Kth smallest element in the array using constant space when array can’t be modified

Difficulty Level : Hard
Last Updated : 11 Aug, 2021

Given an array arr[] of size N having no duplicates and an integer K, the task is to find the K^th smallest element from the array in constant extra space and the array can’t be modified.
Examples:

Input: arr[] = {7, 10, 4, 3, 20, 15}, K = 3
Output: 7
Given array in sorted is {3, 4, 7, 10, 15, 20}
where 7 is the third smallest element.
Input: arr[] = {12, 3, 5, 7, 19}, K = 2
Output: 5

Approach: First we find the min and max element from the array. Then we set low = min, high = max and mid = (low + high) / 2.
Now, perform a modified binary search, and for each mid we count the number of elements less than mid and equal to mid. If countLess < k and countLess + countEqual ≥ k then mid is our answer, else we have to modify our low and high.
Below is the implementation of the above approach:

C++

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to return the kth smallest
// element from the array
int kthSmallest(int* arr, int k, int n)
{
 
    // Minimum and maximum element from the array
    int low = *min_element(arr, arr + n);
    int high = *max_element(arr, arr + n);
 
    // Modified binary search
    while (low <= high) {
 
        int mid = low + (high - low) / 2;
 
        // To store the count of elements from the array
        // which are less than mid and
        // the elements which are equal to mid
        int countless = 0, countequal = 0;
        for (int i = 0; i < n; ++i) {
            if (arr[i] < mid)
                ++countless;
            else if (arr[i] == mid)
                ++countequal;
        }
 
        // If mid is the kth smallest
        if (countless < k
            && (countless + countequal) >= k) {
            return mid;
        }
 
        // If the required element is less than mid
        else if (countless >= k) {
            high = mid - 1;
        }
 
        // If the required element is greater than mid
        else if (countless < k
                 && countless + countequal < k) {
            low = mid + 1;
        }
    }
}
 
// Driver code
int main()
{
    int arr[] = { 7, 10, 4, 3, 20, 15 };
    int n = sizeof(arr) / sizeof(int);
    int k = 3;
 
    cout << kthSmallest(arr, k, n);
 
    return 0;
}

Java

// Java implementation of the approach
import java.util.*;
 
class GFG
{
 
// Function to return the kth smallest
// element from the array
static int kthSmallest(int[] arr, int k, int n)
{
 
    // Minimum and maximum element from the array
    int low = Arrays.stream(arr).min().getAsInt();
    int high = Arrays.stream(arr).max().getAsInt();
 
    // Modified binary search
    while (low <= high)
    {
 
        int mid = low + (high - low) / 2;
 
        // To store the count of elements from the array
        // which are less than mid and
        // the elements which are equal to mid
        int countless = 0, countequal = 0;
        for (int i = 0; i < n; ++i)
        {
            if (arr[i] < mid)
                ++countless;
            else if (arr[i] == mid)
                ++countequal;
        }
 
        // If mid is the kth smallest
        if (countless < k
            && (countless + countequal) >= k)
        {
            return mid;
        }
 
        // If the required element is less than mid
        else if (countless >= k)
        {
            high = mid - 1;
        }
 
        // If the required element is greater than mid
        else if (countless < k
                && countless + countequal < k)
        {
            low = mid + 1;
        }
    }
    return Integer.MIN_VALUE;
}
 
// Driver code
public static void main(String[] args)
{
    int arr[] = { 7, 10, 4, 3, 20, 15 };
    int n = arr.length;
    int k = 3;
 
    System.out.println(kthSmallest(arr, k, n));
}
}
 
// This code is contributed by 29AjayKumar

Python3

# Python3 implementation of the approach
 
# Function to return the kth smallest
# element from the array
def kthSmallest(arr, k, n) :
 
    # Minimum and maximum element from the array
    low = min(arr);
    high = max(arr);
 
    # Modified binary search
    while (low <= high) :
 
        mid = low + (high - low) // 2;
 
        # To store the count of elements from the array
        # which are less than mid and
        # the elements which are equal to mid
        countless = 0; countequal = 0;
         
        for i in range(n) :
             
            if (arr[i] < mid) :
                countless += 1;
                 
            elif (arr[i] == mid) :
                countequal += 1;
 
 
        # If mid is the kth smallest
        if (countless < k and (countless + countequal) >= k) :
            return mid;
         
 
        # If the required element is less than mid
        elif (countless >= k) :
            high = mid - 1;
 
        # If the required element is greater than mid
        elif (countless < k and countless + countequal < k) :
            low = mid + 1;
     
# Driver code
if __name__ == "__main__" :
     
    arr = [ 7, 10, 4, 3, 20, 15 ];
    n = len(arr);
    k = 3;
 
    print(kthSmallest(arr, k, n));
 
# This code is contributed by AnkitRai01

C#

// C# implementation of the approach
using System;
using System.Linq;
 
class GFG
{
 
// Function to return the kth smallest
// element from the array
static int kthSmallest(int[] arr, int k, int n)
{
 
    // Minimum and maximum element from the array
    int low = arr.Min();
    int high = arr.Max();
 
    // Modified binary search
    while (low <= high)
    {
 
        int mid = low + (high - low) / 2;
 
        // To store the count of elements from the array
        // which are less than mid and
        // the elements which are equal to mid
        int countless = 0, countequal = 0;
        for (int i = 0; i < n; ++i)
        {
            if (arr[i] < mid)
                ++countless;
            else if (arr[i] == mid)
                ++countequal;
        }
 
        // If mid is the kth smallest
        if (countless < k
            && (countless + countequal) >= k)
        {
            return mid;
        }
 
        // If the required element is less than mid
        else if (countless >= k)
        {
            high = mid - 1;
        }
 
        // If the required element is greater than mid
        else if (countless < k
                && countless + countequal < k)
        {
            low = mid + 1;
        }
    }
    return int.MinValue;
}
 
// Driver code
public static void Main(String[] args)
{
    int []arr = { 7, 10, 4, 3, 20, 15 };
    int n = arr.Length;
    int k = 3;
 
    Console.WriteLine(kthSmallest(arr, k, n));
}
}
 
// This code is contributed by Rajput-Ji

Javascript

<script>
 
// JavaScript implementation of the approach
 
 
// Function to return the kth smallest
// element from the array
function kthSmallest(arr, k, n) {
 
    let temp = [...arr];
    // Minimum and maximum element from the array
    let low = temp.sort((a, b) => a - b)[0];
    let high = temp[temp.length - 1];
 
    // Modified binary search
    while (low <= high) {
 
        let mid = low + Math.floor((high - low) / 2);
 
        // To store the count of elements from the array
        // which are less than mid and
        // the elements which are equal to mid
        let countless = 0, countequal = 0;
        for (let i = 0; i < n; ++i) {
            if (arr[i] < mid)
                ++countless;
            else if (arr[i] == mid)
                ++countequal;
        }
 
        // If mid is the kth smallest
        if (countless < k
            && (countless + countequal) >= k) {
            return mid;
        }
 
        // If the required element is less than mid
        else if (countless >= k) {
            high = mid - 1;
        }
 
        // If the required element is greater than mid
        else if (countless < k
            && countless + countequal < k) {
            low = mid + 1;
        }
    }
}
 
// Driver code
 
let arr = [7, 10, 4, 3, 20, 15];
let n = arr.length;
let k = 3;
 
document.write(kthSmallest(arr, k, n));
 
 
// This code is contributed by gfgking
 
</script>

Output:

Time Complexity: O(N log(Max – Min)) where Max and Min are the maximum and minimum elements from the array respectively and N is the size of the array.

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