K’th Smallest/Largest Element in Unsorted Array | Set 2 (Expected Linear Time)

• Difficulty Level : Hard
• Last Updated : 10 Jun, 2021

We recommend reading the following post as a prerequisite of this post.

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

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 three different solutions here.

In this post method 5 is discussed which is mainly an extension of method 4 (QuickSelect) discussed in the previous post. The idea is to randomly pick a pivot element. To implement randomized partition, we use a random function, rand() to generate index between l and r, swap the element at randomly generated index with the last element, and finally call the standard partition process which uses last element as pivot.

Following is an implementation of the above Randomized QuickSelect.

C++

 // C++ implementation of randomized quickSelect #include #include #include using namespace std;   int randomPartition(int arr[], int l, int r);   // This function returns k'th smallest element in arr[l..r] using // QuickSort based method. ASSUMPTION: 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 a random element and         // get position of pivot element in sorted array         int pos = randomPartition(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 the number of elements in the 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. This function is used by randomPartition() 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; }   // Picks a random pivot element between l and r and partitions // arr[l..r] around the randomly picked element using partition() int randomPartition(int arr[], int l, int r) {     int n = r-l+1;     int pivot = rand() % n;     swap(&arr[l + pivot], &arr[r]);     return partition(arr, l, r); }   // Driver program to test above methods int main() {     int arr[] = {12, 3, 5, 7, 4, 19, 26};     int n = sizeof(arr)/sizeof(arr), k = 3;     cout << "K'th smallest element is " << kthSmallest(arr, 0, n-1, k);     return 0; }

Java

 // Java program to find k'th smallest element in expected // linear time class KthSmallst {     // 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 a random element and             // get position of pivot element in sorted array             int pos = randomPartition(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;     }       // Utility method to swap arr[i] and arr[j]     void swap(int arr[], int i, int j)     {         int temp = arr[i];         arr[i] = arr[j];         arr[j] = 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. This function     // is used by randomPartition()     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, j);                 i++;             }         }         swap(arr, i, r);         return i;     }       // Picks a random pivot element between l and r and     // partitions arr[l..r] arount the randomly picked     // element using partition()     int randomPartition(int arr[], int l, int r)     {         int n = r-l+1;         int pivot = (int)(Math.random()) * (n-1);         swap(arr, l + pivot, r);         return partition(arr, l, r);     }       // Driver method to test above     public static void main(String args[])     {         KthSmallst ob = new KthSmallst();         int arr[] = {12, 3, 5, 7, 4, 19, 26};         int n = arr.length,k = 3;         System.out.println("K'th smallest element is "+                         ob.kthSmallest(arr, 0, n-1, k));     } } /*This code is contributed by Rajat Mishra*/

Python3

 # Python3 implementation of randomized # quickSelect import random   # This function returns k'th smallest # element in arr[l..r] using QuickSort # based method. ASSUMPTION: ELEMENTS # IN ARR[] ARE DISTINCT 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 a random         # element and get position of pivot         # element in sorted array         pos = randomPartition(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 the number of     # elements in the array     return 999999999999   def swap(arr, a, b):     temp = arr[a]     arr[a] = arr[b]     arr[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. This function # is used by randomPartition() def partition(arr, l, r):     x = arr[r]     i = l     for j in range(l, r):         if (arr[j] <= x):             swap(arr, i, j)             i += 1     swap(arr, i, r)     return i   # Picks a random pivot element between l and r # and partitions arr[l..r] around the randomly # picked element using partition() def randomPartition(arr, l, r):     n = r - l + 1     pivot = int(random.random() * n)     swap(arr, l + pivot, r)     return partition(arr, l, r)   # 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 PranchalK

C#

 // C# program to find k'th smallest // element in expected linear time using System;   class GFG { // 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 a         // random element and get position         // of pivot element in sorted array         int pos = randomPartition(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; }   // Utility method to swap arr[i] and arr[j] void swap(int []arr, int i, int j) {     int temp = arr[i];     arr[i] = arr[j];     arr[j] = 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. This function // is used by randomPartition() 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, j);             i++;         }     }     swap(arr, i, r);     return i; }   // Picks a random pivot element between // l and r and partitions arr[l..r] // around the randomly picked element // using partition() int randomPartition(int []arr, int l, int r) {     int n = r - l + 1;     Random rnd = new Random();     int rand = rnd.Next(0, 1);     int pivot = (int)(rand * (n - 1));     swap(arr, l + pivot, r);     return partition(arr, l, r); }   // Driver Code public static void Main() {     GFG ob = new GFG();     int []arr = {12, 3, 5, 7, 4, 19, 26};     int n = arr.Length,k = 3;     Console.Write("K'th smallest element is "+             ob.kthSmallest(arr, 0, n - 1, k)); } }   // This code is contributed by 29AjayKumar

PHP

 0 && \$k <= \$r - \$l + 1)     {         // Partition the array around a random element and         // get position of pivot element in sorted array         \$pos = randomPartition(\$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 the number of elements in the array     return PHP_INT_MAX; }   function swap(\$a, \$b) {     \$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. // This function is used by randomPartition() function partition(\$arr, \$l, \$r) {     \$x = \$arr[\$r];     \$i = \$l;     for (\$j = \$l; \$j <= \$r - 1; \$j++)     {         if (\$arr[\$j] <= \$x)         {             list(\$arr[\$i], \$arr[\$j])=array(\$arr[\$j],\$arr[\$i]);             //swap(&arr[i], &arr[j]);             \$i++;         }     }     list(\$arr[\$i], \$arr[\$r])=array(\$arr[\$r],\$arr[\$i]);     //swap(&arr[i], &arr[r]);     return \$i; }   // Picks a random pivot element between // l and r and partitions arr[l..r] around // the randomly picked element using partition() function randomPartition(\$arr, \$l, \$r) {     \$n = \$r-\$l+1;     \$pivot = rand() % \$n;           list(\$arr[\$l + \$pivot], \$arr[\$r]) =             array(\$arr[\$r],\$arr[\$l + \$pivot] );           //swap(&arr[l + pivot], &arr[r]);     return partition(\$arr, \$l, \$r); }   // Driver program to test the above methods     \$arr = array(12, 3, 5, 7, 4, 19, 260);     \$n = sizeof(\$arr)/sizeof(\$arr);     \$k = 3;     echo "K'th smallest element is " ,             kthSmallest(\$arr, 0, \$n-1, \$k);         // This code is contributed by ajit. ?>

Javascript



Output:

K'th smallest element is 5

Time Complexity:
The worst case time complexity of the above solution is still O(n2). In the worst case, the randomized function may always pick a corner element. The expected time complexity of above randomized QuickSelect is O(n), see CLRS book or MIT video lecture for proof. The assumption in the analysis is, random number generator is equally likely to generate any number in the input range.