Minimum Window Substring
// https://leetcode.com/problems/minimum-window-substring/
/*
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.
Ex1:
Input: s = "ADOBECODEBANC", t = "ABC"
Output: "BANC"
Explanation: The minimum window substring "BANC" includes 'A', 'B', and 'C' from string t.
Ex2:
Input: s = "a", t = "a"
Output: "a"
Explanation: The entire string s is the minimum window.
Ex3:
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.
*/
#include <string>
#include <iostream>
#include <unordered_map>
using namespace std;
// Time: O(n)
string minWindow(string s, string t) {
if (s.empty()) {
return "";
}
int m = s.size(), n = t.size();
unordered_map<char, int> char_map;
for (char c : t) char_map[c]++;
int total = char_map.size();
int left = 0, right = 0;
int match = 0;
string res = "";
while (right < m) {
if (char_map.count(s[right])) {
char_map[s[right]]--;
// We only count a match when num of a specific char
// equals between s and t.
if (char_map[s[right]] == 0) {
match++;
}
}
while (left < m && match == total) {
if (res.empty() || right-left+1 < res.size()) {
res = s.substr(left, right-left+1);
}
if (char_map.count(s[left])) {
// If already equal, this char is matched before.
// We need to decrement match since we are getting rid of this char.
if (char_map[s[left]] == 0) {
match--;
}
char_map[s[left]]++;
}
left++;
}
right++;
}
return res;
}
int main() {
string s = "ADOBECODEBANC";
string t = "ABC";
cout << minWindow(s, t) << endl;
s = "a";
t = "a";
cout << minWindow(s, t) << endl;
s = "a";
t = "aa";
cout << minWindow(s, t) << endl;
s = "bba";
t = "ab";
cout << minWindow(s, t) << endl;
return 0;
}```
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