Subsets with sum divisible by m

Given an array of size n 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.

Note:

The sum of all array elements in small, i.e., it is within integer range.
m > 0.

Examples:

Input : arr[] = [1, 2, 3, 4], m = 2
Output : 7
Explanation: The subsequences are [1, 3], [1, 3, 4], [1, 2, 3], [1, 2, 3, 4], [2], [2, 4], and [4].
Input : arr[] = [1, 2, 3], m = 3
Output : 3
Explanation: The subsequences are [1, 2], [1, 2, 3], [3].

[Naive Approach] Using Recursion - O(2^n) time and O(n) space

The idea is to recursively generate all the possible subsets. For every subset, compute its sum, and if the sum is multiple of m, increment result by 1. Finally, decrement the result value by 1 as empty subset is also counted in this approach.
Recurrence relation:
subCount(i, sum, m, arr) = subCount(i+1, sum+arr[i], m, arr) + subCount(i+1, sum, m, arr).
Base Case:
For i == n, subCount(i, sum, m, arr) = 1 if sum % m == 0, 0 otherwise.

C++

// C++ program to find Number of 
// subsets with sum divisible by m
#include <bits/stdc++.h>
using namespace std;

// Recursive function which finds the number of subsequences 
// starting from index i and current sum.
int subCountRecur(int i, int sum, int m, vector<int> &arr) {
    
    // Base case: End of array 
    // Check if sum is divisible by m
    if (i == arr.size()) {
        return (sum % m == 0)?1:0;
    }
    
    // Include current element in subset
    int take = subCountRecur(i+1, sum+arr[i], m, arr);
    
    // Exclude current element 
    int noTake = subCountRecur(i+1, sum, m, arr);
    
    return take + noTake;
}

// Function which finds Number of 
// subsets with sum divisible by m
int subCount(vector<int> &arr, int m) {
    
    // Decrement 1 from answer as empty 
    // subsequence is also counted.
    return subCountRecur(0, 0, m, arr) - 1;
}

int main() {
    vector<int> arr = {1, 2, 3, 4};
    int m = 2;
    cout << subCount(arr, m);

    return 0;
}

Java

// Java program to find Number of 
// subsets with sum divisible by m
import java.util.*;

class GfG {

    // Recursive function which finds the number of subsequences 
    // starting from index i and current sum.
    static int subCountRecur(int i, int sum, int m, int[] arr) {
        
        // Base case: End of array 
        // Check if sum is divisible by m
        if (i == arr.length) {
            return (sum % m == 0) ? 1 : 0;
        }
        
        // Include current element in subset
        int take = subCountRecur(i + 1, sum + arr[i], m, arr);
        
        // Exclude current element 
        int noTake = subCountRecur(i + 1, sum, m, arr);
        
        return take + noTake;
    }

    // Function which finds Number of 
    // subsets with sum divisible by m
    static int subCount(int[] arr, int m) {
        
        // Decrement 1 from answer as empty 
        // subsequence is also counted.
        return subCountRecur(0, 0, m, arr) - 1;
    }

    public static void main(String[] args) {
        int[] arr = {1, 2, 3, 4};
        int m = 2;
        System.out.println(subCount(arr, m));
    }
}

Python

# Python program to find Number of 
# subsets with sum divisible by m

# Recursive function which finds the number of subsequences 
# starting from index i and current sum.
def subCountRecur(i, sum, m, arr):
    
    # Base case: End of array 
    # Check if sum is divisible by m
    if i == len(arr):
        return 1 if sum % m == 0 else 0
    
    # Include current element in subset
    take = subCountRecur(i + 1, sum + arr[i], m, arr)
    
    # Exclude current element 
    noTake = subCountRecur(i + 1, sum, m, arr)
    
    return take + noTake

# Function which finds Number of 
# subsets with sum divisible by m
def subCount(arr, m):
    
    # Decrement 1 from answer as empty 
    # subsequence is also counted.
    return subCountRecur(0, 0, m, arr) - 1

if __name__ == "__main__":
    arr = [1, 2, 3, 4]
    m = 2
    print(subCount(arr, m))

