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Maximize Modulo Sum possible from an array

  • Difficulty Level : Hard
  • Last Updated : 06 Jul, 2021

Given an array arr[] consisting of N positive integers, the task is to find the maximum value of ∑(M mod arr[i]), where arr[i] is any array element, for a non-negative integer M.

Examples: 

Input: arr[] = {3, 4, 6} 
Output: 10 
Explanation: For M = 11, (11 mod 3) + (11 mod 4) + (11 mod 6) =10

Input: arr[]={7, 46, 11, 20, 11} 
Output: 90

 

Approach: Follow the steps below to solve the problem:



  • Since A mod B is the remainder when A divided by B, then the maximum value of the expression ∑(M mod arr[i]) is: 

(M mod Arr[0]) + (M mod Arr[1]) +. . . + (M mod Arr[N-1]) = (Arr[0] − 1) + (Arr[1] − 1) + · · · + (Arr[N-1]− 1)

  • Considering K = Arr[0] × Arr[1] × ···· × Arr[n – 1], then (K mod Arr[i]) = 0 for each i in range [0, N – 1]
  • Therefore, ((K − 1) mod Arr[i]) = Arr[i] − 1. Therefore, for M = K – 1, the optimal result can be obtained.

Below is the implementation of the above approach :

C++




// C++ Program to implement
// the above approach
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to calculate maximum modulo
// sum possible from a given array
int MaximumModuloSum(int Arr[], int N)
{
    // Stores the sum
    int sum = 0;
 
    // Traverse the array
    for (int i = 0; i < N; i++) {
        sum += Arr[i] - 1;
    }
 
    return sum;
}
 
// Driver Code
int main()
{
    int arr[] = { 3, 4, 6 };
    int N = sizeof(arr) / sizeof(arr[0]);
    cout << MaximumModuloSum(arr, N);
 
    return 0;
}

Java




// Java Program to implement
// the above approach
 
import java.io.*;
 
class GFG {
    public static int MaximumModuloSum(int Arr[], int N)
    {
       
        // Stores the sum
        int sum = 0;
 
        // Traverse the array
        for (int i = 0; i < N; i++) {
            sum += Arr[i] - 1;
        }
        return sum;
    }
    public static void main(String[] args)
    {
        int arr[] = { 3, 4, 6 };
        int N = 3;
        System.out.println(MaximumModuloSum(arr, N));
    }
}
 
// This code is contributed by aditya7409.

Python3




# Python 3 Program to implement
# the above approach
 
# Function to calculate maximum modulo
# sum possible from a given array
def MaximumModuloSum( Arr, N):
 
    # Stores the sum
    sum = 0;
 
    # Traverse the array
    for i in range( N ):
        sum += Arr[i] - 1;
     
    return sum;
 
# Driver Code
if __name__ == "__main__":
   
    arr = [ 3, 4, 6 ];
    N = len(arr)
    print(MaximumModuloSum(arr, N))
 
    # This code is contributed by chitranayal.

C#




// C# program for the above approach
using System;
class GFG
{
 
  public static int MaximumModuloSum(int[] Arr, int N)
  {
 
    // Stores the sum
    int sum = 0;
 
    // Traverse the array
    for (int i = 0; i < N; i++) {
      sum += Arr[i] - 1;
    }
    return sum;
  }
 
  // Driver code
  static void Main()
  {
    int[] arr = { 3, 4, 6 };
    int N = 3;
    Console.WriteLine(MaximumModuloSum(arr, N));
  }
}
 
// This code is contributed by susmitakundugoaldanga.

Javascript




<script>
 
// Javascript Program to implement
// the above approach
 
// Function to calculate maximum modulo
// sum possible from a given array
    function MaximumModuloSum(Arr, N)
    {
 
        // Stores the sum
        var sum = 0;
 
        // Traverse the array
        for (i = 0; i < N; i++)
        {
            sum += Arr[i] - 1;
        }
        return sum;
    }
 
    // Driver code
    var arr = [ 3, 4, 6 ];
    var N = 3;
    document.write(MaximumModuloSum(arr, N));
 
// This code is contributed by umadevi9616
</script>
Output: 
10

 

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

 

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