Number of subsequences in a string divisible by n

Given a string consisting of digits 0-9, count the number of subsequences in it divisible by m.
Examples:

Input  : str = "1234", n = 4
Output : 4
The subsequences 4, 12, 24 and 124 are 
divisible by 4.
 
Input  : str = "330", n = 6
Output : 4
The subsequences 30, 30, 330 and 0 are 
divisible by n.
Input  : str = "676", n = 6
Output : 3
The subsequences 6, 6 and 66

This problem can be recursively defined. Let the remainder of a string with value x be ‘r’ when divided by n. Adding one more character to this string changes its remainder to (r*10 + newdigit) % n. For every new character, we have two choices, either add it in all current subsequences or ignore it. Thus, we have an optimal substructure. The following shows the brute force version of this:

string str = "330";
int n = 6
// idx is value of current index in str
// rem is current remainder
int count(int idx, int rem)
{
    // If last character reached
    if (idx == n)
        return (rem == 0)? 1 : 0;
    int ans = 0;
    // we exclude it, thus remainder
    // remains the same
    ans += count(idx+1, rem);
    // we include it and thus new remainder
    ans += count(idx+1, (rem*10 + str[idx]-'0')%n);
    return ans;
}

The above recursive solution has overlapping subproblems as shown in below recursion tree.

          input string = "330"
             (0,0) ===> at 0th index with 0 remainder
(exclude 0th /      (include 0th character)
 character) /      
       (1,0)      (1,3) ======> at index 1 with 3 as 
      (E)/  (I)     /(E)       the current remainder
     (2,0)  (2,3)   (2,3)
             |-------|
     These two subproblems overlap  

Thus, we can apply Dynamic Programming. Below is the implementation.

C++

// C++ program to count subsequences of a
// string divisible by n.
#include<bits/stdc++.h>
using namespace std;

// Returns count of subsequences of str
// divisible by n.
int countDivisibleSubseq(string str, int n)
{
    int len = str.length();

    // division by n can leave only n remainder
    // [0..n-1]. dp[i][j] indicates number of
    // subsequences in string [0..i] which leaves
    // remainder j after division by n.
    int dp[len][n];
    memset(dp, 0, sizeof(dp));

    // Filling value for first digit in str
    dp[0][(str[0]-'0')%n]++;

    for (int i=1; i<len; i++)
    {
        // start a new subsequence with index i
        dp[i][(str[i]-'0')%n]++;

        for (int j=0; j<n; j++)
        {
            // exclude i'th character from all the
            // current subsequences of string [0...i-1]
            dp[i][j] += dp[i-1][j];

            // include i'th character in all the current
            // subsequences of string [0...i-1]
            dp[i][(j*10 + (str[i]-'0'))%n] += dp[i-1][j];
        }
    }

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

// Driver code
int main()
{
    string str = "1234";
    int n = 4;
    cout << countDivisibleSubseq(str, n);
    return 0;
}

Java

//Java program to count subsequences of a
// string divisible by n

class GFG {

// Returns count of subsequences of str
// divisible by n.
    static int countDivisibleSubseq(String str, int n) {
        int len = str.length();

        // division by n can leave only n remainder
        // [0..n-1]. dp[i][j] indicates number of
        // subsequences in string [0..i] which leaves
        // remainder j after division by n.
        int dp[][] = new int[len][n];

        // Filling value for first digit in str
        dp[0][(str.charAt(0) - '0') % n]++;

        for (int i = 1; i < len; i++) {
            // start a new subsequence with index i
            dp[i][(str.charAt(i) - '0') % n]++;

            for (int j = 0; j < n; j++) {
                // exclude i'th character from all the
                // current subsequences of string [0...i-1]
                dp[i][j] += dp[i - 1][j];

                // include i'th character in all the current
                // subsequences of string [0...i-1]
                dp[i][(j * 10 + (str.charAt(i) - '0')) % n] += dp[i - 1][j];
            }
        }

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

// Driver code
    public static void main(String[] args) {
        String str = "1234";
        int n = 4;
        System.out.print(countDivisibleSubseq(str, n));
    }
}
// This code is contributed by 29AjayKumar

Python 3

# Python 3 program to count subsequences
# of a string divisible by n.

# Returns count of subsequences of
# str divisible by n.
def countDivisibleSubseq(str, n):
    l = len(str)

    # division by n can leave only n remainder
    # [0..n-1]. dp[i][j] indicates number of
    # subsequences in string [0..i] which leaves
    # remainder j after division by n.
    dp = [[0 for x in range(l)]
             for y in range(n)]

