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QuickSelect (A Simple Iterative Implementation)

  • Difficulty Level : Medium
  • Last Updated : 08 Jun, 2021

Quickselect is a selection algorithm to find the k-th smallest element in an unordered list. It is related to the quick sort sorting algorithm.
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

 

The QuickSelect algorithm is based QuickSort. The difference is, instead of recurring for both sides (after finding pivot), it recurs only for the part that contains the k-th smallest element. The logic is simple, if index of partitioned element is more than k, then we recur for left part. If index is same as k, we have found the k-th smallest element and we return. If index is less than k, then we recur for right part. This reduces the expected complexity from O(n log n) to O(n), with a worst case of O(n^2).
 

function quickSelect(list, left, right, k)

   if left = right
      return list[left]

   Select a pivotIndex between left and right

   pivotIndex := partition(list, left, right, 
                                  pivotIndex)
   if k = pivotIndex
      return list[k]
   else if k < pivotIndex
      right := pivotIndex - 1
   else
      left := pivotIndex + 1

We have discussed a recursive implementation of QuickSelect. In this post, we are going to discuss simple iterative implementation. 
 

C++




// CPP program for iterative implementation of QuickSelect
#include <bits/stdc++.h>
using namespace std;
 
// Standard Lomuto partition function
int partition(int arr[], int low, int high)
{
    int pivot = arr[high];
    int i = (low - 1);
    for (int j = low; j <= high - 1; j++) {
        if (arr[j] <= pivot) {
            i++;
            swap(arr[i], arr[j]);
        }
    }
    swap(arr[i + 1], arr[high]);
    return (i + 1);
}
 
// Implementation of QuickSelect
int kthSmallest(int a[], int left, int right, int k)
{
 
    while (left <= right) {
 
        // Partition a[left..right] around a pivot
        // and find the position of the pivot
        int pivotIndex = partition(a, left, right);
 
        // If pivot itself is the k-th smallest element
        if (pivotIndex == k - 1)
            return a[pivotIndex];
 
        // If there are more than k-1 elements on
        // left of pivot, then k-th smallest must be
        // on left side.
        else if (pivotIndex > k - 1)
            right = pivotIndex - 1;
 
        // Else k-th smallest is on right side.
        else
            left = pivotIndex + 1;
    }
    return -1;
}
 
// Driver program to test above methods
int main()
{
    int arr[] = { 10, 4, 5, 8, 11, 6, 26 };
    int n = sizeof(arr) / sizeof(arr[0]);
    int k = 5;
    cout << "K-th smallest element is "
         << kthSmallest(arr, 0, n - 1, k);
    return 0;
}


Java




// Java program for iterative implementation
// of QuickSelect
class GFG
{
     
    // Standard Lomuto partition function
    static int partition(int arr[],
                         int low, int high)
    {
        int temp;
        int pivot = arr[high];
        int i = (low - 1);
        for (int j = low; j <= high - 1; j++)
        {
            if (arr[j] <= pivot)
            {
                i++;
                temp = arr[i];
                arr[i] = arr[j];
                arr[j] = temp;
            }
        }
         
            temp = arr[i + 1];
            arr[i + 1] = arr[high];
            arr[high] = temp;
             
        return (i + 1);
    }
     
    // Implementation of QuickSelect
    static int kthSmallest(int a[], int left,
                           int right, int k)
    {
        while (left <= right)
        {
     
            // Partition a[left..right] around a pivot
            // and find the position of the pivot
            int pivotIndex = partition(a, left, right);
     
            // If pivot itself is the k-th smallest element
            if (pivotIndex == k - 1)
                return a[pivotIndex];
     
            // If there are more than k-1 elements on
            // left of pivot, then k-th smallest must be
            // on left side.
            else if (pivotIndex > k - 1)
                right = pivotIndex - 1;
     
            // Else k-th smallest is on right side.
            else
                left = pivotIndex + 1;
        }
        return -1;
    }
     
    // Driver Code
    public static void main (String[] args)
    {
        int arr[] = { 10, 4, 5, 8, 11, 6, 26 };
        int n = arr.length;
        int k = 5;
        System.out.println("K-th smallest element is " +
                         kthSmallest(arr, 0, n - 1, k));
    }
}
 
