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)

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