    # Filling value for first digit in str
    dp[int(str[0]) % n][0] += 1

    for i in range(1, l):

       # start a new subsequence with index i
        dp[int(str[i]) % n][i] += 1

        for j in range(n):

          # exclude i'th character from all the
          # current subsequences of string [0...i-1]
          dp[j][i] += dp[j][i-1]

          # include i'th character in all the current
          # subsequences of string [0...i-1]
          dp[(j * 10 + int(str[i])) % n][i] += dp[j][i-1]

    return dp[0][l-1]

# Driver code
if __name__ == "__main__":

    str = "1234"
    n = 4
    print(countDivisibleSubseq(str, n))

# This code is contributed by ita_c

C#

//C# program to count subsequences of a
// string divisible by n

using System;
class GFG {

// Returns count of subsequences of str
// divisible by n.
    static int countDivisibleSubseq(string str, int n) {
        int len = str.Length;

        // division by n can leave only n remainder
        // [0..n-1]. dp[i][j] indicates number of
        // subsequences in string [0..i] which leaves
        // remainder j after division by n.
        int[,] dp = new int[len,n];

        // Filling value for first digit in str
        dp[0,(str[0] - '0') % n]++;

        for (int i = 1; i < len; i++) {
            // start a new subsequence with index i
            dp[i,(str[i] - '0') % n]++;

            for (int j = 0; j < n; j++) {
                // exclude i'th character from all the
                // current subsequences of string [0...i-1]
                dp[i,j] += dp[i - 1,j];

                // include i'th character in all the current
                // subsequences of string [0...i-1]
                dp[i,(j * 10 + (str[i] - '0')) % n] += dp[i - 1,j];
            }
        }

        return dp[len - 1,0];
    }

// Driver code
    public static void Main() {
        String str = "1234";
        int n = 4;
        Console.Write(countDivisibleSubseq(str, n));
    }
}

Javascript

<script>
//Javascript program to count subsequences of a
// string divisible by n

    // Returns count of subsequences of str
    // divisible by n.
    function countDivisibleSubseq(str,n)
    {
        let len = str.length;

        // division by n can leave only n remainder
        // [0..n-1]. dp[i][j] indicates number of
        // subsequences in string [0..i] which leaves
        // remainder j after division by n.
        let dp = new Array(len);
        for(let i = 0; i < len; i++)
        {
            dp[i] = new Array(n);
            for(let j = 0; j < n; j++)
            {
                dp[i][j] = 0;
            }
        }

        // Filling value for first digit in str
        dp[0][(str[0] - '0') % n]++;

        for (let i = 1; i < len; i++) {
            // start a new subsequence with index i
            dp[i][(str[i] - '0') % n]++;

            for (let j = 0; j < n; j++) {
                // exclude i'th character from all the
                // current subsequences of string [0...i-1]
                dp[i][j] += dp[i - 1][j];

                // include i'th character in all the current
                // subsequences of string [0...i-1]
                dp[i][(j * 10 + (str[i] - '0')) % n] += dp[i - 1][j];
            }
        }

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

    // Driver code
    let str = "1234";
    let n = 4;
    document.write(countDivisibleSubseq(str, n));

    // This code is contributed by avanitrachhadiya2155
</script>

Output

4

Time Complexity: O(len * n) 
Auxiliary Space : O(len * n)

Efficient approach : Space optimization

In previous approach the dp[i][j] is depend upon the current and previous row of 2D matrix. So to optimize space we use two vectors temp and dp that keep track of current and previous row of DP.

Implementation Steps:

• The countDivisibleSubseq function counts the number of subsequences in a given string str that are divisible by a given number n.
• It initializes an array dp of size n to store counts.
• It iterates through each digit of the string and updates the counts in dp based on remainders.
• At each iteration, it considers the current digit and the previous counts to calculate the updated counts.
• Finally, it returns the count of subsequences divisible by n stored in dp[0].

Implementation:

C++

#include<bits/stdc++.h>
using namespace std;

int countDivisibleSubseq(string str, int n)
{
    int len = str.length();
    int dp[n];
    memset(dp, 0, sizeof(dp));
    dp[(str[0]-'0')%n]++; // Increment the count of remainder of first digit by n

    for (int i=1; i<len; i++)
    {
        int temp[n];
        memset(temp, 0, sizeof(temp));

        temp[(str[i]-'0')%n]++; // Increment the count of remainder of current digit by n

        for (int j=0; j<n; j++)
        {
            temp[j] += dp[j]; // Carry over the counts from previous digit

            temp[(j*10 + (str[i]-'0'))%n] += dp[j]; // Update the count with the new remainder formed by appending the current digit
        }

        for (int j=0; j<n; j++)
        {
            dp[j] = temp[j]; // Copy the updated counts from temp back to dp for the next iteration
        }
    }

    return dp[0]; // Return the count of subsequences divisible by n
}

int main()
{
    string str = "1234";
    int n = 4;
    cout << countDivisibleSubseq(str, n);
    return 0;
}

