[leetcode] Distinct Subsequences 3


Given a string S and a string T, count the number of distinct subsequences of T in S.

A subsequence of a string is a new string which is formed from the original string by deleting some (can be none) of the characters without disturbing the relative positions of the remaining characters. (ie, "ACE" is a subsequence of "ABCDE" while "AEC" is not).

Here is an example:
S = "rabbbit", T = "rabbit"

Return 3.

//Dynamic programming
//dp[i][j] stores the number of distinct subsequences of first j chars in T and first i chars in S
//if(s[i - 1] == t[j - 1]) dp[i][j] = dp[i - 1][j - 1] + dp[i - 1][j];
//else dp[i][j] = dp[i - 1][j];
//dp[0][0] = 1
//dp[0][1~m] = 0
//dp[1~n][0] = 1

//reference: http://www.cnblogs.com/ganganloveu/p/3836519.html
class Solution {
public:
    int numDistinct(string s, string t) {
        int n = s.size();
        int m = t.size();
        if(n == 0 && m != 0){
            return 0;
        }
        if(n == 0 && m == 0){
            return 1;
        }
        if(m == 0){
            return 1;
        }
        int dp[n + 1][m + 1];
        for(int i = 0; i <= n; i++){
            dp[i][0] = 1;   
        }
        for(int i = 0; i <= m; i++){
            dp[0][i] = 0;
        }
        dp[0][0] = 1;
        for(int i = 1; i <= n; i++){
            for(int j = 1; j <= m; j++){
                if(s[i - 1] == t[j - 1]){
                    dp[i][j] = dp[i - 1][j - 1] + dp[i - 1][j];
                }
                else{
                    dp[i][j] = dp[i - 1][j];
                }
            }
        }
        return dp[n][m];
        
    }
};

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