LeetCode 727. Minimum Window Subsequence

原题链接在这里：https://leetcode.com/problems/minimum-window-subsequence/

题目：

Given strings s1 and s2, return the minimum contiguous substring part of s1, so that s2 is a subsequence of the part.

If there is no such window in s1 that covers all characters in s2, return the empty string "". If there are multiple such minimum-length windows, return the one with the left-most starting index.

Example 1:

Input: s1 = "abcdebdde", s2 = "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 s2 in the window must occur in order.

Example 2:

Input: s1 = "jmeqksfrsdcmsiwvaovztaqenprpvnbstl", s2 = "u"
Output: ""

Constraints:

• 1 <= s1.length <= 2 * 104
• 1 <= s2.length <= 100
• s1 and s2 consist of lowercase English letters.

题解：

Having a pointer j pointing at s2 index, when s1[i] == s2[j], move j++.

When j hits s2.length(), then we find a substring candidate.

Set a end to mark end of this candidate.

We need to move i and j back to find the start of the candidate.

When j hits 0, now i + 1 is the start of the candidate.

if s1.substring(i+ 1, end) length is smaller than minLen, update minWin.

Time Complexity: O(m*n). m = S.length(), n = T.length(). S = "aaaaaa", T = "aas", for each char in S, it is visited 2 * T.length() times.

Space: O(1). 

AC Java:

class Solution {
    public String minWindow(String s1, String s2) {
        int start = -1;
        int s1Len = s1.length();
        int minLen = s1Len + 1;
        int j = 0;
        for(int i = 0; i < s1Len; i++){
            if(s1.charAt(i) == s2.charAt(j)){
                j++;
            }

            if(j == s2.length()){
                int end = i + 1;
                j--;
                while(j >= 0){
                    if(s1.charAt(i) == s2.charAt(j)){
                        j--;
                    }
                    i--;
                }

                i++;
                j++;
                if(end - i < minLen){
                    minLen = end - i;
                    start = i;
                }
            }
        }

        return start == -1 ? "" : s1.substring(start, start + minLen);
    }
}

类似Number of Longest Increasing Subsequence, Minimum Window Substring.