thus, pop that element
      // and add kth largest element into the heap till curr
      // at last all the greater element than kth element will get pop off
      // and at the top of heap there will be kth smallest element
      if(arr[i] < pq[0])
      {
        pq.Sort();
        pq.Reverse();
 
        pq.RemoveAt(0);
        // Push curr element
        pq.Add(arr[i]); 
      }
    }
 
    // Return top of element
    return pq[0];  
  }
 
  // Driver Code
  public static void Main()
  {
    // Given array
    int []vec = {10, 5, 4 , 3 ,48, 15, 6 , 2 , 33, 53, 10};
 
    // Size of array
    int N = vec.Length;
 
    // Given K
    int K = 4;
 
    // Function Call
    Console.WriteLine("Kth Smallest Element: " + kthSmallest(vec, N, K)) ;
  }
}
 
// This code is contributed by sanjoy_62.


Output

Kth Smallest Element is: 5

Time complexity: O(nlogk)
Auxiliary Space: O(logK)

This method is contributed by rupeshsk30.

Method 8(Using Binary Search):
The idea to solve this problem is that the Kth smallest element would be the element at the kth position if the array was sorted in increasing order. Using this logic, we use binary search to predict the index of an element as if the array was sorted but without actually sorting the array. 
Example: {1, 4, 5, 3, 19, 3} & k = 2 
Here we find that element which has exactly k + 1 elements (including itself) lesser to it. Hence, the kth smallest element would be 3 in this case. 
Follow the steps below to implement the above idea:

  1. Find low and high that is the range where our answer can lie. 
  2.  Apply Binary Search on this range. 
  3.  If the selected element which would be mid has less than k elements lesser to it then increase the number that is low = mid + 1.
  4. Otherwise, Decrement the number and try to find a better answer (to understand this please try running on an array containing duplicates).
  5. The Binary Search will end when only one element remains in the answer space which would be our answer.

Below is the implementation of the above approach:

C++




#include <iostream>
#include<bits/stdc++.h>
using namespace std;
 
int count(vector <int>& nums, int& mid)
{//function to calculate number of elements less than equal to mid
        int cnt = 0;
         
        for(int i = 0; i < nums.size(); i++)
           if(nums[i] <= mid)
              cnt++;
         
        return cnt;
}
     
 
int kthSmallest(vector <int> nums, int& k)
{
        int low = INT_MAX;
        int high = INT_MIN;
        //calculate minimum and maximum the array.
        for(int i = 0; i < nums.size(); i++)
        {
            low = min(low, nums[i]);
            high = max(high, nums[i]);
        }
        //Our answer range lies between minimum and maximum element of the array on which Binary Search is Applied
        while(low < high)
        {
            int mid = low + (high - low) / 2;
           /*if the count of number of elements in the array less than equal to mid is less than k
             then increase the number. Otherwise decrement the number and try to find a better answer.
           */
            if(count(nums, mid) < k)
               low = mid + 1;
                
            else
                high = mid;
        }
         
        return low;
}
 
int main()
{
 
    vector <int> nums{1, 4, 5, 3, 19, 3};
    int k = 3;
    cout << "K'th smallest element is " << kthSmallest(nums, k);
    return 0;
}
 
// This code is contributed by garvjuneja98


Java




// Java code for kth smallest element in an array
import java.util.Arrays;
import java.util.Collections;
 
class GFG {
    static int count(int [] nums, int mid)
    {
        // function to calculate number of elements less than equal to mid
            int cnt = 0;
              
            for(int i = 0; i < nums.length; i++)
               if(nums[i] <= mid)
                  cnt++;
              
            return cnt;
    }
          
      
    static int kthSmallest(int [] nums, int k)
    {
            int low = Integer.MAX_VALUE;
            int high = Integer.MIN_VALUE;
            //calculate minimum and maximum the array.
            for(int i = 0; i < nums.length; i++)
            {
                low = Math.min(low, nums[i]);
                high = Math.max(high, nums[i]);
            }
            //Our answer range lies between minimum and maximum element of the array on which Binary Search is Applied
            while(low < high)
            {
                int mid = low + (high - low) / 2;
               /*if the count of number of elements in the array less than equal to mid is less than k
                 then increase the number. Otherwise decrement the number and try to find a better answer.
               */
                if(count(nums, mid) < k)
                   low = mid + 1;
                     
                else
                    high = mid;
            }
              
            return low;
    }
 
    // Driver program to test above methods
    public static void main(String[] args)
    {
        int arr[] = {1, 4, 5, 3, 19, 3};
        int k = 3;
        System.out.print("Kth smallest element is " + kthSmallest(arr, k));
    }
}
 
