# Longest subsequence whose sum is divisible by a given number

Given an array arr[] and an integer M, the task is to find the length of the longest subsequence whose sum is divisible by M. If there is no such sub-sequence then print 0
Examples:

Input: arr[] = {3, 2, 2, 1}, M = 3
Output:
Longest sub-sequence whose sum is
divisible by 3 is {3, 2, 1}
Input: arr[] = {2, 2}, M = 3
Output:

Approach: A simple way to solve this will be to generate all the possible sub-sequences and then find the largest among them divisible whose sum is divisible by M. However, for smaller values of M, a dynamic programming based approach can be used.
Let’s look at the recurrence relation first.

dp[i][curr_mod] = max(dp[i + 1][curr_mod], dp[i + 1][(curr_mod + arr[i]) % m] + 1)

Let’s understand the states of DP now. Here, dp[i][curr_mod] stores the longest subsequence of subarray arr[i…N-1] such that the sum of this subsequence and curr_mod is divisible by M. At each step, either index i can be chosen updating curr_mod or it can be ignored.
Also, note that only SUM % m needs to be stored instead of the entire sum as this information is sufficient to complete the states of DP.
Below is the implementation of the above approach:

## C++

 // C++ implementation of the approach #include using namespace std; #define maxN 20 #define maxM 64   // To store the states of DP int dp[maxN][maxM]; bool v[maxN][maxM];   // Function to return the length // of the longest subsequence // whose sum is divisible by m int findLen(int* arr, int i, int curr,             int n, int m) {     // Base case     if (i == n) {         if (!curr)             return 0;         else             return -1;     }       // 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     int l = findLen(arr, i + 1, curr, n, m);     int r = findLen(arr, i + 1,                     (curr + arr[i]) % m, n, m);     dp[i][curr] = l;     if (r != -1)         dp[i][curr] = max(dp[i][curr], r + 1);     return dp[i][curr]; }   // Driver code int main() {     int arr[] = { 3, 2, 2, 1 };     int n = sizeof(arr) / sizeof(int);     int m = 3;       cout << findLen(arr, 0, 0, n, m);       return 0; }

## Java

 // Java implementation of the approach class GFG {   static int maxN = 20; static int maxM = 64;   // To store the states of DP static int [][]dp = new int[maxN][maxM]; static boolean [][]v = new boolean[maxN][maxM];   // Function to return the length // of the longest subsequence // whose sum is divisible by m static int findLen(int[] arr, int i,                    int curr, int n, int m) {     // Base case     if (i == n)     {         if (curr == 0)             return 0;         else             return -1;     }       // 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     int l = findLen(arr, i + 1, curr, n, m);     int r = findLen(arr, i + 1,                    (curr + arr[i]) % m, n, m);     dp[i][curr] = l;     if (r != -1)         dp[i][curr] = Math.max(dp[i][curr], r + 1);     return dp[i][curr]; }   // Driver code public static void main(String []args) {     int arr[] = { 3, 2, 2, 1 };     int n = arr.length;     int m = 3;       System.out.println(findLen(arr, 0, 0, n, m)); } }   // This code is contributed by 29AjayKumar

## Python3

 # Python3 implementation of the approach import numpy as np   maxN = 20 maxM = 64   # To store the states of DP dp = np.zeros((maxN, maxM)); v = np.zeros((maxN, maxM));   # Function to return the length # of the longest subsequence # whose sum is divisible by m def findLen(arr, i, curr, n, m) :           # Base case     if (i == n) :         if (not curr) :             return 0;         else :             return -1;       # 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     l = findLen(arr, i + 1, curr, n, m);     r = findLen(arr, i + 1,                (curr + arr[i]) % m, n, m);           dp[i][curr] = l;     if (r != -1) :         dp[i][curr] = max(dp[i][curr], r + 1);               return dp[i][curr];   # Driver code if __name__ == "__main__" :       arr = [ 3, 2, 2, 1 ];     n = len(arr);     m = 3;       print(findLen(arr, 0, 0, n, m));   # This code is contributed by AnkitRai

