Given an array **arr[]** of **N** integers and an integer **M**, the task is to find the number of non-empty subsequences such that the sum of the subsequence is divisible by **M**.

**Examples:**

Input:arr[] = {1, 2, 3}, M = 1

Output:7

Number of non-empty subsets of this array are 7.

Since m = 1, m will divide all the possible subsets.

Input:arr[] = {1, 2, 3}, M = 2

Output:3

**Approach:** A dynamic programming based approach with **O(N * SUM)** time complexity where **N** is the length of the array and **SUM** is the sum of all the array elements has been discussed in this article.

In this article, an improvement over the previous approach will be discussed. Instead of the sum as one of the states of DP, (sum % m) will be used as one of the states of the DP as it is sufficient. Thus, the time complexity boils down to O(m * N).

Recurrence relation:

dp[i][curr] = dp[i + 1][(curr + arr[i]) % m] + dp[i + 1][curr]

Let’s define the states now, **dp[i][curr]** simply means number of subsets of the sub-array **arr[i…N-1]** such that **(sum of its element + curr) % m = 0**.

Below is the implementation of the above approach:

## C++

`// C++ implementation of the approach ` `#include <bits/stdc++.h> ` `using` `namespace` `std; ` `#define maxN 20 ` `#define maxM 10 ` ` ` `// To store the states of DP ` `int` `dp[maxN][maxM]; ` `bool` `v[maxN][maxM]; ` ` ` `// Function to find the required count ` `int` `findCnt(` `int` `* arr, ` `int` `i, ` `int` `curr, ` `int` `n, ` `int` `m) ` `{ ` ` ` `// Base case ` ` ` `if` `(i == n) { ` ` ` `if` `(curr == 0) ` ` ` `return` `1; ` ` ` `else` ` ` `return` `0; ` ` ` `} ` ` ` ` ` `// If the state has been solved before ` ` ` `// return the value of the state ` ` ` `if` `(v[i][curr]) ` ` ` `return` `dp[i][curr]; ` ` ` ` ` `// Setting the state as solved ` ` ` `v[i][curr] = 1; ` ` ` ` ` `// Recurrence relation ` ` ` `return` `dp[i][curr] = findCnt(arr, i + 1, ` ` ` `curr, n, m) ` ` ` `+ findCnt(arr, i + 1, ` ` ` `(curr + arr[i]) % m, ` ` ` `n, m); ` `} ` ` ` `// Driver code ` `int` `main() ` `{ ` ` ` `int` `arr[] = { 3, 3, 3, 3 }; ` ` ` `int` `n = ` `sizeof` `(arr) / ` `sizeof` `(` `int` `); ` ` ` `int` `m = 6; ` ` ` ` ` `cout << findCnt(arr, 0, 0, n, m) - 1; ` ` ` ` ` `return` `0; ` `} ` |

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## Java

`// Java implementation of the approach ` `class` `GFG ` `{ ` `static` `int` `maxN = ` `20` `; ` `static` `int` `maxM = ` `10` `; ` ` ` `// To store the states of DP ` `static` `int` `[][]dp = ` `new` `int` `[maxN][maxM]; ` `static` `boolean` `[][]v = ` `new` `boolean` `[maxN][maxM]; ` ` ` `// Function to find the required count ` `static` `int` `findCnt(` `int` `[] arr, ` `int` `i, ` ` ` `int` `curr, ` `int` `n, ` `int` `m) ` `{ ` ` ` `// Base case ` ` ` `if` `(i == n) ` ` ` `{ ` ` ` `if` `(curr == ` `0` `) ` ` ` `return` `1` `; ` ` ` `else` ` ` `return` `0` `; ` ` ` `} ` ` ` ` ` `// If the state has been solved before ` ` ` `// return the value of the state ` ` ` `if` `(v[i][curr]) ` ` ` `return` `dp[i][curr]; ` ` ` ` ` `// Setting the state as solved ` ` ` `v[i][curr] = ` `true` `; ` ` ` ` ` `// Recurrence relation ` ` ` `return` `dp[i][curr] = findCnt(arr, i + ` `1` `, ` ` ` `curr, n, m) + ` ` ` `findCnt(arr, i + ` `1` `, ` ` ` `(curr + arr[i]) % m, ` ` ` `n, m); ` `} ` ` ` `// Driver code ` `public` `static` `void` `main(String[] args) ` `{ ` ` ` `int` `arr[] = { ` `3` `, ` `3` `, ` `3` `, ` `3` `}; ` ` ` `int` `n = arr.length; ` ` ` `int` `m = ` `6` `; ` ` ` ` ` `System.out.println(findCnt(arr, ` `0` `, ` `0` `, n, m) - ` `1` `); ` `} ` `} ` ` ` `// This code is contributed by 29AjayKumar ` |

