# Minimum Window Substring

Minimum Window Substring hash table

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).

For example,
T = "ABC"
Minimum window is "BANC".

Note: If there is no such window in S that covers all characters in T, return the empty string “”.

If there are multiple such windows, you are guaranteed that there will always be only one unique minimum window in S.

Minimum Window Substring Solution

public class MinimumWindowSubstring {
public String minWindow(String s, String t) {
HashMap<Character, Integer> map = new HashMap<>();

for(char c : s.toCharArray()) {
map.put(c, 0);
}

for(char c : t.toCharArray()) {
if(map.containsKey(c)) {
map.put(c, map.get(c)+ 1);
} else {
return "";
}
}

int start = 0;
int end = 0;
int minStart = 0;
int minLength = Integer.MAX_VALUE;
int counter = t.length();

while(end < s.length()) {
char c1 = s.charAt(end);

if(map.get(c1) > 0) {
counter--;
}

map.put(c1, map.get(c1) - 1);
end++;

while(counter == 0) {
if(minLength > end - start) {
minLength = end - start;
minStart = start;
}

char c2 = s.charAt(start);
map.put(c2, map.get(c2) + 1);

if(map.get(c2) > 0) {
counter++;
}

start++;
}
}

return minLength == Integer.MAX_VALUE ? "" : s.substring(minStart, minStart + minLength);
}
}