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# Maximize the sum of modulus with every Array element

• Last Updated : 14 Jan, 2022

Given an array A[] consisting of N positive integers, the task is to find the maximum possible value of:

F(M) = M % A[0] + M % A[1] + …. + M % A[N -1] where M can be any integer value

Examples:

Input: arr[] = {3, 4, 6}
Output: 10
Explanation:
The maximum sum occurs for M = 11.
(11 % 3) + (11 % 4) + (11 % 6) = 2 + 3 + 5 = 10
Input: arr[] = {2, 5, 3}
Output:
Explanation:
The maximum sum occurs for M = 29.
(29 % 2) + (29 % 5) + (29 % 3) = 1 + 4 + 2 = 7.

Approach:
Follow the steps below to solve the problem:

1. Calculate the LCM of all array elements.
2. If M is equal to the LCM of the array, then F(M) = 0 i.e. the minimum possible value of the F(M). This is because, M % a[i] will always be 0 for every ith index.
3. For M = LCM of array elements – 1, F(M) is maximized. This is because, M % a[i] is equal to a[i] – 1 for every ith index, which is the maximum possible.
4. Hence, the maximum possible value of F(M) can be Sum of array elements – N.

Below is the implementation of the above approach:

## C++

 `// C++ program to find the` `// maximum sum of modulus` `// with every array element` `#include ` `using` `namespace` `std;`   `// Function to return the` `// maximum sum of modulus` `// with every array element` `int` `maxModulosum(``int` `a[], ``int` `n)` `{` `    ``int` `sum = 0;`   `    ``// Sum of array elements` `    ``for` `(``int` `i = 0; i < n; i++) {` `        ``sum += a[i];` `    ``}`   `    ``// Return the answer` `    ``return` `sum - n;` `}`   `// Driver Program` `int` `main()` `{` `    ``int` `a[] = { 3, 4, 6 };` `    ``int` `n = ``sizeof``(a) / ``sizeof``(a[0]);` `    ``cout << maxModulosum(a, n);`   `    ``return` `0;` `}`

## Java

 `// Java program to find the maximum` `// sum of modulus with every array` `// element` `import` `java.io.*; `   `class` `GFG{ `   `// Function to return the maximum` `// sum of modulus with every array` `// element` `static` `int` `maxModulosum(``int` `a[], ``int` `n)` `{` `    ``int` `sum = ``0``;` `    `  `    ``// Sum of array elements` `    ``for``(``int` `i = ``0``; i < n; i++)` `    ``{` `       ``sum += a[i];` `    ``}` `    `  `    ``// Return the answer` `    ``return` `sum - n;` `}` `    `  `// Driver Code ` `public` `static` `void` `main (String[] args) ` `{ ` `    ``int` `a[] = ``new` `int``[]{ ``3``, ``4``, ``6` `};` `    ``int` `n = a.length;` `    `  `    ``System.out.println(maxModulosum(a, n)); ` `} ` `} `   `// This code is contributed by Shubham Prakash`

## Python3

 `# Python3 program to find the` `# maximum sum of modulus` `# with every array element`   `# Function to return the` `# maximum sum of modulus` `# with every array element` `def` `maxModulosum(a, n):`   `    ``sum1 ``=` `0``;`   `    ``# Sum of array elements` `    ``for` `i ``in` `range``(``0``, n):` `        ``sum1 ``+``=` `a[i];` `    `  `    ``# Return the answer` `    ``return` `sum1 ``-` `n;`   `# Driver Code` `a ``=` `[ ``3``, ``4``, ``6` `];` `n ``=` `len``(a);` `print``(maxModulosum(a, n));`   `# This code is contributed by Code_Mech`

## C#

 `// C# program to find the maximum` `// sum of modulus with every array` `// element` `using` `System;` `class` `GFG{ `   `// Function to return the maximum` `// sum of modulus with every array` `// element` `static` `int` `maxModulosum(``int` `[]a, ``int` `n)` `{` `    ``int` `sum = 0;` `    `  `    ``// Sum of array elements` `    ``for``(``int` `i = 0; i < n; i++)` `    ``{` `        ``sum += a[i];` `    ``}` `    `  `    ``// Return the answer` `    ``return` `sum - n;` `}` `    `  `// Driver Code ` `public` `static` `void` `Main(String[] args) ` `{ ` `    ``int` `[]a = ``new` `int``[]{ 3, 4, 6 };` `    ``int` `n = a.Length;` `    `  `    ``Console.Write(maxModulosum(a, n)); ` `} ` `} `   `// This code is contributed ` `// by shivanisinghss2110`

## Javascript

 ``

Output:

`10`

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

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