Skip to content
Related Articles

Related Articles

Improve Article
Longest subsequence whose sum is divisible by a given number
  • Difficulty Level : Hard
  • Last Updated : 14 May, 2021

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 <bits/stdc++.h>
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




<script>
 
// Javascript implementation of the approach
var maxN = 20
var maxM = 64
 
// To store the states of DP
var dp = Array.from(Array(maxN), ()=> Array(maxM).fill(0));
var v = Array.from(Array(maxN), ()=> Array(maxM).fill(false));
 
// Function to return the length
// of the longest subsequence
// whose sum is divisible by m
function findLen(arr, i, curr, n, 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
    var l = findLen(arr, i + 1, curr, n, m);
    var 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
var arr = [3, 2, 2, 1];
var n = arr.length;
var m = 3;
document.write( findLen(arr, 0, 0, n, m));
 
</script>
Output: 
3

 

Time Complexity: O(N * M)
 

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.  To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.

In case you wish to attend live classes with industry experts, please refer Geeks Classes Live 




My Personal Notes arrow_drop_up
Recommended Articles
Page :