Given two strings s and t of lengths m and n respectively, return the minimum window 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.

Example 1:

Input: s = "ADOBECODEBANC", t = "ABC" Output: "BANC" Explanation: The minimum window substring "BANC" includes 'A', 'B', and 'C' from string t.

Example 2:

Input: s = "a", t = "a" Output: "a" Explanation: The entire string s is the minimum window.

Example 3:

Input: s = "a", t = "aa" Output: "" Explanation: Both 'a's from t must be included in the window. Since the largest window of s only has one 'a', return empty string.

Constraints:

• m == s.length
• n == t.length
• 1 <= m, n <= 105
• s and t consist of uppercase and lowercase English letters.

Follow up: Could you find an algorithm that runs in O(m + n) time?

The problem asks us to find a substring of s, such that s includes every character in t(including duplicates). The shorter, the better. This means the result should have the following characteristics:

• The length of s must be no smaller than t. Otherwise, it's impossible to find such a substring.
• result.length >= t.length
• If char a occurs x times in t, it must occur x or more times in result.
• It's ok if char a occurs more times in s than it does in t.

Based on the analysis, we can use the hash map + variable technique to check if a sliding window contains all characters in t:

• Use a hash map/array to record the number of each character in t: countT[128]
• Use the same thing to record the number of each character in the sliding window: countS[128]
• Use a variable unmatched to track the number of characters that doesn't match between the sliding window and t.
  • At the beginning, any unique char in t should be considered as unmatched.
  • When the window slides, whenever countS[c] < countT[c], it's considered as unmatched.
• If a char doesn't exist in t, it's irrelevant. Even if it exists in the sliding window, we should ignore it: don't record its amount; don't change unmatched; just increment the index.