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| 1 | +class Solution { |
| 2 | +public: |
| 3 | + string minWindow(string s, string t) { |
| 4 | + |
| 5 | + if(s.size() < t.size()) return ""; |
| 6 | + |
| 7 | + int hashPattern[256] = {0}, hashString[256] = {0}; |
| 8 | + for(auto ch : t) { |
| 9 | + hashPattern[ch]++; |
| 10 | + } |
| 11 | + |
| 12 | + int count = 0, start = 0, startIndex = -1, minWindowLength = INT_MAX; |
| 13 | + for(int i=0; i<s.size(); i++) { |
| 14 | + |
| 15 | + //increment occurence of char in string_map |
| 16 | + hashString[s[i]]++; |
| 17 | + |
| 18 | + // Here we check if the current char is in the pattern as well. |
| 19 | + // Since we calculate running frequency of char in string_map as we encounter in the for loop, we do a <= comparison of hashString[s[i]] and hashPattern[s[i]]. It means the count of current char in string_map should be either equal count of current char in hash_map or less than count of current char in hash_map |
| 20 | + //if s[i] is in pattern_map and hashString[s[i]] == hashPattern[s[i]] or hashString[s[i]] < hashPattern[s[i]] |
| 21 | + if(hashPattern[s[i]] != 0 and hashString[s[i]] <= hashPattern[s[i]]) { |
| 22 | + count++; |
| 23 | + } |
| 24 | + |
| 25 | + |
| 26 | + if(count == t.size()) { //if we find the pattern in the string |
| 27 | + |
| 28 | + //minimize the window from start |
| 29 | + |
| 30 | + while((hashString[s[start]] > hashPattern[s[start]]) or hashPattern[s[start]] == 0) { |
| 31 | + |
| 32 | + //if current char at start is occuring more times than in pattern, we can minimize the window. |
| 33 | + //remove the current char at start from string_map and increment start; thereby reducing size of window. |
| 34 | + if(hashString[s[start]] > hashPattern[s[start]]) hashString[s[start]]--; |
| 35 | + |
| 36 | + //if the current char at start is not present in the pattern, just increment the start... |
| 37 | + start++; |
| 38 | + } |
| 39 | + |
| 40 | + //get the minimum of all the windows seen so far |
| 41 | + int currentWindowLength = i - start + 1; |
| 42 | + if(currentWindowLength < minWindowLength) { |
| 43 | + minWindowLength = currentWindowLength; |
| 44 | + startIndex = start; |
| 45 | + } |
| 46 | + } |
| 47 | + } |
| 48 | + |
| 49 | + if(startIndex == -1) return ""; |
| 50 | + else return s.substr(startIndex, minWindowLength); |
| 51 | + |
| 52 | + } |
| 53 | +}; |
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