Distinct Subsequences solution report, subsequences

Distinct Subsequences solution report, subsequences

Question: give two strings S and T to determine the number of times T appears 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"ABCDE"While"AEC"Is not ).

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

Return3.

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Thought 1: recursion (TLE)

If the current character is the same, the number of matching methods after S and T are added to the result

If the current character is different, the pointer of S is moved back for Recursive calculation.

class Solution {private:    int cnt;    int len_s;    int len_t;public:    Solution():cnt(0){}    void Count(string S,string T, int idx_ss, int idx_ts){        if(idx_ts == len_t){            cnt++;            return;        }        int i,j,k;        for (i=idx_ss; i<len_s; i++) {            if (S[i] == T[idx_ts]) {                Count(S, T, i + 1, idx_ts + 1);            }        }    }        int numDistinct(string S, string T) {        len_s = S.length();        len_t = T.length();        Count(S, T, 0, 0);        return cnt;    }};

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Idea 2: DP

If the current character is the same, the dp [I] [j] result is equivalent to using S [I] (dp [I-1] [J-1]) sum of the number of [I] (dp [I-1] [j]) Methods

If the current character is different, dp [I] [j] = dp [I-1] [j]

class Solution {private:    int len_s;    int len_t;public:    int Count(string S,string T){        int i,j;        int dp[len_s][len_t];        memset(dp, 0, sizeof(dp));                if (S[0]==T[0]) {            dp[0][0] = 1;        }                for(i=1;i<len_s;i++){            dp[i][0] = dp[i-1][0];            if (T[0]==S[i]) {                dp[i][0]++;            }        }                        for (i=1; i<len_s; i++) {            for (j=1; j<len_t && j<=i; j++) {                if (S[i]!=T[j]) {                    dp[i][j] = dp[i-1][j];                    //cout<<dp[i-1][j]<<endl;                }                else{                    dp[i][j] = dp[i-1][j-1] + dp[i-1][j];                    //dp[i-1][j-1]: use S[i], as S[i]==T[j]                    //dp[i-1][j]  : don't use S[i]                    //cout<<dp[i][j]<<endl;                }            }        }        return dp[len_s-1][len_t-1];    }        int numDistinct(string S, string T) {        len_s = S.length();        len_t = T.length();        return Count(S, T);    }};