/**
* 76. Minimum Window Substring
* https://leetcode.com/problems/minimum-window-substring/description/
*
* Given a string S and a string T, find the minimum window in S which will contain all the characters in T in complexity O(n).
Example:
Input: S = "ADOBECODEBANC", T = "ABC"
Output: "BANC"
Note:
If there is no such window in S that covers all characters in T, return the empty string "".
If there is such window, you are guaranteed that there will always be only one unique minimum window in S.
*/
/**
* @param {string} s
* @param {string} t
* @return {string}
*/
var minWindow = function (s, t) {
if (s == "" || t == "") return "";
let matchCount = 0, result = "";
let arrT = [], arrS = [];
//set up arrT and arrS
for (let c of t) {
if (arrT[c] == NaN || arrT[c] == undefined)
arrT[c] = 1;
else
arrT[c]++;
}
for (let c of s) {
if (arrS[c] == undefined)
arrS[c] = 0;
}
//console.log(arrS);
let left = findNextStr(0, s, arrT);
if (left == s.length)
return "";
let right = left;
//start scan.
//use right pointer to check each letter
while (right < s.length) {
let rightChar = s.charAt(right);
if (arrS[rightChar] < arrT[rightChar])
matchCount++;
arrS[rightChar]++;
while (left < s.length && matchCount == t.length) {
if (result == "" || result.length > right - left + 1) {
//如果result为空或者找到比当前result更短的
result = s.substring(left, right + 1);
}
//start to move left pointer
let leftChar = s.charAt(left);
if (arrS[leftChar] <= arrT[leftChar])
matchCount--;
arrS[leftChar]--;
//left pointer move to next vaild character
left = findNextStr(left + 1, s, arrT);
}
//right pointer also move to next vaild character
//right = findNextStr(right + 1, s, arrT);
//i changed to right++, fast than right = findNextStr(right + 1, s, arrT);
right++;
}
//console.log(result);
return result
};
var findNextStr = function (start, s, arrT) {
while (start < s.length) {
let c = s.charAt(start);
if (arrT[c] > 0)
return start;
start++;
}
};