// C# program to find Number of 
// subsets with sum divisible by m
using System;

class GfG {

    // Recursive function which finds the number of subsequences 
    // starting from index i and current sum.
    static int subCountRecur(int i, int sum, int m, int[] arr) {
        
        // Base case: End of array 
        // Check if sum is divisible by m
        if (i == arr.Length) {
            return (sum % m == 0) ? 1 : 0;
        }
        
        // Include current element in subset
        int take = subCountRecur(i + 1, sum + arr[i], m, arr);
        
        // Exclude current element 
        int noTake = subCountRecur(i + 1, sum, m, arr);
        
        return take + noTake;
    }

    // Function which finds Number of 
    // subsets with sum divisible by m
    static int subCount(int[] arr, int m) {
        
        // Decrement 1 from answer as empty 
        // subsequence is also counted.
        return subCountRecur(0, 0, m, arr) - 1;
    }

    static void Main(string[] args) {
        int[] arr = {1, 2, 3, 4};
        int m = 2;
        Console.WriteLine(subCount(arr, m));
    }
}

JavaScript

// JavaScript program to find Number of 
// subsets with sum divisible by m

// Recursive function which finds the number of subsequences 
// starting from index i and current sum.
function subCountRecur(i, sum, m, arr) {

    // Base case: End of array 
    // Check if sum is divisible by m
    if (i === arr.length) {
        return (sum % m === 0) ? 1 : 0;
    }

    // Include current element in subset
    let take = subCountRecur(i + 1, sum + arr[i], m, arr);

    // Exclude current element 
    let noTake = subCountRecur(i + 1, sum, m, arr);

    return take + noTake;
}

// Function which finds Number of 
// subsets with sum divisible by m
function subCount(arr, m) {

    // Decrement 1 from answer as empty 
    // subsequence is also counted.
    return subCountRecur(0, 0, m, arr) - 1;
}

let arr = [1, 2, 3, 4];
let m = 2;
console.log(subCount(arr, m));

Output

[Better Approach - 1] Using Top-Down DP (Memoization) – O(sumn) time and O(sumn) space

Optimal Substructure: Number of subsequences with sum divisible by m starting at index i and sum depends on the subproblems subCount(i+1, sum+arr[i], m) and subCount(i+1, sum, m). By solving these subproblems, we can determine the result for the current state.
Overlapping Subproblems: Recursive calls often compute the same (i, sum) states multiple times. To avoid recomputation, we store results in a 2D memo table.

Since recursion changes with i and sum, create a 2D memo table of size n * (sum + 1).
memo[i][j] stores the number of subsequences starting from index i with current sum j.
Before computing a state, check if it is already stored in memo. If yes, return it.
Subtract 1 from the final result to exclude the empty subsequence.

C++

// C++ program to find Number of 
// subsets with sum divisible by m
#include <bits/stdc++.h>
using namespace std;

// Memoized function which finds the number of subsequences 
// starting from index i and current sum.
int subCountRecur(int i, int sum, int m, vector<int> &arr, 
vector<vector<int>> &memo) {
    
    // Base case: End of array 
    // Check if sum is divisible by m
    if (i == arr.size()) {
        return (sum % m == 0) ? 1 : 0;
    }
    
    // If this state has already been computed, 
    // return the stored result
    if (memo[i][sum] != -1) {
        return memo[i][sum];
    }
    
    // Include current element in subset
    int take = subCountRecur(i+1, sum+arr[i], m, arr, memo);
    
    // Exclude current element 
    int noTake = subCountRecur(i+1, sum, m, arr, memo);
    
    // Store and return the result
    return memo[i][sum] = take + noTake;
}

// Function which finds Number of 
// subsets with sum divisible by m
int subCount(vector<int> &arr, int m) {
    int n = arr.size();
    
    // Calculate total sum of array
    int totalSum = 0;
    for (int num : arr) {
        totalSum += num;
    }
    
