Given an array of **n** elements and an integer m. The task is to find the maximum value of the sum of its subarray modulo m i.e find the sum of each subarray mod m and print the maximum value of this modulo operation.

Examples:

Input : arr[] = { 3, 3, 9, 9, 5 } m = 7 Output : 6 All sub-arrays and their value: { 9 } => 9%7 = 2 { 3 } => 3%7 = 3 { 5 } => 5%7 = 5 { 9, 5 } => 14%7 = 2 { 9, 9 } => 18%7 = 4 { 3, 9 } => 12%7 = 5 { 3, 3 } => 6%7 = 6 { 3, 9, 9 } => 21%7 = 0 { 3, 3, 9 } => 15%7 = 1 { 9, 9, 5 } => 23%7 = 2 { 3, 3, 9, 9 } => 24%7 = 3 { 3, 9, 9, 5 } => 26%7 = 5 { 3, 3, 9, 9, 5 } => 29%7 = 1 Input : arr[] = {10, 7, 18} m = 13 Output : 12 The subarray {7, 18} has maximum sub-array sum modulo 13.

**Method 1 (Brute Force):**

Use brute force to find all the subarrays of the given array and find sum of each subarray mod m and keep track of maximum.

**Method 2 (efficient approach):**

The idea is to compute prefix sum of array. We find maximum sum ending with every index and finally return overall maximum. To find maximum sum ending at index at index, we need to find the starting point of maximum sum ending with i. Below steps explain how to find the starting point.

Let prefix sum for index i be prefix_{i}, i.e., prefix_{i}= (arr[0] + arr[1] + .... arr[i] ) % m Let maximum sum ending with i be, maxSum_{i}. Let this sum begins with index j. maxSum_{i}= (prefix_{i}- prefix_{j}+ m) % m From above expression it is clear that the value of maxSum_{i}becomes maximum when prefix_{j}is greater than prefix_{i}and closest to prefix_{i}

We mainly have two operations in above algorithm.

- Store all prefixes.
- For current prefix, prefix
_{i}, find the smallest value greater than or equal to prefix_{i}+ 1.

For above operations, a self-balancing-binary-search-trees like AVL Tree, Red-Black Tree, etc are best suited. In below implementation we use set in STL which implements a self-balancing-binary-search-tree.

Below is C++ implementation of this approach:

`// C++ program to find sub-array having maximum ` `// sum of elements modulo m. ` `#include<bits/stdc++.h> ` `using` `namespace` `std; ` ` ` `// Return the maximum sum subarray mod m. ` `int` `maxSubarray(` `int` `arr[], ` `int` `n, ` `int` `m) ` `{ ` ` ` `int` `x, prefix = 0, maxim = 0; ` ` ` ` ` `set<` `int` `> S; ` ` ` `S.insert(0); ` ` ` ` ` `// Traversing the array. ` ` ` `for` `(` `int` `i = 0; i < n; i++) ` ` ` `{ ` ` ` `// Finding prefix sum. ` ` ` `prefix = (prefix + arr[i])%m; ` ` ` ` ` `// Finding maximum of prefix sum. ` ` ` `maxim = max(maxim, prefix); ` ` ` ` ` `// Finding iterator pointing to the first ` ` ` `// element that is not less than value ` ` ` `// "prefix + 1", i.e., greater than or ` ` ` `// equal to this value. ` ` ` `auto` `it = S.lower_bound(prefix+1); ` ` ` ` ` `if` `(it != S.end()) ` ` ` `maxim = max(maxim, prefix - (*it) + m ); ` ` ` ` ` `// Inserting prefix in the set. ` ` ` `S.insert(prefix); ` ` ` `} ` ` ` ` ` `return` `maxim; ` `} ` ` ` `// Driver Program ` `int` `main() ` `{ ` ` ` `int` `arr[] = { 3, 3, 9, 9, 5 }; ` ` ` `int` `n = ` `sizeof` `(arr)/` `sizeof` `(arr[0]); ` ` ` `int` `m = 7; ` ` ` `cout << maxSubarray(arr, n, m) << endl; ` ` ` `return` `0; ` `} ` |

*chevron_right*

*filter_none*

Output:

6

**Reference: **

http://stackoverflow.com/questions/31113993/maximum-subarray-sum-modulo-m

This article is contributed by **Anuj Chauhan**. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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