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++

// 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; }

## Java

// Java program to check if there is a subset // with sum divisible by m. import java.util.Arrays; class GFG { // Returns true if there is a subset // of arr[] with sum divisible by m static boolean 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 boolean DP[]=new boolean[m]; Arrays.fill(DP, false); // 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. boolean temp[] = new boolean[m]; Arrays.fill(temp, false); // 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 public static void main(String arg[]) { int arr[] = {1, 7}; int n = arr.length; int m = 5; if(modularSum(arr, n, m)) System.out.print("YES\n"); else System.out.print("NO\n"); } } //This code is contributed by Anant Agarwal.

## Python3

# Python3 program to check if there is # a subset with sum divisible by m. # Returns true if there is a subset # of arr[] with sum divisible by m def modularSum(arr, n, 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 DP = [False for i in range(m)] # we'll loop through all the elements of arr[] for i in range(n): # 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. temp = [False for i in range(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 j in range(m): # 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 j in range(m): 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 arr = [1, 7] n = len(arr) m = 5 print("YES") if(modularSum(arr, n, m)) else print("NO") # This code is contributed by Anant Agarwal.

## C#

// C# program to check if there is // a subset with sum divisible by m. using System; class GFG { // Returns true if there is a subset // of arr[] with sum divisible by m static 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=new bool[m]; for (int l=0;l<DP.Length;l++) DP[l]=false; // 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=new bool[m]; for (int l=0;l<temp.Length;l++) temp[l]=false; // 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 public static void Main() { int []arr = {1, 7}; int n = arr.Length; int m = 5; if(modularSum(arr, n, m)) Console.Write("YES\n"); else Console.Write("NO\n"); } } //This code is contributed by Anant Agarwal.

Output:

NO

Time Complexity : O(m^2)

Auxiliary Space : O(m)

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