76. Minimum Window Substring

Hard

Hash Table String Sliding Window

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.

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 <= 10^5
s and t consist of uppercase and lowercase English letters.

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

func minWindow(s string, t string) string {
	tCounts := map[byte]int{}
	for i := range t {
		tCounts[t[i]]++
	}
	tTotal := len(tCounts)

	result := ""

	l := 0
	minL := math.MaxInt
	subCounts := map[byte]int{}
	subTotal := 0

	for r := 0; r < len(s); r++ {
		c := s[r]

		subCounts[c]++
		if subCounts[c] == tCounts[c] {
			subTotal++
		}

		for subTotal == tTotal {
			if r-l < minL {
				minL = r - l
				result = s[l : r+1]
			}

			subCounts[s[l]]--
			if subCounts[s[l]] < tCounts[s[l]] {
				subTotal--
			}

			l++
		}
	}

	return result
}

💡 Hints (4)

Hint 1

Use two pointers to create a window of letters in s, which would have all the characters from t.

Hint 2

Expand the right pointer until all the characters of t are covered.

Hint 3

Once all the characters are covered, move the left pointer and ensure that all the characters are still covered to minimize the subarray size.

Hint 4

Continue expanding the right and left pointers until you reach the end of s.