Open In App

Maximum subarray sum modulo m

Improve
Improve
Like Article
Like
Save
Share
Report

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 the overall maximum. To find maximum sum ending at an 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, 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 the implementation of this approach: 

C++




// 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 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;
}


Java




// Java program to find sub-array
// having maximum sum of elements modulo m.
import java.util.*;
class GFG
{
 
  // Return the maximum sum subarray mod m.
  static int maxSubarray(int[] arr, int n, int m)
  {
    int prefix = 0;
    int maxim = 0;
    Set<Integer> S = new HashSet<Integer>();
    S.add(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 = Math.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.
      int it = 0;
 
      for (int j : S)
      {
        if (j >= prefix + 1)
          it = j;
      }
      if (it != 0)
      {
        maxim = Math.max(maxim, prefix - it + m);
      }
 
      // adding prefix in the set.
      S.add(prefix);
    }
    return maxim;
  }
 
  // Driver code
  public static void main(String[] args)
  {
 
    // Driver Code
    int[] arr = new int[] { 3, 3, 9, 9, 5 };
    int n = 5;
    int m = 7;
    System.out.print(maxSubarray(arr, n, m));
  }
}
 
// This code is contributed by pratham76.


Python3




# Python3 program to find sub-array
# having maximum sum of elements modulo m.
 
# Return the maximum sum subarray mod m.
def maxSubarray(arr, n, m):
 
    prefix = 0
    maxim = 0
 
    S = set()
    S.add(0)    
 
    # Traversing the array.
    for i in range(n):
 
        # 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.
        it = 0
        for i in S:
            if i >= prefix + 1:
                it = i
        if (it != 0) :
                maxim = max(maxim, prefix - it + m )
 
        # adding prefix in the set.
        S.add(prefix)
 
    return maxim
 
# Driver Code
arr = [3, 3, 9, 9, 5]
n = 5
m = 7
print(maxSubarray(arr, n, m))
 
# This code is contributed by
# Shubham Singh(SHUBHAMSINGH10)


C#




// C# program to find sub-array
// having maximum sum of elements modulo m.
using System;
using System.Collections;
using System.Collections.Generic;
 
class GFG
{
 
  // Return the maximum sum subarray mod m.
  static int maxSubarray(int[] arr, int n, int m)
  {
     
    int prefix = 0;
    int maxim = 0;
    HashSet<int> S = new HashSet<int>();
    S.Add(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 = Math.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.
      int it = 0;
 
      foreach(int j in S)
      {
        if (j >= prefix + 1)
          it = j;
      }
      if (it != 0)
      {
        maxim = Math.Max(maxim, prefix - it + m);
      }
 
      // adding prefix in the set.
      S.Add(prefix);
    }
    return maxim;
  }
 
  // Driver code
  public static void Main(string[] args)
  {
 
    int[] arr = new int[] { 3, 3, 9, 9, 5 };
    int n = 5;
    int m = 7;
    Console.Write(maxSubarray(arr, n, m));
  }
}
 
// This code is contributed by rutvik_56.


Javascript




<script>
// Javascript program to find sub-array
// having maximum sum of elements modulo m.
 
// Return the maximum sum subarray mod m.
function maxSubarray(arr,n,m)
{
    let prefix = 0;
    let maxim = 0;
    let S = new Set();
    S.add(0);
  
    // Traversing the array.
    for (let i = 0; i < n; i++)
    {
  
      // Finding prefix sum.
      prefix = (prefix + arr[i]) % m;
  
      // Finding maximum of prefix sum.
      maxim = Math.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.
      let it = 0;
  
      for (let j of S.values())
      {
        if (j >= prefix + 1)
          it = j;
      }
      if (it != 0)
      {
        maxim = Math.max(maxim, prefix - it + m);
      }
  
      // adding prefix in the set.
      S.add(prefix);
    }
    return maxim;
}
 
 // Driver Code
let arr=[3, 3, 9, 9, 5];
let n = 5;
let m = 7;
document.write(maxSubarray(arr, n, m));
 
// This code is contributed by avanitrachhadiya2155
</script>


Output

6

Time Complexity: O(n*log(n))
Auxiliary Space:  O(n)

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



Last Updated : 30 Dec, 2022
Like Article
Save Article
Previous
Next
Share your thoughts in the comments
Similar Reads