Given an array arr[] of size N having no duplicates 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: 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 elements 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 arrayint 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 codeint 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 approachimport java.util.*;class GFG {// Function to return the kth smallest// element from the arraystatic 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 codepublic 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 approachusing System;using System.Linq; class GFG {// Function to return the kth smallest// element from the arraystatic 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 codepublic 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 arrayfunction 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 codelet 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
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.
Auxiliary Space: O(1), no extra space is required, so it is a constant.
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