Maximum subarray sum modulo m

3.7

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 prefixi, i.e., 
prefixi = (arr[0] + arr[1] + .... arr[i] ) % m

Let maximum sum ending with i be, maxSumi. 
Let this sum begins with index j.

maxSumi = (prefixi - prefixj + m) % m
                   
From above expression it is clear that the
value of maxSumi becomes maximum when 
prefixj is greater than prefixi 
and closest to prefixi

We mainly have two operations in above algorithm.

  1. Store all prefixes.
  2. For current prefix, prefixi, find the smallest value greater than or equal to prefixi + 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;
}

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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