# 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 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 ` `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

## 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 "````); ` `        ``else` `            ``System.out.print(````"NO "````); ` `    ``} ` `} ` ` `  `//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

## PHP

 ` ``\$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` `= ``Array_fill``(0, ``\$m``, false); ` ` `  `    ``// we'll loop through all the elements of arr[]  ` `    ``for` `(``\$i` `= 0; ``\$i` `< ``\$n``; ``\$i``++)  ` `    ``{  ` `        ``// anytime we encounter a sum divisible  ` `        ``// by m, we are done  ` `        ``if` `(``\$DP``)  ` `            ``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` `= ``array_fill``(0, ``\$m``, 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` `(``\$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` `(``\$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``;  ` `}  ` ` `  `// Driver Code ` `\$arr` `= ``array``(1, 7);  ` `\$n` `= sizeof(``\$arr``);  ` `\$m` `= 5;  ` ` `  `if` `(modularSum(``\$arr``, ``\$n``, ``\$m``) == true ) ` `    ``echo` ```"YES "````; ` `else` `    ``echo` ```"NO "````;  ` ` `  `// This code is contributed by Ryuga ` `?> `

Output:

```NO
```

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