    // Initialize memoization table with -1
    vector<vector<int>> memo(arr.size(), vector<int>(totalSum + 1, -1));
    
    // Decrement 1 from answer as empty 
    // subsequence is also counted.
    return subCountRecur(0, 0, m, arr, memo) - 1;
}

int main() {
    vector<int> arr = {1, 2, 3, 4};
    int m = 2;
    cout << subCount(arr, m);
    return 0;
}

Java

// Java program to find Number of 
// subsets with sum divisible by m
import java.util.*;

class GfG {

    // Memoized function which finds the number of subsequences 
    // starting from index i and current sum.
    static int subCountRecur(int i, int sum, int m, 
    int[] arr, int[][] memo) {
        
        // Base case: End of array 
        // Check if sum is divisible by m
        if (i == arr.length) {
            return (sum % m == 0) ? 1 : 0;
        }

        // If this state has already been computed, 
        // return the stored result
        if (memo[i][sum] != -1) {
            return memo[i][sum];
        }

        // Include current element in subset
        int take = subCountRecur(i + 1, sum + arr[i], m, arr, memo);

        // Exclude current element 
        int noTake = subCountRecur(i + 1, sum, m, arr, memo);

        // Store and return the result
        return memo[i][sum] = take + noTake;
    }

    // Function which finds Number of 
    // subsets with sum divisible by m
    static int subCount(int[] arr, int m) {
        int n = arr.length;

        // Calculate total sum of array
        int totalSum = 0;
        for (int num : arr) {
            totalSum += num;
        }

        // Initialize memoization table with -1
        int[][] memo = new int[n][totalSum + 1];
        for (int i = 0; i < n; i++) {
            Arrays.fill(memo[i], -1);
        }

        // Decrement 1 from answer as empty 
        // subsequence is also counted.
        return subCountRecur(0, 0, m, arr, memo) - 1;
    }

    public static void main(String[] args) {
        int[] arr = {1, 2, 3, 4};
        int m = 2;
        System.out.println(subCount(arr, m));
    }
}

Python

# Python program to find Number of 
# subsets with sum divisible by m

# Memoized function which finds the number of subsequences 
# starting from index i and current sum.
def subCountRecur(i, sum, m, arr, memo):
    
    # Base case: End of array 
    # Check if sum is divisible by m
    if i == len(arr):
        return 1 if sum % m == 0 else 0

    # If this state has already been computed, 
    # return the stored result
    if memo[i][sum] != -1:
        return memo[i][sum]

    # Include current element in subset
    take = subCountRecur(i + 1, sum + arr[i], m, arr, memo)

    # Exclude current element 
    noTake = subCountRecur(i + 1, sum, m, arr, memo)

    # Store and return the result
    memo[i][sum] = take + noTake
    return memo[i][sum]

# Function which finds Number of 
# subsets with sum divisible by m
def subCount(arr, m):
    n = len(arr)

    # Calculate total sum of array
    totalSum = sum(arr)

    # Initialize memoization table with -1
    memo = [[-1 for _ in range(totalSum + 1)] for _ in range(n)]

    # Decrement 1 from answer as empty 
    # subsequence is also counted.
    return subCountRecur(0, 0, m, arr, memo) - 1

if __name__ == "__main__":
    arr = [1, 2, 3, 4]
    m = 2
    print(subCount(arr, m))

// C# program to find Number of 
// subsets with sum divisible by m
using System;

class GfG {

    // Memoized function which finds the number of subsequences 
    // starting from index i and current sum.
    static int subCountRecur(int i, int sum, int m, 
    int[] arr, int[,] memo) {
        
        // Base case: End of array 
        // Check if sum is divisible by m
        if (i == arr.Length) {
            return (sum % m == 0) ? 1 : 0;
        }

        // If this state has already been computed, 
        // return the stored result
        if (memo[i, sum] != -1) {
            return memo[i, sum];
        }

        // Include current element in subset
        int take = subCountRecur(i + 1, sum + arr[i], m, arr, memo);