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## Python3

`# Python3 implementation of the approach ` `maxN ` `=` `20` `maxM ` `=` `10` ` ` `# To store the states of DP ` `dp ` `=` `[[` `0` `for` `i ` `in` `range` `(maxN)] ` ` ` `for` `i ` `in` `range` `(maxM)] ` `v ` `=` `[[` `0` `for` `i ` `in` `range` `(maxN)] ` ` ` `for` `i ` `in` `range` `(maxM)] ` ` ` `# Function to find the required count ` `def` `findCnt(arr, i, curr, n, m): ` ` ` ` ` `# Base case ` ` ` `if` `(i ` `=` `=` `n): ` ` ` `if` `(curr ` `=` `=` `0` `): ` ` ` `return` `1` ` ` `else` `: ` ` ` `return` `0` ` ` ` ` `# If the state has been solved before ` ` ` `# return the value of the state ` ` ` `if` `(v[i][curr]): ` ` ` `return` `dp[i][curr] ` ` ` ` ` `# Setting the state as solved ` ` ` `v[i][curr] ` `=` `1` ` ` ` ` `# Recurrence relation ` ` ` `dp[i][curr] ` `=` `findCnt(arr, i ` `+` `1` `, ` ` ` `curr, n, m) ` `+` `\ ` ` ` `findCnt(arr, i ` `+` `1` `, ` ` ` `(curr ` `+` `arr[i]) ` `%` `m, n, m) ` ` ` `return` `dp[i][curr] ` ` ` `# Driver code ` `arr ` `=` `[` `3` `, ` `3` `, ` `3` `, ` `3` `] ` `n ` `=` `len` `(arr) ` `m ` `=` `6` ` ` `print` `(findCnt(arr, ` `0` `, ` `0` `, n, m) ` `-` `1` `) ` ` ` `# This code is contributed by Mohit Kumar ` |

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## C#

`// C# implementation of the approach ` `using` `System; ` ` ` `class` `GFG ` `{ ` ` ` `static` `int` `maxN = 20; ` ` ` `static` `int` `maxM = 10; ` ` ` ` ` `// To store the states of DP ` ` ` `static` `int` `[,]dp = ` `new` `int` `[maxN, maxM]; ` ` ` `static` `bool` `[,]v = ` `new` `bool` `[maxN, maxM]; ` ` ` ` ` `// Function to find the required count ` ` ` `static` `int` `findCnt(` `int` `[] arr, ` `int` `i, ` ` ` `int` `curr, ` `int` `n, ` `int` `m) ` ` ` `{ ` ` ` `// Base case ` ` ` `if` `(i == n) ` ` ` `{ ` ` ` `if` `(curr == 0) ` ` ` `return` `1; ` ` ` `else` ` ` `return` `0; ` ` ` `} ` ` ` ` ` `// If the state has been solved before ` ` ` `// return the value of the state ` ` ` `if` `(v[i, curr]) ` ` ` `return` `dp[i, curr]; ` ` ` ` ` `// Setting the state as solved ` ` ` `v[i, curr] = ` `true` `; ` ` ` ` ` `// Recurrence relation ` ` ` `return` `dp[i, curr] = findCnt(arr, i + 1, ` ` ` `curr, n, m) + ` ` ` `findCnt(arr, i + 1, ` ` ` `(curr + arr[i]) % m, n, m); ` ` ` `} ` ` ` ` ` `// Driver code ` ` ` `public` `static` `void` `Main() ` ` ` `{ ` ` ` `int` `[]arr = { 3, 3, 3, 3 }; ` ` ` `int` `n = arr.Length; ` ` ` `int` `m = 6; ` ` ` ` ` `Console.WriteLine(findCnt(arr, 0, 0, n, m) - 1); ` ` ` `} ` `} ` ` ` `// This code is contributed by kanugargng ` |

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**Output:**

7

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