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

Last Updated : 23 Jul, 2025

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

We recommend reading the following post as a prerequisite to this post. K’th Smallest/Largest Element in Unsorted Array | Set 1

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 4, it is discussed which is mainly an extension of method 3 (QuickSelect) discussed in the previous post.

The flowchart is as follows:

C++

// C++ program of above implementation

#include <bits/stdc++.h>
using namespace std;

// Standard partition process of QuickSort().
// It considers the last element as pivot and
// oves 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;
}

int randomPartition(int* arr, int l, int r)
{
    int n = r - l + 1;
    int pivot = (rand() % 100 + 1) % n;
    swap(arr[l + pivot], arr[r]);
    return partition(arr, l, 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 = 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_MAX;
}

// Driver Code
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);
}

Java

// Java program of above implementation

import java.util.Random;

public class GFG {

// This function returns k'th smallest element in arr[l..r] using
// QuickSort based method. ASSUMPTION: ALL ELEMENTS IN ARR[] ARE DISTINCT
    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 = 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 number of elements in array
        return Integer.MAX_VALUE;
    }

    static 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
    static 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;
    }

    static int randomPartition(int arr[], int l, int r) {
        int n = r - l + 1;
        int pivot = new Random().nextInt(n);
        swap(arr, l + pivot, r);
        return partition(arr, l, r);
    }

// Driver program to test above methods
    public static void main(String args[]) {
        int arr[] = {12, 3, 5, 7, 4, 19, 26};
        int n = arr.length, k = 3;
        System.out.println("K'th smallest element is " + kthSmallest(arr, 0, n - 1, k));
    }
}

/*This code is contributed by 29AjayKumar*/

Python3

# Python3 implementation of above implementation

# This function returns k'th smallest element 
# in arr[l..r] using QuickSort based method.
# ASSUMPTION: ALL ELEMENTS IN ARR[] ARE DISTINCT
from random import randint

def randomPartition(arr, l, r):
    n = r - l + 1
    pivot = randint(1, 100) % n
    arr[l + pivot], arr[r] = arr[l + pivot], arr[r]
    return partition(arr, l, r)

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 = 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 10**9

# 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
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 Mohit Kumar

// C# program of above implementation

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 
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 = 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; 
} 

static 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 
// oves all smaller element to left of it 
// and greater elements to right 
static 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; 
} 

static int randomPartition(int []arr, int l, int r)
{ 
    int n = r - l + 1; 
    int pivot = new Random().Next(1); 
    swap(arr, l + pivot, r); 
    return partition(arr, l, r); 
} 

// Driver Code
public static void Main()
{ 
    int []arr = {12, 3, 5, 7, 4, 19, 26}; 
    int n = arr.Length, k = 3; 
    Console.WriteLine("K'th smallest element is " + 
                    kthSmallest(arr, 0, n - 1, k)); 
} 
} 

// this code is contributed by 29AjayKumar

JavaScript

<script>

// Javascript program of above implementation
    
// 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
        let 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 Number.MAX_VALUE;
}

function swap(arr, i, j)
{
    let 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
function partition(arr, l, r)
{
    let x = arr[r], i = l;
    for(let j = l; j <= r - 1; j++) 
    {
        if (arr[j] <= x) 
        {
            swap(arr, i, j);
            i++;
        }
    }
    swap(arr, i, r);
    return i;
}

function randomPartition(arr, l, r)
{
    let n = r - l + 1;
    let pivot = (Math.floor(Math.random() * 101)) % n;
    swap(arr, l + pivot, r);
    
    return partition(arr, l, r);
}

// Driver code
let arr = [ 12, 3, 5, 7, 4, 19, 26 ];
let n = arr.length, k = 3;

document.write("K'th smallest element is " + 
               kthSmallest(arr, 0, n - 1, k));
    
</script>

Output

K'th smallest element is 5

Time Complexity : O(N * log N).
Auxiliary Space: O(1).

References: https://www.geeksforgeeks.org/dsa/kth-smallest-largest-element-in-unsorted-array/

Analysis of Algorithms

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