// This code is contributed by AnkitRai01


Python3




# Python3 program for iterative implementation
# of QuickSelect
 
# Standard Lomuto partition function
def partition(arr, low, high) :
 
    pivot = arr[high]
    i = (low - 1)
    for j in range(low, high) :
        if arr[j] <= pivot :
            i += 1
            arr[i], arr[j] = arr[j], arr[i]
         
    arr[i + 1], arr[high] = arr[high], arr[i + 1]
    return (i + 1)
 
# Implementation of QuickSelect
def kthSmallest(a, left, right, k) :
 
    while left <= right :
 
        # Partition a[left..right] around a pivot
        # and find the position of the pivot
        pivotIndex = partition(a, left, right)
 
        # If pivot itself is the k-th smallest element
        if pivotIndex == k - 1 :
            return a[pivotIndex]
 
        # If there are more than k-1 elements on
        # left of pivot, then k-th smallest must be
        # on left side.
        elif pivotIndex > k - 1 :
            right = pivotIndex - 1
 
        # Else k-th smallest is on right side.
        else :
            left = pivotIndex + 1
     
    return -1
 
# Driver Code
arr = [ 10, 4, 5, 8, 11, 6, 26 ]
n = len(arr)
k = 5
print("K-th smallest element is",
       kthSmallest(arr, 0, n - 1, k))
 
# This code is contributed by
# divyamohan123


C#




// C# program for iterative implementation
// of QuickSelect
using System;
                     
class GFG
{
     
    // Standard Lomuto partition function
    static int partition(int []arr,
                         int low, int high)
    {
        int temp;
        int pivot = arr[high];
        int i = (low - 1);
        for (int j = low; j <= high - 1; j++)
        {
            if (arr[j] <= pivot)
            {
                i++;
                temp = arr[i];
                arr[i] = arr[j];
                arr[j] = temp;
            }
        }
         
        temp = arr[i + 1];
        arr[i + 1] = arr[high];
        arr[high] = temp;
             
        return (i + 1);
    }
     
    // Implementation of QuickSelect
    static int kthSmallest(int []a, int left,
                           int right, int k)
    {
        while (left <= right)
        {
     
            // Partition a[left..right] around a pivot
            // and find the position of the pivot
            int pivotIndex = partition(a, left, right);
     
            // If pivot itself is the k-th smallest element
            if (pivotIndex == k - 1)
                return a[pivotIndex];
     
            // If there are more than k-1 elements on
            // left of pivot, then k-th smallest must be
            // on left side.
            else if (pivotIndex > k - 1)
                right = pivotIndex - 1;
     
            // Else k-th smallest is on right side.
            else
                left = pivotIndex + 1;
        }
        return -1;
    }
     
    // Driver Code
    public static void Main (String[] args)
    {
        int []arr = { 10, 4, 5, 8, 11, 6, 26 };
        int n = arr.Length;
        int k = 5;
        Console.WriteLine("K-th smallest element is " +
                        kthSmallest(arr, 0, n - 1, k));
    }
}
 
// This code is contributed by 29AjayKumar


Javascript




<script>
    // Javascript program for iterative implementation of QuickSelect
     
    // Standard Lomuto partition function
    function partition(arr, low, high)
    {
        let temp;
        let pivot = arr[high];
        let i = (low - 1);
        for (let j = low; j <= high - 1; j++)
        {
            if (arr[j] <= pivot)
            {
                i++;
                temp = arr[i];
                arr[i] = arr[j];
                arr[j] = temp;
            }
        }
           
        temp = arr[i + 1];
        arr[i + 1] = arr[high];
        arr[high] = temp;
               
        return (i + 1);
    }
       
    // Implementation of QuickSelect
    function kthSmallest(a, left, right, k)
    {
        while (left <= right)
        {
       
            // Partition a[left..right] around a pivot
            // and find the position of the pivot
            let pivotIndex = partition(a, left, right);
       
            // If pivot itself is the k-th smallest element
            if (pivotIndex == k - 1)
                return a[pivotIndex];
       
            // If there are more than k-1 elements on
            // left of pivot, then k-th smallest must be
            // on left side.
            else if (pivotIndex > k - 1)
                right = pivotIndex - 1;
       
            // Else k-th smallest is on right side.
            else
                left = pivotIndex + 1;
        }
        return -1;
    }
     
    let arr = [ 10, 4, 5, 8, 11, 6, 26 ];
    let n = arr.length;
    let k = 5;
    document.write("K-th smallest element is " + kthSmallest(arr, 0, n - 1, k));
     
     
    // This code is contributed by divyeshrabadiya07.
</script>


Output: 

K-th smallest element is 10

 

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