## C#

 // C# implementation of the approach using System;                       class GFG {       static int maxN = 20; static int maxM = 64;   // To store the states of DP static int [,]dp = new int[maxN, maxM]; static Boolean [,]v = new Boolean[maxN, maxM];   // Function to return the length // of the longest subsequence // whose sum is divisible by m static int findLen(int[] arr, int i,                    int curr, int n, int m) {     // Base case     if (i == n)     {         if (curr == 0)             return 0;         else             return -1;     }       // 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     int l = findLen(arr, i + 1, curr, n, m);     int r = findLen(arr, i + 1,                    (curr + arr[i]) % m, n, m);     dp[i, curr] = l;     if (r != -1)         dp[i, curr] = Math.Max(dp[i, curr], r + 1);     return dp[i, curr]; }   // Driver code public static void Main(String []args) {     int []arr = { 3, 2, 2, 1 };     int n = arr.Length;     int m = 3;       Console.WriteLine(findLen(arr, 0, 0, n, m)); } }   // This code is contributed by 29AjayKumar

## Javascript



Output

3

Time Complexity: O(N * M)
Auxiliary Space: O(N * M).

Efficient approach : Using DP Tabulation method ( Iterative approach )

The approach to solve this problem is same but DP tabulation(bottom-up) method is better then Dp + memoization(top-down) because memoization method needs extra stack space of recursion calls.

Steps to solve this problem :

• Create a DP to store the solution of the subproblems.
• Initialize the DP with base cases by initializing the values of DP with 0 and -1.
• Now Iterate over subproblems to get the value of current problem form previous computation of subproblems stored in DP
• Return the final solution stored in dp[0][0].

Below is the implementation of the above approach:

## C++

 #include using namespace std; #define maxN 20 #define maxM 64   // Function to return the length of the longest subsequence // whose sum is divisible by m int findLen(int* arr, int n, int m) {     // To store the states of DP     int dp[n + 1][maxM];       // Base case     for (int curr = 0; curr < m; ++curr) {         if (curr == 0)             dp[n][curr] = 0;         else             dp[n][curr] = -1;     }       // Tabulation     for (int i = n - 1; i >= 0; --i) {         for (int curr = 0; curr < m; ++curr) {             // Recurrence relation             int l = dp[i + 1][curr];             int r = dp[i + 1][(curr + arr[i]) % m];             dp[i][curr] = l;             if (r != -1)                 dp[i][curr] = max(dp[i][curr], r + 1);         }     }       if (dp[0][0] == -1)         return 0;     else         return dp[0][0]; }   // Driver code int main() {     int arr[] = { 3, 2, 2, 1 };     int n = sizeof(arr) / sizeof(int);     int m = 3;         // Function call     cout << findLen(arr, n, m);       return 0; } // -- by bhardwajji

## Java

 import java.util.Arrays;   public class Main {     static final int maxN = 20;     static final int maxM = 64;       // Function to return the length of the longest     // subsequence whose sum is divisible by m     static int findLen(int[] arr, int n, int m)     {         // To store the states of DP         int[][] dp = new int[n + 1][maxM];           // Base case         for (int curr = 0; curr < m; ++curr) {             if (curr == 0)                 dp[n][curr] = 0;             else                 dp[n][curr] = -1;         }           // Tabulation         for (int i = n - 1; i >= 0; --i) {             for (int curr = 0; curr < m; ++curr) {                 // Recurrence relation                 int l = dp[i + 1][curr];                 int r = dp[i + 1][(curr + arr[i]) % m];                 dp[i][curr] = l;                 if (r != -1)                     dp[i][curr]                         = Math.max(dp[i][curr], r + 1);             }         }           if (dp[0][0] == -1)             return 0;         else             return dp[0][0];     }       // Driver code     public static void main(String[] args)     {         int[] arr = { 3, 2, 2, 1 };         int n = arr.length;         int m = 3;           // Function call         System.out.println(findLen(arr, n, m));     } }

## Python

 def find_len(arr, n, m):     # Define the maximum values for maxN and maxM     maxN = 20     maxM = 64       # To store the states of DP     dp = [[0 for _ in range(maxM)] for _ in range(n + 1)]       # Base case     for curr in range(m):         if curr == 0:             dp[n][curr] = 0         else:             dp[n][curr] = -1       # Tabulation     for i in range(n - 1, -1, -1):         for curr in range(m):             # Recurrence relation             l = dp[i + 1][curr]             r = dp[i + 1][(curr + arr[i]) % m]             dp[i][curr] = l             if r != -1:                 dp[i][curr] = max(dp[i][curr], r + 1)       if dp[0][0] == -1:         return 0     else:         return dp[0][0]     if __name__ == "__main__":     arr = [3, 2, 2, 1]     n = len(arr)     m = 3       # Function call     print(find_len(arr, n, m))

