Minimum Window Substring

Problem statement

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?

My solution

/**
 * @param {string} s
 * @param {string} t
 * @return {string}
 */
var minWindow = function(s, t) {
    if (!s || !t) {
        return ""
    }
    
    let result = ""
    let letterMap = new Map();
    let left = 0;
    let count = 0;
    let minValue = Number.MAX_SAFE_INTEGER;
    
    // Fill the Map
    for (const char of t.split("")) {
        if (!letterMap.has(char)) {
            letterMap.set(char, 0)
        }
        letterMap.set(char, letterMap.get(char) + 1)
    }
    
    for (let right = 0; right < s.length; right++) {
        const currChar = s.charAt(right);
        
        if (letterMap.has(currChar) !== undefined) {
            letterMap.set(currChar, letterMap.get(currChar) - 1);
            if (letterMap.get(currChar) >= 0) {
                count++
            }
        }
        
        while (count === t.length) {
            // console.log(">>> count", count, currChar)
            if (minValue > right - left + 1) {
                minValue = right - left + 1;
                result = s.substring(left, right + 1);
            }
            if (letterMap.has(s.charAt(left)) !== undefined) {
                letterMap.set(s.charAt(left), +letterMap.get(s.charAt(left)) + 1);

                // console.log("<<<",s.charAt(left), letterMap.get(s.charAt(left)))
                if (letterMap.get(s.charAt(left)) > 0) {
                    count--
                }
            }
            left++
        }
    }
    
    return result
};