Maximize Array sum by replacing any K elements by its modulo with any positive integer

Given an array of positive integer arr[], and a number K. the task is to maximize the sum of the array by replacing any K elements of the array by taking modulus with any positive integer which is less than arr[i] i.e, (arr[i] = arr[i]%X where X ≤ arr[i]).

Examples:

Input: arr[] = {5, 7, 18, 12, 11, 3}, K = 4
Output: 41
Explanation: The replacement should be {5%3, 7%4, 18, 12, 11%6, 3%2}

Input: arr[] = {8, 2, 28, 12, 7, 9}, K = 4
Output: 55
Explanation: The replacement should be {8%5, 2%2, 28, 12, 7%4, 9}

Approach: For every element arr[i] in the array arr[], module it with (arr[i]/2 +1) which will give the highest possible value of arr[i] after the operation. Following are the steps to solve the problem

• Sort the array arr[].
• Iterate over the range [0, K) using the variable i and perform the following tasks:
  • For every element arr[i], module it with (arr[i]/2 +1) and update the result.
• Find the sum of the updated array and output it.

Below is the implementation of the above approach.

// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;

// Function to find the maximum possible sum
int find(int arr[], int K, int N)
{
    // Sorting the array
    sort(arr, arr + N);
    int sum = 0;

    // Loop to take update K
    for (int i = 0; i < K; i++) {

        // Smallest number in array
        arr[i] %= (arr[i] / 2) + 1;
    }

    // Loop to find sum
    for (int i = 0; i < N; i++) {
        sum += arr[i];
    }
    return sum;
}

// Driver Code
int main()
{
    int arr[] = { 5, 7, 18, 12, 11, 3 };
    int K = 4;
    int N = sizeof(arr) / sizeof(arr[0]);

    cout << find(arr, K, N);
    return 0;
}

// C program for the above approach
#include <stdio.h>
#include <stdlib.h>
#include <limits.h>
#include <limits.h>

void sort(int arr[], int n)
{
  int i, j, temp;
  for (i = 0; i < n - 1; i++)
  {
    for (j = 0; j < n - i - 1; j++)
    {
      if (arr[j] > arr[j + 1])
      {
        temp = arr[j];
        arr[j] = arr[j + 1];
        arr[j + 1] = temp;
      }
    }
  }
}

// Function to find the maximum possible sum
int find(int arr[], int K, int N)
{
  // Sorting the array
  sort(arr, N);
  int sum = 0;

  // Loop to take update K
  for (int i = 0; i < K; i++)
  {

    // Smallest number in array
    arr[i] %= (arr[i] / 2) + 1;
  }

  // Loop to find sum
  for (int i = 0; i < N; i++)
  {
    sum += arr[i];
  }
  return sum;
}

int main()
{
  int arr[] = {5, 7, 18, 12, 11, 3};
  int K = 4;
  int N = sizeof(arr) / sizeof(arr[0]);

  printf("%d", find(arr, K, N));
  return 0;
}

// This code is contributed by abhinavprkash.

// Java program for the above approach
import java.util.Arrays;
class GFG {

  // Function to find the maximum possible sum
  static int find(int arr[], int K, int N) {
    // Sorting the array
    Arrays.sort(arr);
    int sum = 0;

    // Loop to take update K
    for (int i = 0; i < K; i++) {

      // Smallest number in array
      arr[i] %= (arr[i] / 2) + 1;
    }

    // Loop to find sum
    for (int i = 0; i < N; i++) {
      sum += arr[i];
    }
    return sum;
  }

  // Driver Code
  public static void main(String args[]) {
    int arr[] = { 5, 7, 18, 12, 11, 3 };
    int K = 4;
    int N = arr.length;

    System.out.println(find(arr, K, N));
  }
}

// This code is contributed by saurabh_jaiswal.

# python3 program for the above approach

# Function to find the maximum possible sum
def find(arr, K, N):

    # Sorting the array
    arr.sort()
    sum = 0

    # Loop to take update K
    for i in range(0, K):

        # Smallest number in array
        arr[i] %= (arr[i] // 2) + 1

    # Loop to find sum
    for i in range(0, N):
        sum += arr[i]

    return sum

# Driver Code
if __name__ == "__main__":

    arr = [5, 7, 18, 12, 11, 3]
    K = 4
    N = len(arr)

    print(find(arr, K, N))

# This code is contributed by rakeshsahni

// C# program for the above approach
using System;
class GFG {

  // Function to find the maximum possible sum
  static int find(int[] arr, int K, int N)
  {
    // Sorting the array
    Array.Sort(arr);
    int sum = 0;

    // Loop to take update K
    for (int i = 0; i < K; i++) {

      // Smallest number in array
      arr[i] %= (arr[i] / 2) + 1;
    }

    // Loop to find sum
    for (int i = 0; i < N; i++) {
      sum += arr[i];
    }
    return sum;
  }

  // Driver Code
  public static void Main()
  {
    int[] arr = { 5, 7, 18, 12, 11, 3 };
    int K = 4;
    int N = arr.Length;

    Console.Write(find(arr, K, N));
  }
}

// This code is contributed by ukasp.

    <script>
        // JavaScript code for the above approach 


        // Function to find the maximum possible sum
        function find(arr, K, N) {
            // Sorting the array
            arr.sort(function (a, b) { return a - b })
            let sum = 0;

            // Loop to take update K
            for (let i = 0; i < K; i++) {

                // Smallest number in array
                arr[i] %= Math.floor(arr[i] / 2) + 1;
            }

            // Loop to find sum
            for (let i = 0; i < N; i++) {
                sum += arr[i];
            }
            return sum;
        }

        // Driver Code

        let arr = [5, 7, 18, 12, 11, 3];
        let K = 4;
        let N = arr.length;

        document.write(find(arr, K, N));

  // This code is contributed by Potta Lokesh
    </script>

41

Time Complexity: O(N * logN) where N is the size of the array
Auxiliary Space: O(1)