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[0] replaced by 4 
k = 2, arr = {4, 4, 6, 4, 15}, arr[3] replaced by 4 
k = 3, arr = {4, 4, 6, 4, 4}, arr[4] 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[3] replaced by 6 
k = 2, arr = {1, 4, 6, 6, 6}, arr[4] 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 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:
 

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// C++ program of the approach
#include <bits/stdc++.h>
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[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 = 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[0]);
 
    // Given K replacements
    int K = 3;
 
    // Function Call
    cout << maxMinDifference(arr, N, K);
    return 0;
}
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// 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
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# 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
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// 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[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 = 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
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Output: 
2




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

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