LeetCode-in-All
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
class Solution {
String minWindow(String s, String t) {
List<int> map = List.filled(128, 0);
for (int i = 0; i < t.length; i++) {
map[t.codeUnitAt(i)]++;
}
int count = t.length;
int begin = 0;
int end = 0;
int d = double.maxFinite
.toInt();
int head = 0;
while (end < s.length) {
if (map[s.codeUnitAt(end++)]-- > 0) {
count--;
}
while (count == 0) {
if (end - begin < d) {
d = end - begin;
head = begin;
}
if (map[s.codeUnitAt(begin++)]++ == 0) {
count++;
}
}
}
return d == double.maxFinite.toInt() ? "" : s.substring(head, head + d);
}
}