Distinct Subsequences


Problem Statement :


Given two strings s and t, return the number of distinct subsequences of s which equals t.

The test cases are generated so that the answer fits on a 32-bit signed integer.

 

Example 1:

Input: s = "rabbbit", t = "rabbit"
Output: 3
Explanation:
As shown below, there are 3 ways you can generate "rabbit" from s.
rabbbit
rabbbit
rabbbit
Example 2:

Input: s = "babgbag", t = "bag"
Output: 5
Explanation:
As shown below, there are 5 ways you can generate "bag" from s.
babgbag
babgbag
babgbag
babgbag
babgbag
 

Constraints:

1 <= s.length, t.length <= 1000
s and t consist of English letters.


Solution :



title-img 


                            Solution in C :

int numDistinct(char * s, char * t){
    int M = 1e9 + 7 ;
    int m = strlen(s) ;
    int n = strlen(t) ;
    if(m < n)
        return 0 ;
    if(m == n)
        return (strcmp(s, t) == 0) ;
        
    int** dp = malloc((m+1) * sizeof(int*)) ;
    for(int i = 0; i <= m; i++){
        dp[i] = calloc(n+1, sizeof(int)) ;
        dp[i][0] = 1 ;
    }
    
    for(int i = 1;i <= m; i++){
        for(int j = 1; j <= n; j++){
            if(s[i-1] == t[j-1])
                dp[i][j] =( dp[i-1][j-1] + dp[i-1][j] ) % M;
            else
                dp[i][j] = dp[i-1][j] ;
        }
    }
    int ans = dp[m][n] ;
    for(int i = 0; i <= m; i++)
        free(dp[i]) ;
    free(dp) ;
        
    return ans ;
}
                        


                        Solution in C++ :

class Solution {
public:
    int recursiveWithoutMemoization(string& s, string& t, int s_ind, int t_ind) {
        if (t_ind == t.length()) {
            return 1;
        }

        if (s_ind == s.length()) {
            return 0;
        }

        int take = 0, notTake = 0;

        if (s[s_ind] == t[t_ind]) {
            take = recursiveWithoutMemoization(s, t, s_ind + 1, t_ind + 1);
        }

        notTake = recursiveWithoutMemoization(s, t, s_ind + 1, t_ind);

        return take + notTake;
    }

    int numDistinct(string s, string t) {
        return recursiveWithoutMemoization(s, t, 0, 0);
    }
};
                    


                        Solution in Java :

class Solution {
    public int recursiveWithoutMemoization(String s, String t, int s_ind, int t_ind) {
        if (t_ind == t.length()) {
            return 1;
        }
        if (s_ind == s.length()) {
            return 0;
        }

        int take = 0, notTake = 0;

        if (s.charAt(s_ind) == t.charAt(t_ind)) {
            take = recursiveWithoutMemoization(s, t, s_ind + 1, t_ind + 1);
        }

        notTake = recursiveWithoutMemoization(s, t, s_ind + 1, t_ind);

        return take + notTake;
    }

    public int numDistinct(String s, String t) {
        return recursiveWithoutMemoization(s, t, 0, 0);
    }
}
                    


                        Solution in Python : 
                            
class Solution:
    def recursiveWithoutMemoization(self, s: str, t: str, s_ind: int, t_ind: int) -> int:
        if t_ind == len(t):
            return 1
        if s_ind == len(s):
            return 0

        take, notTake = 0, 0

        if s[s_ind] == t[t_ind]:
            take = self.recursiveWithoutMemoization(s, t, s_ind + 1, t_ind + 1)

        notTake = self.recursiveWithoutMemoization(s, t, s_ind + 1, t_ind)

        return take + notTake

    def numDistinct(self, s: str, t: str) -> int:
        return self.recursiveWithoutMemoization(s, t, 0, 0)


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