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# 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.#includeusing 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

## 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 codeif __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



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++

 #includeusing 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

## 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 codestr = "1234"n = 4print(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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