        // Exclude current element 
        int noTake = subCountRecur(i + 1, sum, m, arr, memo);

        // Store and return the result
        return memo[i, sum] = take + noTake;
    }

    // Function which finds Number of 
    // subsets with sum divisible by m
    static int subCount(int[] arr, int m) {
        int n = arr.Length;

        // Calculate total sum of array
        int totalSum = 0;
        foreach (int num in arr) {
            totalSum += num;
        }

        // Initialize memoization table with -1
        int[,] memo = new int[n, totalSum + 1];
        for (int i = 0; i < n; i++) {
            for (int j = 0; j <= totalSum; j++) {
                memo[i, j] = -1;
            }
        }

        // Decrement 1 from answer as empty 
        // subsequence is also counted.
        return subCountRecur(0, 0, m, arr, memo) - 1;
    }

    static void Main(string[] args) {
        int[] arr = {1, 2, 3, 4};
        int m = 2;
        Console.WriteLine(subCount(arr, m));
    }
}

JavaScript

// JavaScript program to find Number of 
// subsets with sum divisible by m

// Memoized function which finds the number of subsequences 
// starting from index i and current sum.
function subCountRecur(i, sum, m, arr, memo) {

    // Base case: End of array 
    // Check if sum is divisible by m
    if (i === arr.length) {
        return (sum % m === 0) ? 1 : 0;
    }

    // If this state has already been computed, 
    // return the stored result
    if (memo[i][sum] !== -1) {
        return memo[i][sum];
    }

    // Include current element in subset
    let take = subCountRecur(i + 1, sum + arr[i], m, arr, memo);

    // Exclude current element 
    let noTake = subCountRecur(i + 1, sum, m, arr, memo);

    // Store and return the result
    memo[i][sum] = take + noTake;
    return memo[i][sum];
}

// Function which finds Number of 
// subsets with sum divisible by m
function subCount(arr, m) {
    let n = arr.length;

    // Calculate total sum of array
    let totalSum = arr.reduce((a, b) => a + b, 0);

    // Initialize memoization table with -1
    let memo = new Array(n).fill(0).map(() => new Array(totalSum + 1).fill(-1));

    // Decrement 1 from answer as empty 
    // subsequence is also counted.
    return subCountRecur(0, 0, m, arr, memo) - 1;
}

let arr = [1, 2, 3, 4];
let m = 2;
console.log(subCount(arr, m));

Output

[Better Approach - 2] Using Bottom-Up DP (Tabulation) – O(sumn) time and O(sumn) space

The idea is to fill the DP table based on future values. For each index i and current sum j, we either include arr[i] or exclude it to compute the number of subsequences whose sum is divisible by m. The table is filled in an iterative manner from i = n-1 down to 0, and for each sum from 0 to total sum.
The dynamic programming relation is as follows:
dp[i][j] = dp[i+1][j] + dp[i+1][j + arr[i]]
At the base case i = n, we set dp[n][j] = 1 if j % m == 0, else 0.
Finally, subtract 1 from dp[0][0] to exclude the empty subsequence.

C++

// C++ program to find Number of 
// subsets with sum divisible by m
#include <bits/stdc++.h>
using namespace std;

// Function which finds Number of 
// subsets with sum divisible by m
int subCount(vector<int> &arr, int m) {
    int n = arr.size();
    
    // Calculate total sum of array
    int totalSum = 0;
    for (int num : arr) {
        totalSum += num;
    }

    vector<vector<int>> dp(n + 1, vector<int>(totalSum + 1, 0));
    
    // Precompute the sums which are divisible
    // by m.
    for (int j=0; j<=totalSum; j++) {
        if (j%m == 0) dp[n][j] = 1;
    }
    
    // Current index
    for (int i=n-1; i>=0; i--) {
        
        // Current sum 
        for (int j=0; j<=totalSum; j++) {
            
            dp[i][j] = dp[i+1][j] + dp[i+1][j+arr[i]];
        }
    }
    
    return dp[0][0] - 1;
}

int main() {
    vector<int> arr = {1, 2, 3, 4};
    int m = 2;
    cout << subCount(arr, m);
    return 0;
}

