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# Minimize difference between maximum and minimum of Array by at most K replacements

Given an array arr[] and an integer K, that task is to choose at most K elements of the array and replace it by any number. Find the minimum difference between the maximum and minimum value of the array after performing at most K replacement.

Examples:

Input: arr[] = {1, 4, 6, 11, 15}, k = 3
Output:
Explanation:
k = 1, arr = {4, 4, 6, 11, 15}, arr replaced by 4
k = 2, arr = {4, 4, 6, 4, 15}, arr replaced by 4
k = 3, arr = {4, 4, 6, 4, 4}, arr replaced by 4
Max – Min = 6 – 4 = 2

Input: arr[] = {1, 4, 6, 11, 15}, k = 2
Output:
Explanation:
k = 1, arr = {1, 4, 6, 6, 15}, arr replaced by 6
k = 2, arr = {1, 4, 6, 6, 6}, arr replaced by 6
Max – Min = 6 – 1 = 5

Approach: The idea is to use the concept of Two Pointers. Below are the steps:

1. Sort the given array.
2. Maintain two pointers, one pointing to the last element of the array and the other to the Kth element of the array.
3. Iterate over the array K + 1 times and each time find the difference of the elements pointed by the two pointers.
4. Every time on finding the difference, keep track of the minimum possible difference in a variable and return that value at the end.

Below is the implementation of the above approach:

## C++

 `// C++ program of the approach` `#include ` `using` `namespace` `std;`   `// Function to find minimum difference` `// between the maximum and the minimum` `// elements arr[] by at most K replacements` `int` `maxMinDifference(``int` `arr[], ``int` `n, ``int` `k)` `{` `    ``// Check if turns are more than` `    ``// or equal to n-1 then simply` `    ``// return zero` `    ``if` `(k >= n - 1)` `        ``return` `0;`   `    ``// Sort the array` `    ``sort(arr, arr + n);`   `    ``// Set difference as the` `    ``// maximum possible difference` `    ``int` `ans = arr[n - 1] - arr;`   `    ``// Iterate over the array to` `    ``// track the minimum difference` `    ``// in k turns` `    ``for` `(``int` `i = k, j = n - 1;` `         ``i >= 0; --i, --j) {`   `        ``ans = min(arr[j] - arr[i], ans);` `    ``}`   `    ``// Return the answer` `    ``return` `ans;` `}`   `// Driver Code` `int` `main()` `{` `    ``// Given array arr[]` `    ``int` `arr[] = { 1, 4, 6, 11, 15 };` `    ``int` `N = ``sizeof``(arr) / ``sizeof``(arr);`   `    ``// Given K replacements` `    ``int` `K = 3;`   `    ``// Function Call` `    ``cout << maxMinDifference(arr, N, K);` `    ``return` `0;` `}`

## Java

 `// Java program of the approach` `import` `java.io.*;` `import` `java.util.Arrays;` `class` `GFG{`   `// Function to find minimum difference` `// between the maximum and the minimum` `// elements arr[] by at most K replacements` `static` `int` `maxMinDifference(``int` `arr[], ``int` `n, ``int` `k)` `{` `    ``// Check if turns are more than` `    ``// or equal to n-1 then simply` `    ``// return zero` `    ``if` `(k >= n - ``1``)` `        ``return` `0``;`   `    ``// Sort the array` `    ``Arrays.sort(arr);`   `    ``// Set difference as the` `    ``// maximum possible difference` `    ``int` `ans = arr[n - ``1``] - arr[``0``];`   `    ``// Iterate over the array to` `    ``// track the minimum difference` `    ``// in k turns` `    ``for` `(``int` `i = k, j = n - ``1``;` `             ``i >= ``0``; --i, --j) ` `    ``{` `        ``ans = Math.min(arr[j] - arr[i], ans);` `    ``}`   `    ``// Return the answer` `    ``return` `ans;` `}`   `// Driver Code` `public` `static` `void` `main(String[] args)` `{` `    ``// Given array arr[]` `    ``int` `arr[] = { ``1``, ``4``, ``6``, ``11``, ``15` `};` `    ``int` `N = arr.length;`   `    ``// Given K replacements` `    ``int` `K = ``3``;`   `    ``// Function Call` `    ``System.out.print(maxMinDifference(arr, N, K));` `}` `}`   `// This code is contributed by shivanisinghss2110`

## Python3

 `# Python3 program for the above approach`   `# Function to find minimum difference` `# between the maximum and the minimum` `# elements arr[] by at most K replacements ` `def` `maxMinDifference(arr, n, k):`   `    ``# Check if turns are more than` `    ``# or equal to n-1 then simply` `    ``# return zero` `    ``if``(k >``=` `n ``-` `1``):` `        ``return` `0`   `    ``# Sort the array` `    ``arr.sort()`   `    ``# Set difference as the` `    ``# maximum possible difference` `    ``ans ``=` `arr[n ``-` `1``] ``-` `arr[``0``]`   `    ``# Iterate over the array to` `    ``# track the minimum difference` `    ``# in k turns` `    ``i ``=` `k` `    ``j ``=` `n ``-` `1` `    ``while` `i >``=` `0``:` `        ``ans ``=` `min``(arr[j] ``-` `arr[i], ans)` `        ``i ``-``=` `1` `        ``j ``-``=` `1`   `    ``# Return the answer` `    ``return` `ans`   `# Driver code` `if` `__name__ ``=``=` `'__main__'``:`   `    ``# Given array arr[]` `    ``arr ``=` `[ ``1``, ``4``, ``6``, ``11``, ``15` `]` `    ``N ``=` `len``(arr)`   `    ``# Given K replacements` `    ``K ``=` `3`   `    ``# Function Call` `    ``print``(maxMinDifference(arr, N, K))`   `# This code is contributed by Shivam Singh`

## C#

 `// C# program of the approach` `using` `System;` `class` `GFG{` ` `  `// Function to find minimum difference` `// between the maximum and the minimum` `// elements arr[] by at most K replacements` `static` `int` `maxMinDifference(``int` `[]arr, ``int` `n, ``int` `k)` `{` `    ``// Check if turns are more than` `    ``// or equal to n-1 then simply` `    ``// return zero` `    ``if` `(k >= n - 1)` `        ``return` `0;` ` `  `    ``// Sort the array` `    ``Array.Sort(arr);` ` `  `    ``// Set difference as the` `    ``// maximum possible difference` `    ``int` `ans = arr[n - 1] - arr;` ` `  `    ``// Iterate over the array to` `    ``// track the minimum difference` `    ``// in k turns` `    ``for` `(``int` `i = k, j = n - 1;` `             ``i >= 0; --i, --j) ` `    ``{` `        ``ans = Math.Min(arr[j] - arr[i], ans);` `    ``}` ` `  `    ``// Return the answer` `    ``return` `ans;` `}` ` `  `// Driver Code` `public` `static` `void` `Main(``string``[] args)` `{` `    ``// Given array arr[]` `    ``int` `[] arr = ``new`  `int``[] { 1, 4, 6, 11, 15 };` `    ``int` `N = arr.Length;` ` `  `    ``// Given K replacements` `    ``int` `K = 3;` ` `  `    ``// Function Call` `    ``Console.Write(maxMinDifference(arr, N, K));` `}` `}` ` `  `// This code is contributed by Ritik Bansal`

## Javascript

 ``

Output:

`2`

Time Complexity: O(N*log2N)
Auxiliary Space: O(1)

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