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Minimize decrements to make each Array elements same or 0

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  • Last Updated : 29 Mar, 2022
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Given an array arr[] consisting of N positive integers. In one operation any number of the array can be decremented by 1. The task is to find the minimum number of operations required to make all of the elements equal or 0.

Examples:

Input: arr[] = {4, 1, 6, 6}
Output: 5
Explanation: Remove 1 from arr{1}, array becomes, {4, 0, 6, 6}, makes 1 operation. 
Remove 2 from arr[2], arr[] = {4, 0, 4, 6}, in 2 operations. 
Remove 2 from arr[3], arr[] = {4, 0, 4, 4} in 2 operations.
So, Total operations = 1 + 2 + 2 = 5.

Input: arr = {1, 1, 2, 10}
Output: 4
Explanation: Remove 1 from arr[0], 1 from arr[1] and 2 from arr[2]. Total operations = 1 + 1 + 2 = 4

 

Approach: The problem can be solved with help of greedy approach using below idea: 

  • If each element of the array is taken as the value that must be equal to the rest of the array elements,
  • Then total removal for any index i would be the (sum_array – (N – i)*arr[i])
  • So the approach would be just to find the above value for each index and return the minimum among them.

Follow the steps mentioned below to solve the problem:

  • Sort the given array.
  • Calculate the sum of all the array elements and store it in a variable, say, total_sum.
  • Create a variable minimum_operation = INT_MAX, run a loop and update minimum_operation  = min(minimum_operation, total_sum  – (N – i)*arr[i])
  • Return minimum_operation as the answer.

Below is the implementation of the above approach.

C++




// C++ code to implement the approach
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to find minimum operation
int minimumRemoval(int arr[], int n)
{
 
    // Variable to store sum of array
    int total_sum = 0;
 
    // Variable to store answer
    int minimum_operation = INT_MAX;
 
    // Sort the given array
    sort(arr, arr + n);
 
    for (int index = 0; index < n; index++) {
        total_sum += arr[index];
    }
 
    // Find minimum operation by keeping
    // each array  value as the remaining
    // elements value
    for (int index = 0; index < n; index++) {
 
        int curr_operation
            = total_sum
              - ((n - index) * arr[index]);
 
        if (curr_operation < minimum_operation) {
            minimum_operation = curr_operation;
        }
    }
 
    // Return minimum_operation
    return minimum_operation;
}
 
// Driver code
int main()
{
    int arr[4] = { 4, 1, 6, 6 };
    int N = 4;
 
    // Function call
    cout << minimumRemoval(arr, N);
    return 0;
}


Java




// Java code to implement the approach
import java.util.*;
 
class GFG{
 
// Function to find minimum operation
static int minimumRemoval(int arr[], int n)
{
 
    // Variable to store sum of array
    int total_sum = 0;
 
    // Variable to store answer
    int minimum_operation = Integer.MAX_VALUE;
 
    // Sort the given array
    Arrays.sort(arr);
 
    for (int index = 0; index < n; index++) {
        total_sum += arr[index];
    }
 
    // Find minimum operation by keeping
    // each array  value as the remaining
    // elements value
    for (int index = 0; index < n; index++) {
 
        int curr_operation
            = total_sum
              - ((n - index) * arr[index]);
 
        if (curr_operation < minimum_operation) {
            minimum_operation = curr_operation;
        }
    }
 
    // Return minimum_operation
    return minimum_operation;
}
 
// Driver code
public static void main(String[] args)
{
    int arr[] = { 4, 1, 6, 6 };
    int N = 4;
 
    // Function call
    System.out.print(minimumRemoval(arr, N));
}
}
 
// This code is contributed by 29AjayKumar


Python3




# Python code to implement the approach
import sys
 
# Function to find minimum operation
def minimumRemoval(arr, n) :
 
    # Variable to store sum of array
    total_sum = 0
 
    # Variable to store answer
    minimum_operation = sys.maxsize
 
    # Sort the given array
    arr.sort()
 
    for index in range(0, n) :
        total_sum += arr[index]
     
 
    # Find minimum operation by keeping
    # each array value as the remaining
    # elements value
    for index in range(0, n) :
 
        curr_operation = (total_sum
            - ((n - index) * arr[index]))
 
        if (curr_operation < minimum_operation) :
            minimum_operation = curr_operation
         
     
 
    # Return minimum_operation
    return minimum_operation
 
 
# Driver code
 
arr = [ 4, 1, 6, 6 ]
N = 4
 
# Function call
print(minimumRemoval(arr, N))
 
# This code is contributed by code_hunt.


C#




// C# code to implement the approach
using System;
class GFG {
 
  // Function to find minimum operation
  static int minimumRemoval(int[] arr, int n)
  {
 
    // Variable to store sum of array
    int total_sum = 0;
 
    // Variable to store answer
    int minimum_operation = Int32.MaxValue;
 
    // Sort the given array
    Array.Sort(arr);
 
    for (int index = 0; index < n; index++) {
      total_sum += arr[index];
    }
 
    // Find minimum operation by keeping
    // each array  value as the remaining
    // elements value
    for (int index = 0; index < n; index++) {
 
      int curr_operation
        = total_sum - ((n - index) * arr[index]);
 
      if (curr_operation < minimum_operation) {
        minimum_operation = curr_operation;
      }
    }
 
    // Return minimum_operation
    return minimum_operation;
  }
 
  // Driver code
  public static void Main()
  {
    int[] arr = { 4, 1, 6, 6 };
    int N = 4;
 
    // Function call
    Console.Write(minimumRemoval(arr, N));
  }
}
 
// This code is contributed by Samim Hossain Mondal.


Javascript




<script>
    // JavaScript code to implement the approach
    const INT_MAX = 2147483647;
 
    // Function to find minimum operation
    const minimumRemoval = (arr, n) => {
 
        // Variable to store sum of array
        let total_sum = 0;
 
        // Variable to store answer
        let minimum_operation = INT_MAX;
 
        // Sort the given array
        arr.sort();
 
        for (let index = 0; index < n; index++) {
            total_sum += arr[index];
        }
 
        // Find minimum operation by keeping
        // each array value as the remaining
        // elements value
        for (let index = 0; index < n; index++) {
 
            let curr_operation
                = total_sum
                - ((n - index) * arr[index]);
 
            if (curr_operation < minimum_operation) {
                minimum_operation = curr_operation;
            }
        }
 
        // Return minimum_operation
        return minimum_operation;
    }
 
    // Driver code
    let arr = [4, 1, 6, 6];
    let N = 4;
 
    // Function call
    document.write(minimumRemoval(arr, N));
 
    // This code is contributed by rakeshsahni
 
</script>


 
 

Output

5

 

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

 


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