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

• Difficulty Level : Medium
• Last Updated : 12 May, 2021

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;       // 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       # 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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