Java

// Java program to find Number of 
// subsets with sum divisible by m
import java.util.*;

class Main {

    // Function which finds Number of 
    // subsets with sum divisible by m
    static int subCount(int[] arr, int m) {
        int n = arr.length;
        
        // Calculate total sum of array
        int totalSum = 0;
        for (int num : arr) {
            totalSum += num;
        }

        int[][] dp = new int[n + 1][totalSum + 1];
        
        // Precompute the sums which are divisible
        // by m.
        for (int j = 0; j <= totalSum; j++) {
            if (j % m == 0) dp[n][j] = 1;
        }
        
        // Current index
        for (int i = n - 1; i >= 0; i--) {
            
            // Current sum 
            for (int j = 0; j <= totalSum; j++) {
                
                // Exclude arr[i]
                dp[i][j] = dp[i + 1][j];
                
                // Include arr[i] 
                if (j + arr[i] <= totalSum) {
                    dp[i][j] += dp[i + 1][j + arr[i]];
                }
            }
        }
        
        return dp[0][0] - 1;
    }

    public static void main(String[] args) {
        int[] arr = {1, 2, 3, 4};
        int m = 2;
        System.out.println(subCount(arr, m));
    }
}

Python

# Python program to find Number of 
# subsets with sum divisible by m

# Function which finds Number of 
# subsets with sum divisible by m
def subCount(arr, m):
    n = len(arr)
    
    # Calculate total sum of array
    totalSum = 0
    for num in arr:
        totalSum += num

    dp = [[0] * (totalSum + 1) for _ in range(n + 1)]
    
    # Precompute the sums which are divisible
    # by m.
    for j in range(totalSum + 1):
        if j % m == 0:
            dp[n][j] = 1
    
    # Current index
    for i in range(n - 1, -1, -1):
        
        # Current sum 
        for j in range(totalSum + 1):
            
            # Exclude arr[i]
            dp[i][j] = dp[i + 1][j]
            
            # Include arr[i] 
            if j + arr[i] <= totalSum:
                dp[i][j] += dp[i + 1][j + arr[i]]
    
    return dp[0][0] - 1

if __name__ == "__main__":
    arr = [1, 2, 3, 4]
    m = 2
    print(subCount(arr, m))

// C# program to find Number of 
// subsets with sum divisible by m
using System;

class GfG {

    // Function which finds Number of 
    // subsets with sum divisible by m
    static int subCount(int[] arr, int m) {
        int n = arr.Length;
        
        // Calculate total sum of array
        int totalSum = 0;
        foreach (int num in arr) {
            totalSum += num;
        }

        int[,] dp = new int[n + 1, totalSum + 1];
        
        // Precompute the sums which are divisible
        // by m.
        for (int j = 0; j <= totalSum; j++) {
            if (j % m == 0) dp[n, j] = 1;
        }
        
        // Current index
        for (int i = n - 1; i >= 0; i--) {
            
            // Current sum 
            for (int j = 0; j <= totalSum; j++) {
                
                // Exclude arr[i]
                dp[i, j] = dp[i + 1, j];
                
                // Include arr[i] 
                if (j + arr[i] <= totalSum) {
                    dp[i, j] += dp[i + 1, j + arr[i]];
                }
            }
        }
        
        return dp[0, 0] - 1;
    }

    static void Main(string[] args) {
        int[] arr = {1, 2, 3, 4};
        int m = 2;
        Console.WriteLine(subCount(arr, m));
    }
}

JavaScript

// JavaScript program to find Number of 
// subsets with sum divisible by m

// Function which finds Number of 
// subsets with sum divisible by m
function subCount(arr, m) {
    let n = arr.length;

    // Calculate total sum of array
    let totalSum = 0;
    for (let num of arr) {
        totalSum += num;
    }

    let dp = new Array(n + 1).fill(0).map(() => new Array(totalSum + 1).fill(0));

