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76. Minimum Window Substring

Hard

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.

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?

Solution

object Solution {
    def minWindow(s: String, t: String): String = {
        val map = new Array[Int](128)
        t.foreach(char => map(char - 'A') += 1)

        var count = t.length
        var begin = 0
        var end = 0
        var d = Int.MaxValue
        var head = 0

        while (end < s.length) {
            if (map(s.charAt(end) - 'A') > 0) {
                count -= 1
            }
            map(s.charAt(end) - 'A') -= 1
            end += 1

            while (count == 0) {
                if (end - begin < d) {
                    d = end - begin
                    head = begin
                }
                if (map(s.charAt(begin) - 'A') == 0) {
                    count += 1
                }
                map(s.charAt(begin) - 'A') += 1
                begin += 1
            }
        }

        if (d == Int.MaxValue) "" else s.substring(head, head + d)
    }
}

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