// This code is contributed by CodeWithMini


Python3




# Python code for kth smallest element in an array
import sys
 
# function to calculate number of elements
# less than equal to mid
def count(nums, mid):
    cnt = 0
    for i in range(len(nums)):
        if nums[i] <= mid:
            cnt += 1
    return cnt
 
def kthSmallest(nums, k):
    low = sys.maxsize
    high = -sys.maxsize - 1
     
    # calculate minimum and maximum the array.
    for i in range(len(nums)):
        low = min(low, nums[i])
        high = max(high, nums[i])
 
        # Our answer range lies between minimum and maximum element
        # of the array on which Binary Search is Applied
    while low < high:
        mid = low + (high - low) // 2
        # if the count of number of elements in the array less than equal
        # to mid is less than k then increase the number. Otherwise decrement
        # the number and try to find a better answer.
        if count(nums, mid) < k:
            low = mid + 1
        else:
            high = mid
    return low
 
nums = [1, 4, 5, 3, 19, 3]
k = 3
print("K'th smallest element is", kthSmallest(nums, k))
 
# This code is contributed by Tapesh(tapeshdua420)


C#




// C# program to implement
// the above approach
using System;
using System.Collections.Generic;
 
class GFG
{
  static int count(int []nums, int mid)
  {
    // function to calculate number of elements less than equal to mid
    int cnt = 0;
 
    for(int i = 0; i < nums.Length; i++)
      if(nums[i] <= mid)
        cnt++;
 
    return cnt;
  }
 
 
  static int kthSmallest(int []nums, int k)
  {
    int low = Int32.MaxValue;
    int high = Int32.MinValue;
 
    // calculate minimum and maximum the array.
    for(int i = 0; i < nums.Length; i++)
    {
      low = Math.Min(low, nums[i]);
      high = Math.Max(high, nums[i]);
    }
    // Our answer range lies between minimum
    // and maximum element of the array on which Binary Search is Applied
    while(low < high)
    {
      int mid = low + (high - low) / 2;
      /*if the count of number of elements in the
               array less than equal to mid is less than k
                 then increase the number. Otherwise
                 decrement the number and try to find a better answer.
               */
      if(count(nums, mid) < k)
        low = mid + 1;
 
      else
        high = mid;
    }
 
    return low;
  }
 
  // Driver Code
  public static void Main()
  {
 
    // Given array
    int []vec = {1, 4, 5, 3, 19, 3};
 
    // Given K
    int K = 3;
 
    // Function Call
    Console.WriteLine("Kth Smallest Element: " + kthSmallest(vec, K)) ;
  }
}
 
// This code is contributed by CodeWithMini


Javascript




<script>
    // Javascript program to find the K’th
    // Smallest/Largest Element in Unsorted Array
    function count(nums, mid)
    {
    // function to calculate number of elements less than equal to mid
            var cnt = 0;
              
            for(var i = 0; i < nums.length; i++)
               if(nums[i] <= mid)
                  cnt++;
              
            return cnt;
    }
     
    function  kthSmallest(nums,k){
        var low = Number. MAX_VALUE;
        var high = Number. MIN_VALUE;
        // calculate minimum and maximum the array.
        for(var i = 0; i < nums.length; i++)
        {
            low = Math.min(low, nums[i]);
            high = Math.max(high, nums[i]);
        }
        // Our answer range lies between minimum and
        // maximum element of the array on which Binary Search is Applied
        while(low < high)
        {
            var mid = Math.floor(low + ((high - low) / 2));
           /*if the count of number of elements in the array
           less than equal to mid is less than k
             then increase the number. Otherwise
             decrement the number and try to find a better answer.
           */
            if(count(nums, mid) < k)
               low = mid + 1;
                 
            else
                high = mid;
        }
          
        return low;
    }
     
    var k=3;
    var nums = [1, 4, 5, 3, 19, 3];
    document.write("K'th smallest element is " + kthSmallest(nums, k));
     
    // This code is contributed by shruti456rawal
</script>


Output

K'th smallest element is 3

Time complexity: O((mx-mn)(log(mx-mn))), where mn be minimum and mx be maximum.
Auxiliary Space: O(1)
K’th Smallest/Largest Element in Unsorted Array | Set 2 (Expected Linear Time) 
K’th Smallest/Largest Element in Unsorted Array | Set 3 (Worst-Case Linear Time)


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