Published Feb 15, 2022
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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.
Input: s = "ADOBECODEBANC", t = "ABC"
Output: "BANC"
Explanation: The minimum window substring "BANC" includes 'A', 'B', and 'C' from string t.
Input: s = "a", t = "a"
Output: "a"
Explanation: The entire string s is the minimum window.
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.
- m == s.length
- n == t.length
- 1 <= m, n <= 105
- s and t consist of uppercase and lowercase English letters.
O(m + n) time?function minWindow(s: string, t: string): string {
// special case
if(t === "") return "";
// have two hash to store number of characters
const countS: {[index: string]: number} = {};
const countT: {[index: string]: number} = {};
// initialize countT
for(let c of t){
countT[c] = (c in countT) ? (countT[c] + 1) : 0
}
// if we have what we need, counting if each character has met the condition
let have: number = 0;
const need: number = Object.keys(countT).length;
// result, and its length
let resultL: number = -1, resultR: number = -1, resultLen: number = Infinity;
let l: number = 0;
// let's iterate through s and do our update
for(let r = 0; r < s.length; r++){
const c: string = s[r]
// update character count in countS
countS[c] = (c in countS) ? (countS[c] + 1) : 0;
// update our have count
if(c in countT && countS[c] === countT[c]){
have++
}
// we will update result & shrink the result string while the have === need
while(have === need){
// update result
if(r - l + 1 < resultLen){
resultL = l;
resultR = r;
resultLen = r - l + 1;
}
// shrink window
countS[s[l]]--;
// update have
if(s[l] in countT && countS[s[l]] < countT[s[l]]){
have--;
}
l++
}
}
if(resultLen === Infinity){
return ''
} else {
return s.slice(resultL, resultR + 1)
}
};