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

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

Given an array arr[] of size N and an integer K, the task is to find the Kth 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:
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:

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 ``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

 ``
Output:

`7`

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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