    // Precompute the sums which are divisible
    // by m.
    for (let j = 0; j <= totalSum; j++) {
        if (j % m === 0) dp[n][j] = 1;
    }

    // Current index
    for (let i = n - 1; i >= 0; i--) {
        
        // Current sum 
        for (let j = 0; j <= totalSum; j++) {
            
            // Exclude arr[i]
            dp[i][j] = dp[i + 1][j];
            
            // Include arr[i] 
            if (j + arr[i] <= totalSum) {
                dp[i][j] += dp[i + 1][j + arr[i]];
            }
        }
    }

    return dp[0][0] - 1;
}

let arr = [1, 2, 3, 4];
let m = 2;
console.log(subCount(arr, m));

Output

[Expected Approach - 1] Using Space Optimized DP – O(sum*n) time and O(sum) space

In previous approach of dynamic programming we have derived the relation between states as given below:
dp[i][j] = dp[i+1][j] + dp[i+1][j + arr[i]]
If we observe carefully, for calculating current dp[i][j] state we only need values from the next row dp[i+1][...]. Hence, we can optimize space by using a single 1D array and updating it in reverse to simulate the row transitions.

C++

// C++ program to find Number of 
// subsets with sum divisible by m
#include <bits/stdc++.h>
using namespace std;

// Function which finds Number of 
// subsets with sum divisible by m
int subCount(vector<int> &arr, int m) {
    int n = arr.size();
    
    // Calculate total sum of array
    int totalSum = 0;
    for (int num : arr) {
        totalSum += num;
    }

    // dp array for tabulation
    vector<int> dp(totalSum + 1, 0);
    
    // Base case: sums which are divisible by m
    for (int j = 0; j <= totalSum; j++) {
        if (j % m == 0) dp[j] = 1;
    }
    
    // Process each element in the array
    for (int i = n - 1; i >= 0; i--) {
        
        // Current sum 
        for (int j = 0; j<=totalSum; j++) {
            
            // Include arr[i] if possible
            if (j + arr[i] <= totalSum) {
                dp[j] += dp[j + arr[i]];
            }
        }
    }
    
    return dp[0] - 1;
}

int main() {
    vector<int> arr = {1, 2, 3, 4};
    int m = 2;
    cout << subCount(arr, m);
    return 0;
}

Java

// Java program to find Number of 
// subsets with sum divisible by m
import java.util.*;

class GfG {

    // Function which finds Number of 
    // subsets with sum divisible by m
    static int subCount(int[] arr, int m) {
        int n = arr.length;
        
        // Calculate total sum of array
        int totalSum = 0;
        for (int num : arr) {
            totalSum += num;
        }

        // dp array for tabulation
        int[] dp = new int[totalSum + 1];
        
        // Base case: sums which are divisible by m
        for (int j = 0; j <= totalSum; j++) {
            if (j % m == 0) dp[j] = 1;
        }
        
        // Process each element in the array
        for (int i = n - 1; i >= 0; i--) {
            
            // Current sum 
            for (int j = 0; j <= totalSum; j++) {
                
                // Include arr[i] if possible
                if (j + arr[i] <= totalSum) {
                    dp[j] += dp[j + arr[i]];
                }
            }
        }
        
        return dp[0] - 1;
    }

    public static void main(String[] args) {
        int[] arr = {1, 2, 3, 4};
        int m = 2;
        System.out.println(subCount(arr, m));
    }
}

Python

# Python program to find Number of 
# subsets with sum divisible by m

# Function which finds Number of 
# subsets with sum divisible by m
def subCount(arr, m):
    n = len(arr)
    
    # Calculate total sum of array
    totalSum = 0
    for num in arr:
        totalSum += num

    # dp array for tabulation
    dp = [0] * (totalSum + 1)
    
    # Base case: sums which are divisible by m
    for j in range(totalSum + 1):
        if j % m == 0:
            dp[j] = 1
    
