Sliding Window - Question 1

Leetcode 76. Minimum Window Substring

Given two strings s and t of lengths m and n respectively, return the minimum window substring of s such that every character in t (including duplicates) is included in the window. If there is no such substring, return the empty string "".

The testcases will be generated such that the answer is unique.

A substring is a contiguous sequence of characters within the string.

Constraints:

m == s.length

n == t.length

1 <= m, n <= 10^5

s and t consist of uppercase and lowercase English letters.

Analysis:

The first step to do is to count the letters in string t, since the relative order of chars does not matter.

After the counting, we can use two pointers method on string s: starting with beginning of the string s, the second pointer can continue to move until the sub-string [i, j) has all the chars in t, where i the index of the first pointer, and j is the index of the second pointer.

We are looking for the sub-string with the minimum length, so after we one sub-string [i, j), we can try to move the first index, to make the length of the sub string smaller, if possible.

So the question become: when the index can be moved right?

As long as the updated sub-string [i, j) still has all the chars in t, we can keep move the first pointer right. In detail, for each char, as long as the counting is still larger than that in t, then it is okay to move the first pointer.

See the code below:

class Solution {
public:
    string minWindow(string s, string t) {
        int len1 = s.size(), len2 = t.size();
        if(len1 < len2) return "";
        int start = 0, len = 0;
        vector<int> v1(256, 0), v2(256, 0);
        for(auto &a : t) ++v2[a];
        int i = 0, j = 0, ct = 0;
        while(i < len1 && j < len1) {
            ++v1[s[j]];
            if(v1[s[j]] <= v2[s[j]]) ++ct;
            if(ct == len2) {
                while(i < j && v2[s[i]] < v1[s[i]]) --v1[s[i++]];
                if(len == 0 || len > j - i + 1) {
                    len = j - i + 1;
                    start  = i;
                }
            }
            ++j;
        }
        return s.substr(start, len);
    }
};

Upper Layer