Given an array arr[] and an integer K, the task is to count the number of elements to be removed from the array such that the difference of the maximum and the minimum number left does not exceed K.
Examples:
Input: K = 1, arr[] = {1, 2, 3, 4, 5}
Output: 3
Explanation:
Removal of {5, 4, 3} modifies array to {1, 2} where the maximum difference is 1(= K).
Input: K = 3, arr[] = {1, 2, 3, 4, 5}
Output: 1
Explanation:
Removal of {5} modifies array to {1, 2, 3, 4} where the maximum difference is 3(= K).
Approach:
The task is to find the difference between the maximum and minimum array element which should not exceed K.
- Sort the array in ascending order and initialize a variable to a minimum value.
- Iterate over the array to generate all possible pairs and check if the difference between any pair is less than or equal to K.
- Update the minimum number of removals for each pair.
- Finally, print the minimum removals obtained.
Below is the implementation of the above approach:
C++
// C++ Program to implement// the above approach#include <bits/stdc++.h>using namespace std;// Function to count the number of// elements to be removed from the// array based on the given conditionint min_remove(int arr[], int n, int k){ // Sort the array sort(arr, arr + n); /// Initialize the variable int ans = INT_MAX; // Iterate for all possible pairs for (int i = 0; i < n; i++) { for (int j = i; j < n; j++) { // Check the difference // between the numbers if (arr[j] - arr[i] <= k) { // Update the minimum removals ans = min(ans, n - j + i - 1); } } } // Return the answer return ans;}// Driver Codeint main(){ int k = 3; int arr[] = { 1, 2, 3, 4, 5 }; int n = sizeof arr / sizeof arr[0]; cout << min_remove(arr, n, k); return 0;} |
Java
// Java Program to implement// the above approachimport java.util.*;class GFG{// Function to count the number of// elements to be removed from the// array based on the given conditionstatic int min_remove(int arr[], int n, int k){ // Sort the array Arrays.sort(arr); /// Initialize the variable int ans = Integer.MAX_VALUE; // Iterate for all possible pairs for (int i = 0; i < n; i++) { for (int j = i; j < n; j++) { // Check the difference // between the numbers if (arr[j] - arr[i] <= k) { // Update the minimum removals ans = Math.min(ans, n - j + i - 1); } } } // Return the answer return ans;}// Driver Codepublic static void main(String[] args){ int k = 3; int arr[] = { 1, 2, 3, 4, 5 }; int n = arr.length; System.out.print(min_remove(arr, n, k));}}// This code is contributed by sapnasingh4991 |
Python3
# Python3 program to implement# the above approachimport sys# Function to count the number of# elements to be removed from the# array based on the given conditiondef min_remove(arr, n, k): # Sort the array arr.sort() # Initialize the variable ans = sys.maxsize # Iterate for all possible pairs for i in range(n): for j in range(i, n): # Check the difference # between the numbers if (arr[j] - arr[i] <= k): # Update the minimum removals ans = min(ans, n - j + i - 1) # Return the answer return ans# Driver Codeif __name__ == "__main__": k = 3 arr = [ 1, 2, 3, 4, 5 ] n = len(arr) print (min_remove(arr, n, k))# This code is contributed by chitranayal |
C#
// C# Program to implement// the above approachusing System;class GFG{// Function to count the number of// elements to be removed from the// array based on the given conditionstatic int min_remove(int []arr, int n, int k){ // Sort the array Array.Sort(arr); /// Initialize the variable int ans = int.MaxValue; // Iterate for all possible pairs for (int i = 0; i < n; i++) { for (int j = i; j < n; j++) { // Check the difference // between the numbers if (arr[j] - arr[i] <= k) { // Update the minimum removals ans = Math.Min(ans, n - j + i - 1); } } } // Return the answer return ans;}// Driver Codepublic static void Main(String[] args){ int k = 3; int []arr = { 1, 2, 3, 4, 5 }; int n = arr.Length; Console.Write(min_remove(arr, n, k));}}// This code is contributed by sapnasingh4991 |
Javascript
<script>// JavaScript program for the above approach// Function to count the number of// elements to be removed from the// array based on the given conditionfunction min_remove(arr, n, k){ // Sort the array arr.sort(); /// Initialize the variable let ans = Number.MAX_VALUE; // Iterate for all possible pairs for (let i = 0; i < n; i++) { for (let j = i; j < n; j++) { // Check the difference // between the numbers if (arr[j] - arr[i] <= k) { // Update the minimum removals ans = Math.min(ans, n - j + i - 1); } } } // Return the answer return ans;} // Driver Code let k = 3; let arr = [ 1, 2, 3, 4, 5 ]; let n = arr.length; document.write(min_remove(arr, n, k)); </script> |
Output:
1
Time Complexity: O(N2)
Auxiliary Space: O(1)