    # Process each element in the array
    for i in range(n - 1, -1, -1):
        
        # Current sum 
        for j in range(totalSum + 1):
            
            # Include arr[i] if possible
            if j + arr[i] <= totalSum:
                dp[j] += dp[j + arr[i]]
    
    return dp[0] - 1

if __name__ == "__main__":
    arr = [1, 2, 3, 4]
    m = 2
    print(subCount(arr, m))

// C# program to find Number of 
// subsets with sum divisible by m
using System;

class GfG {

    // Function which finds Number of 
    // subsets with sum divisible by m
    static int subCount(int[] arr, int m) {
        int n = arr.Length;
        
        // Calculate total sum of array
        int totalSum = 0;
        foreach (int num in arr) {
            totalSum += num;
        }

        // dp array for tabulation
        int[] dp = new int[totalSum + 1];
        
        // Base case: sums which are divisible by m
        for (int j = 0; j <= totalSum; j++) {
            if (j % m == 0) dp[j] = 1;
        }
        
        // Process each element in the array
        for (int i = n - 1; i >= 0; i--) {
            
            // Current sum 
            for (int j = 0; j <= totalSum; j++) {
                
                // Include arr[i] if possible
                if (j + arr[i] <= totalSum) {
                    dp[j] += dp[j + arr[i]];
                }
            }
        }
        
        return dp[0] - 1;
    }

    static void Main(string[] args) {
        int[] arr = {1, 2, 3, 4};
        int m = 2;
        Console.WriteLine(subCount(arr, m));
    }
}

JavaScript

// JavaScript program to find Number of 
// subsets with sum divisible by m

// Function which finds Number of 
// subsets with sum divisible by m
function subCount(arr, m) {
    let n = arr.length;

    // Calculate total sum of array
    let totalSum = 0;
    for (let num of arr) {
        totalSum += num;
    }

    // dp array for tabulation
    let dp = new Array(totalSum + 1).fill(0);

    // Base case: sums which are divisible by m
    for (let j = 0; j <= totalSum; j++) {
        if (j % m === 0) dp[j] = 1;
    }

    // Process each element in the array
    for (let i = n - 1; i >= 0; i--) {
        
        // Current sum 
        for (let j = 0; j <= totalSum; j++) {
            
            // Include arr[i] if possible
            if (j + arr[i] <= totalSum) {
                dp[j] += dp[j + arr[i]];
            }
        }
    }

    return dp[0] - 1;
}

let arr = [1, 2, 3, 4];
let m = 2;
console.log(subCount(arr, m));

Output

[Expected Approach - 2] Further Space Optimization - O(sum*n) time and O(m) space

The idea is to reduce the state space by observing that we only care whether a subsequence sum is divisible by m, so instead of tracking the exact sum, we can track the remainder of the sum modulo m. We define dp[i][curr] as the number of subsequences starting from index i with a current sum having remainder curr modulo m. The recurrence becomes dp[i][curr] = dp[i + 1][(curr + arr[i]) % m] + dp[i + 1][curr], representing the choices to include or exclude arr[i]. This allows us to reduce the DP table size from O(n × sum) to O(n × m), optimizing both time and space.

Refer to Number of subsets with sum divisible by M | Set 2 for detailed explanation and code.

Subsets with sum divisible by m

[Naive Approach] Using Recursion - O(2^n) time and O(n) space

[Better Approach - 1] Using Top-Down DP (Memoization) – O(sum*n) time and O(sum*n) space

[Better Approach - 2] Using Bottom-Up DP (Tabulation) – O(sum*n) time and O(sum*n) space

[Expected Approach - 1] Using Space Optimized DP – O(sum*n) time and O(sum) space

[Expected Approach - 2] Further Space Optimization - O(sum*n) time and O(m) space

Explore

[Better Approach - 1] Using Top-Down DP (Memoization) – O(sumn) time and O(sumn) space

[Better Approach - 2] Using Bottom-Up DP (Tabulation) – O(sumn) time and O(sumn) space