## C#

 using System;   class GFG {     const int maxN = 20;     const int maxM = 64;       // Function to return the length of the longest     // subsequence whose sum is divisible by m     static int FindLen(int[] arr, int n, int m)     {         // To store the states of DP         int[, ] dp = new int[n + 1, maxM];           // Base case         for (int curr = 0; curr < m; ++curr) {             if (curr == 0)                 dp[n, curr] = 0;             else                 dp[n, curr] = -1;         }           // Tabulation         for (int i = n - 1; i >= 0; --i) {             for (int curr = 0; curr < m; ++curr) {                 // Recurrence relation                 int l = dp[i + 1, curr];                 int r = dp[i + 1, (curr + arr[i]) % m];                 dp[i, curr] = l;                 if (r != -1)                     dp[i, curr]                         = Math.Max(dp[i, curr], r + 1);             }         }           if (dp[0, 0] == -1)             return 0;         else             return dp[0, 0];     }       // Driver code     static void Main(string[] args)     {         int[] arr = { 3, 2, 2, 1 };         int n = arr.Length;         int m = 3;           // Function call         Console.WriteLine(FindLen(arr, n, m));     } }

## Javascript

 // Function to return the length of the longest subsequence // whose sum is divisible by m function findLen(arr, n, m) {     // To store the states of DP     const dp = new Array(n + 1);     for (let i = 0; i <= n; i++) {         dp[i] = new Array(m);     }       // Base case     for (let curr = 0; curr < m; curr++) {         if (curr === 0) {             dp[n][curr] = 0;         } else {             dp[n][curr] = -1;         }     }       // Tabulation     for (let i = n - 1; i >= 0; i--) {         for (let curr = 0; curr < m; curr++) {             // Recurrence relation             let l = dp[i + 1][curr];             let r = dp[i + 1][(curr + arr[i]) % m];             dp[i][curr] = l;             if (r !== -1) {                 dp[i][curr] = Math.max(dp[i][curr], r + 1);             }         }     }       if (dp[0][0] === -1) {         return 0;     } else {         return dp[0][0];     } }   // Driver code const arr = [3, 2, 2, 1]; const n = arr.length; const m = 3;   // Function call console.log(findLen(arr, n, m));

Output

3

Time Complexity: O(N * M)
Auxiliary Space: O(N * M).

Space Optimization Approach: From tabulation, we can observe that we only require previous computed result to compute the current dp values. So, we can only maintain two 1D arrays instead of a 2D array. The whole logic of the code is same, but we can optimize the space complexity.

Below is the implementation of the above approach.

## C++

 #include using namespace std; #define maxN 20 #define maxM 64   // Function to return the length of the longest subsequence // whose sum is divisible by m int findLen(int* arr, int n, int m) {     // To store the states of DP       vector dp(maxM),new_dp(maxM);       // Initialize the DP array     for (int curr = 0; curr < m; ++curr) {         dp[curr] = -1;     }     dp[0] = 0;       // Tabulation with space optimization     for (int i = n-1; i >=0; --i) {           // compute new_dp from prev dp           new_dp=dp;         for (int curr = 0; curr < m; ++curr) {             // Recurrence relation             int l = dp[curr];// not pick             int r = dp[(curr - arr[i] + m) % m]; // pick             new_dp[curr] = l;             if (r != -1)                 new_dp[curr] = max(new_dp[curr], r + 1);         }           // use new_dp as prev dp           // to compute next states         dp=new_dp;     }       if (dp[0] == -1)         return 0;     else         return dp[0]; }   // Driver code int main() {     int arr[] = { 3, 2, 2, 1 };     int n = sizeof(arr) / sizeof(int);     int m = 3;       // Function call     cout << findLen(arr, n, m);       return 0; }

Time Complexity: O(N * M)

Auxiliary Space: O(M + M), as we use two M size arrays.

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