727. Minimum Window Subsequence

Given strings S and T, find the minimum (contiguous) substring W of S, so that T is a subsequence of W.

If there is no such window in S that covers all characters in T, return the empty string "". If there are multiple such minimum-length windows, return the one with the left-most starting index.

Example 1:

Input: 
S = "abcdebdde", T = "bde"
Output: "bcde"
Explanation: 
"bcde" is the answer because it occurs before "bdde" which has the same length.
"deb" is not a smaller window because the elements of T in the window must occur in order.

 

class Solution {
public:
    string minWindow(string s, string t) {
        int ns = s.size(), nt= t.size();
        int dp[ns+1][nt+1] = {};
        const int mxx = ns + 1;
        //for(int i=0;i<=ns;i++) dp[i][0]=i;
        
        for (int i = 0 ; i <= ns; ++i) {
            for (int j = 1; j <= nt; ++j) {
                dp[i][j] = mxx;
                if (i) {
                    dp[i][j] = min(dp[i][j], 1 + dp[i-1][j]);
                    if (s[i-1] == t[j-1]) dp[i][j]  = min(dp[i][j], 1 + dp[i-1][j-1]);
                }
            }
        }
        
        int ans = ns + 1, x = -1;
        for (int i = 0; i <=ns; ++i) 
            if (dp[i][nt] < ans) {
            x = i;
            ans = dp[i][nt];
        }
        
        if (x < 0) return "";
        return s.substr(x-ans,ans);
    }
};