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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.lengthn == t.length1 <= m, n <= 105sandtconsist of uppercase and lowercase English letters.
Follow up: Could you find an algorithm that runs in O(m + n) time?
Solution
function minWindow(s: string, t: string): string { //NOSONAR
const map: Map<string, number> = new Map()
for (const char of t) {
if (map.has(char)) {
map.set(char, map.get(char) + 1)
} else {
map.set(char, 1)
}
}
let minStartIdx = -Infinity,
minEndIdx = Infinity
let remainingChars = t.length
let startIdx = 0
for (let endIdx = 0; endIdx < s.length; endIdx++) {
const endChar = s[endIdx]
if (map.has(endChar)) {
map.set(endChar, map.get(endChar) - 1)
if (map.get(endChar) >= 0) {
remainingChars--
}
}
while (remainingChars === 0) {
if (minEndIdx - minStartIdx > endIdx - startIdx) {
minStartIdx = startIdx
minEndIdx = endIdx
}
const startChar = s[startIdx]
if (map.has(startChar)) {
map.set(startChar, map.get(startChar) + 1)
}
if (map.get(startChar) > 0) {
remainingChars++
}
startIdx++
}
}
return minStartIdx === -Infinity ? '' : s.slice(minStartIdx, minEndIdx + 1)
}
export { minWindow }