Maximum subarray sum modulo m in C++

In this problem, we are given an array of size n and an integer m. our task is to create a program that will find the maximum subarray sum modulo m in C++.

Program description − Here, we will find the maximum value obtained by dividing the sum of all elements of subarray divided by m.

Let’s take an example to understand the problem,

Input − array = {4, 9 ,2} m = 6

Output − 5

Explanation − All subarrays and their remainders on dividing

{4}: 4%6 = 4
{9}: 9%6 = 3
{2}: 2%6 = 2
{4, 9}: 13%6 = 1
{9, 2}: 11%6 = 5
{4, 9, 2}: 15%6 = 3

To solve this problem, we compute prefixSumModulo Array. And the calculating maxSum of each index using the formula, maxSum at i = (prefixi + prefixj + m)%m

Example

Program to illustrate our solution,

Live Demo

#include<bits/stdc++.h>
using namespace std;
int calcMaxSum(int arr[], int n, int m) {
   int x, prefix = 0, maxSumMod = 0;
   set<int> sums;
   sums.insert(0);
   for (int i = 0; i < n; i++){
      prefix = (prefix + arr[i])%m;
      maxSumMod = max(maxSumMod, prefix);
      auto it = sums.lower_bound(prefix+1);
      if (it != sums.end())
         maxSumMod = max(maxSumMod, prefix - (*it) + m );
      sums.insert(prefix);
   }
   return maxSumMod;
}
int main() {
   int arr[] = {4, 9, 2};
   int n = sizeof(arr)/sizeof(arr[0]);
   int m = 5;
   cout<<"Maximum subarray sum modulo "<<m<<" is "<<calcMaxSum(arr, n, m) < endl;
   return 0;
}