Java

// Java program to count subsequences
// of a string divisible by n.

public class GFG {
    public static int countDivisibleSubseq(String str, int n) {
        int length = str.length();
        int[] dp = new int[n]; // Create an array of size n to store counts

        // Increment the count of remainder of first digit by n
        dp[Integer.parseInt(String.valueOf(str.charAt(0))) % n] += 1;

        for (int i = 1; i < length; i++) {
            int[] temp = new int[n]; // Create a temporary array of size n

            // Increment the count of remainder of current digit by n
            temp[Integer.parseInt(String.valueOf(str.charAt(i))) % n] += 1;

            for (int j = 0; j < n; j++) {
                temp[j] += dp[j]; // Carry over the counts from the previous digit

                // Calculate the new remainder
                int newRemainder = (j * 10 + Integer.parseInt(String.valueOf(str.charAt(i)))) % n;
                // Update the count with the new remainder
                temp[newRemainder] += dp[j];
            }

            dp = temp;
        }

        return dp[0];
    }
//Driver code
    public static void main(String[] args) {
        String str = "1234";
        int n = 4;
        System.out.println(countDivisibleSubseq(str, n));
    }
}

Python3

# Python 3 program to count subsequences
# of a string divisible by n.


def countDivisibleSubseq(str, n):
    length = len(str)
    dp = [0] * n  # Create an array of size n
    # Increment the count of remainder of first digit by n
    dp[int(str[0]) % n] += 1

    for i in range(1, length):
        temp = [0] * n  # Create a temporary array of size n
        # Increment the count of remainder of current digit by n
        temp[int(str[i]) % n] += 1

        for j in range(n):
            temp[j] += dp[j]  # Carry over the counts from the previous digit

            # Calculate the new remainder
            new_remainder = (j * 10 + int(str[i])) % n
            # Update the count with the new remainder
            temp[new_remainder] += dp[j]

        dp = temp

    return dp[0]


# Driver code
str = "1234"
n = 4
print(countDivisibleSubseq(str, n))

C#

using System;

class GFG {
    static int CountDivisibleSubseq(string str, int n)
    {
        int len = str.Length;
        int[] dp = new int[n];
        Array.Fill(dp, 0);
        dp[(str[0] - '0')
           % n]++; // Increment the count of remainder of
                   // first digit by n

        for (int i = 1; i < len; i++) {
            int[] temp = new int[n];
            Array.Fill(temp, 0);

            temp[(str[i] - '0')
                 % n]++; // Increment the count of remainder
                         // of current digit by n

            for (int j = 0; j < n; j++) {
                temp[j] += dp[j]; // Carry over the counts
                                  // from previous digit

                temp[(j * 10 + (str[i] - '0')) % n]
                    += dp[j]; // Update the count with the
                              // new remainder formed by
                              // appending the current digit
            }

            for (int j = 0; j < n; j++) {
                dp[j] = temp[j]; // Copy the updated counts
                                 // from temp back to dp for
                                 // the next iteration
            }
        }

        return dp[0]; // Return the count of subsequences
                      // divisible by n
    }

    static void Main()
    {
        string str = "1234";
        int n = 4;
        Console.WriteLine(CountDivisibleSubseq(str, n));
    }
}

Javascript

function countDivisibleSubseq(str, n) {
    const len = str.length;
    const dp = new Array(n).fill(0);

    // Increment the count of remainder of first digit by n
    dp[Number(str[0]) % n]++;

    for (let i = 1; i < len; i++) {
        const temp = new Array(n).fill(0);

         // Increment the count of remainder of current digit by n
        temp[Number(str[i]) % n]++;

        for (let j = 0; j < n; j++) {
            temp[j] += dp[j]; // Carry over the counts from previous digit

            // Update the count with the new remainder
            // formed by appending the current digit
            temp[(j * 10 + Number(str[i])) % n] += dp[j];
        }

        for (let j = 0; j < n; j++) {
            // Copy the updated counts from
            // temp back to dp for the next iteration
            dp[j] = temp[j];
        }
    }

    return dp[0]; // Return the count of subsequences divisible by n
}

const str = "1234";
const n = 4;
console.log(countDivisibleSubseq(str, n));

Output

4

Time Complexity: O(len * n) 
Auxiliary Space : O(n)

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