LC 727. Minimum Window Subsequence 【lock，hard】

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

Note:

All the strings in the input will only contain lowercase letters.
The length of S will be in the range [1, 20000].
The length of T will be in the range [1, 100].

Runtime: 44 ms, faster than 73.35% of C++ online submissions for Minimum Window Subsequence.

网上的DP解法。dp定义是能够匹配T[0,i]的最大的S的index.

举个例子，S = babad, T = bad,

i = 0时，只有当j=0，两个相等，所以此时dp = [0, -1, -1]

然后，dp = [0,1,-1]，然后, dp = [2, 1, -1], -> dp[2,2,-1] -> dp[2,2,2]

因为时从后往前更新，只有当T的每一个字符都匹配了才能把最开头的index传递到最后。中间即使有些匹配到，如果没有全部匹配也传递不了。

比如 S = babd, T = bad,

dp = [-1,-1,-1] -> dp[0,-1,-1] -> dp[0,1,-1] -> dp[2,1,-1] -> dp[2,1,1] 结果还是1.

class Solution {
public:
    string minWindow(string S, string T) {
        vector<int> dp(T.length(), -1);
        string ans = "";
        for (int i = 0; i < S.length(); i++) {
            for (int j = T.length() - 1; j >= 0; j--) {
                if (S[i] == T[j]) {
                    if (j == 0) dp[j] = i;
                    else dp[j] = dp[j - 1];
                }
            }
            int init = dp[T.length() - 1];
            if (init != -1 && (ans == "" || i - init + 1 < ans.size())) {
                ans = S.substr(init, i - init + 1);
            }
        }
        return ans;
    }
};

这题还有双指针法，过段时间更新。