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 <= 105s and t consist of uppercase and lowercase English letters.
Follow up: Could you find an algorithm that runs in O(m + n) time?
class Solution {
public:
string minWindow(string s, string t) {
if(s.length()<t.length())
return "";
map<char, int> m;
int minSum=INT_MAX;
for(char c: t)
m[c]++;
int lt=0, rt=0, ct=0;
string res = s;
for(;rt<s.size();rt++){
if(m[s[rt]]>0)
ct++;
m[s[rt]]--;
if(ct==t.length()){
while(m[s[lt]]<0 && lt<rt) // rm elems from slidin win bcz -ve => we've more than reqd
m[s[lt]]++, lt++;
if(res.length()>rt-lt+1)
res=s.substr(lt,rt-lt+1);
// cout << res << "$" <<endl;
}
// cout << s.substr(lt, rt-lt+1) << "\t";
}
for(auto k:m)
if(k.second>0)
return "";
return res;
}
};class Solution {
public:
string minWindow(string s, string t) {
if(s.length()<t.length())
return "";
map<char, int> m;
int minSum=INT_MAX;
for(char c: t)
m[c]++;
int lt=0, rt=0, ct=0;
string res = s;
for(;rt<s.size();rt++){
if(m[s[rt]]>0)
ct++;
m[s[rt]]--;
if(ct==t.length()){
while(m[s[lt]]<0 && lt<rt) // rm elems from slidin win bcz -ve => we've more than reqd
m[s[lt]]++, lt++;
if(res.length()>rt-lt+1)
res=s.substr(lt,rt-lt+1);
// cout << res << "$" <<endl;
}
// cout << s.substr(lt, rt-lt+1) << "\t";
}
for(auto k:m)
if(k.second>0)
return "";
return res;
}
};