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 remainder of a string with value x be ‘r’ when divided with 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. 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 implementation of same in C++.

`// C++ program to count subsequences of a ` `// string divisble by n. ` `#include<bits/stdc++.h> ` `using` `namespace` `std; ` ` ` `// Returns count of subsequences of str ` `// divisble 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; ` `} ` |

Output:

4

Time Complexity : O(len * n)

Auxiliary Space : O(len * n)

This article is contributed by **Ekta Goel**. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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