Distinct Subsequences
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DP:
class Solution {public: int numDistinct(string S, string T) { int n=S.size(), m=T.size(); if(m>n) return 0; vector<vector<int> > res(m+1, vector<int>(n+1,0)); for (int i=0; i<n+1; i++) res[0][i]=1; for (int i=1; i<=m; i++) { for (int j=1; j<=n; j++) { if (T[i-1]!=S[j-1]) res[i][j]=res[i][j-1]; else res[i][j]=res[i][j-1]+res[i-1][j-1]; } } return res[m][n]; }};
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- Distinct Subsequences
- Distinct Subsequences
- Distinct Subsequences
- Distinct Subsequences
- Distinct Subsequences
- Distinct Subsequences
- Distinct Subsequences
- Distinct Subsequences
- Distinct Subsequences
- Distinct Subsequences
- Distinct Subsequences
- Distinct Subsequences
- Distinct Subsequences
- Distinct Subsequences
- Distinct Subsequences
- Distinct Subsequences
- Distinct Subsequences
- Distinct Subsequences
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