Subset with sum divisible by m

Given a set of non-negative distinct integers, and a value m, determine if there is a subset of the given set with sum divisible by m.
Input Constraints
Size of set i.e., n <= 1000000, m <= 1000

Examples:

Input : arr[] = {3, 1, 7, 5};
        m = 6;
Output : YES

Input : arr[] = {1, 6};
        m = 5;
Output : NO

This problem is a variant of subset sum problem. In subset sum problem we check if given sum subset exist or not, here we need to find if there exist some subset with sum divisible by m or not. Seeing input constraint, it looks like typical DP solution will work in O(nm) time. But in tight time limits in competitive programming, the solution may work. Also auxiliary space is high for DP table, but here is catch.

If n > m there will always be a subset with sum divisible by m (which is easy to prove with pigeonhole principle). So we need to handle only cases of n <= m .

For n <= m we create a boolean DP table which will store the status of each value from 0 to m-1 which are possible subset sum (modulo m) which have been encountered so far.

Now we loop through each element of given array arr[], and we add (modulo m) j which have DP[j] = true and store all the such (j+arr[i])%m possible subset sum in a boolean array temp, and at the end of iteration over j, we update DP table with temp. Also we add arr[i] to DP ie.. DP[arr[i]%m] = true.

In the end if DP[0] is true then it means YES there exist a subset with sum which is divisible by m, else NO.

// C++ program to check if there is a subset
// with sum divisible by m.
#include <bits/stdc++.h>
using namespace std;

// Returns true if there is a subset
// of arr[] with sum divisible by m
bool modularSum(int arr[], int n, int m)
{
    if (n > m)
        return true;

    // This array will keep track of all
    // the possible sum (after modulo m)
    // which can be made using subsets of arr[]
    // initialising boolean array with all false
    bool DP[m];
    memset(DP, false, m);

    // we'll loop through all the elements of arr[]
    for (int i=0; i<n; i++)
    {
        // anytime we encounter a sum divisible
        // by m, we are done
        if (DP[0])
            return true;

        // To store all the new encountered sum (after
        // modulo). It is used to make sure that arr[i]
        // is added only to those entries for which DP[j] 
        // was true before current iteration.  
        bool temp[m];
        memset(temp,false,m);

        // For each element of arr[], we loop through
        // all elements of DP table from 1 to m and
        // we add current element i. e., arr[i] to
        // all those elements which are true in DP
        // table
        for (int j=0; j<m; j++)
        {
            // if an element is true in DP table
            if (DP[j] == true)
            {
                if (DP[(j+arr[i]) % m] == false)

                    // We update it in temp and update
                    // to DP once loop of j is over
                    temp[(j+arr[i]) % m] = true;
            }
        }

        // Updating all the elements of temp
        // to DP table since iteration over
        // j is over
        for (int j=0; j<m; j++)
            if (temp[j])
                DP[j] = true;


        // Also since arr[i] is a single element
        // subset, arr[i]%m is one of the possible
        // sum
        DP[arr[i]%m] = true;
    }

    return DP[0];
}

// Driver code
int main()
{
    int arr[] = {1, 7};
    int n = sizeof(arr)/sizeof(arr[0]);
    int m = 5;

    modularSum(arr, n, m) ?  cout << "YES\n" :
                             cout << "NO\n";

    return 0;
}

Output:

NO

Time Complexity : O(m^2)
Auxiliary Space : O(m)

This article is contributed by Pratik Chhajer. 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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