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Minimize difference between maximum and minimum array elements by exactly K removals

  • Difficulty Level : Medium
  • Last Updated : 18 May, 2021

Given an array arr[] consisting of N positive integers and an integer K, the task is to minimize the difference between the maximum and minimum element in the given array after removing exactly K elements.

Examples:

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Input: arr[] = {5, 1, 6, 7, 12, 10}, K = 3
Output: 2
Explanation:
Remove elements 12, 10 and 1 from the given array.
The array modifies to {5, 6, 7}.
The difference between the minimum and maximum element is 7 – 5 = 2.



Input: arr[] = {14, 5, 61, 10, 21, 12, 54}, K = 4
Output: 4
Explanation:
Remove elements 61, 54, 5 and 21 from the given array.
The array modifies to {14, 10, 12}.
The difference between the minimum and maximum element is 14 – 10 = 4.

Approach: The idea to solve the given problem is that the difference will be minimized by removing either the minimum element in an array or the maximum element in the array. Follow the steps below to solve the problem:

  • Sort the array arr[] in ascending order.
  • Initialize variables left = 0 and right = (N – 1).
  • Iterate for K times and change the maximum or minimum to 0 according to the following condition:
    • If arr[right – 1] – arr[left] < arr[right] – arr[left + 1], then change arr[right] to 0 and decrement right pointer by 1.
    • Else, change arr[left] as 0 and increment the left pointer by 1.
  • After the above steps, the difference between the elements at the left and right index is the required minimum difference.

Below is the implementation of the above approach:

C++




// C++ program for the above approach
#include <bits/stdc++.h>
#include <iostream>
using namespace std;
 
// Function to minimize the difference
// of the maximum and minimum array
// elements by removing K elements
void minimumRange(int arr[], int N, int K)
{
    // Base Condition
    if (K >= N)
    {
        cout << 0;
        return;
    }
 
    // Sort the array
    sort(arr, arr + N);
 
    // Initialize left and right pointers
    int left = 0, right = N - 1, i;
 
    // Iterate for K times
    for (i = 0; i < K; i++)
    {
 
        // Removing right element
        // to reduce the difference
        if (arr[right - 1] - arr[left]
            < arr[right] - arr[left + 1])
            right--;
 
        // Removing the left element
        // to reduce the difference
        else
            left++;
    }
 
    // Print the minimum difference
    cout << arr[right] - arr[left];
}
 
// Driver Code
int main()
{
    int arr[] = { 5, 10, 12, 14, 21, 54, 61 };
    int N = sizeof(arr) / sizeof(arr[0]);
 
    int K = 4;
 
    // Function Call
    minimumRange(arr, N, K);
 
    return 0;
}

Java




// Java program for the above approach
import java.util.*;
 
class GFG{
 
// Function to minimize the difference
// of the maximum and minimum array
// elements by removing K elements
static void minimumRange(int arr[], int N,
                         int K)
{
     
    // Base Condition
    if (K >= N)
    {
        System.out.print(0);
        return;
    }
     
    // Sort the array
    Arrays.sort(arr);
     
    // Initialize left and right pointers
    int left = 0, right = N - 1, i;
 
    // Iterate for K times
    for(i = 0; i < K; i++)
    {
         
        // Removing right element
        // to reduce the difference
        if (arr[right - 1] - arr[left] <
            arr[right] - arr[left + 1])
            right--;
             
        // Removing the left element
        // to reduce the difference
        else
            left++;
    }
     
    // Print the minimum difference
    System.out.print(arr[right] - arr[left]);
}
 
// Driver Code
public static void main(String[] args)
{
    int arr[] = { 5, 10, 12, 14, 21, 54, 61 };
    int N = arr.length;
    int K = 4;
 
    // Function Call
    minimumRange(arr, N, K);
}
}
 
// This code is contributed by 29AjayKumar

Python3




# Python3 program for the above approach
 
# Function to minimize the difference
# of the maximum and minimum array
# elements by removing K elements
def minimumRange(arr, N, K) :
 
    # Base Condition
    if (K >= N) :
        print(0, end = '');
        return;
 
    # Sort the array
    arr.sort();
 
    # Initialize left and right pointers
    left = 0; right = N - 1;
 
    # Iterate for K times
    for i in range(K) :
 
        # Removing right element
        # to reduce the difference
        if (arr[right - 1] - arr[left] < arr[right] - arr[left + 1]) :
            right -= 1;
 
        # Removing the left element
        # to reduce the difference
        else :
            left += 1;
 
    # Print the minimum difference
    print(arr[right] - arr[left], end = '');
 
# Driver Code
if __name__ == "__main__" :
 
    arr = [ 5, 10, 12, 14, 21, 54, 61 ];
    N = len(arr);
 
    K = 4;
 
    # Function Call
    minimumRange(arr, N, K);
 
    # This code is contributed by AnkitRai01

C#




// C# program for the above approach
using System;
class GFG{
 
// Function to minimize the difference
// of the maximum and minimum array
// elements by removing K elements
static void minimumRange(int []arr, int N,
                         int K)
{
     
    // Base Condition
    if (K >= N)
    {
        Console.Write(0);
        return;
    }
     
    // Sort the array
    Array.Sort(arr);
     
    // Initialize left and right pointers
    int left = 0, right = N - 1, i;
 
    // Iterate for K times
    for(i = 0; i < K; i++)
    {
         
        // Removing right element
        // to reduce the difference
        if (arr[right - 1] - arr[left] <
            arr[right] - arr[left + 1])
            right--;
             
        // Removing the left element
        // to reduce the difference
        else
            left++;
    }
     
    // Print the minimum difference
    Console.Write(arr[right] - arr[left]);
}
 
// Driver Code
public static void Main(String[] args)
{
    int []arr = { 5, 10, 12, 14, 21, 54, 61 };
    int N = arr.Length;
    int K = 4;
 
    // Function Call
    minimumRange(arr, N, K);
}
}
 
// This code is contributed by 29AjayKumar

Javascript




<script>
 
// JavaScript program for the above approach
 
// Function to minimize the difference
// of the maximum and minimum array
// elements by removing K elements
function minimumRange(arr, N, K)
{
    // Base Condition
    if (K >= N)
    {
        document.write( 0);
        return;
    }
 
    // Sort the array
    arr.sort((a,b)=> a-b);
 
    // Initialize left and right pointers
    var left = 0, right = N - 1, i;
 
    // Iterate for K times
    for (i = 0; i < K; i++)
    {
 
        // Removing right element
        // to reduce the difference
        if (arr[right - 1] - arr[left]
            < arr[right] - arr[left + 1])
            right--;
 
        // Removing the left element
        // to reduce the difference
        else
            left++;
    }
 
    // Print the minimum difference
    document.write( arr[right] - arr[left]);
}
 
// Driver Code
var arr = [5, 10, 12, 14, 21, 54, 61];
var N = arr.length;
var K = 4;
 
// Function Call
minimumRange(arr, N, K);
 
</script>
Output: 
4

 

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




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