LeetCode-Minimum Window Substring

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,

S =

"ADOBECODEBANC"

T =

"ABC"

Minimum window is

"BANC"

.

Note:

If there is no such window in S that covers all characters in T, return the emtpy string

""

.

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

class Solution {
public:
int sub(char c){
if(c>='a'&&c<='z'){
return c-'a';
}
else if(c>='A'&&c<='Z')
{
return c-'A'+26;
}
else return -1;
}
string minWindow(string S, string T) {
// Note: The Solution object is instantiated only once and is reused by each test case.
vector<int> count,count2;
count.resize(52,0);
count2.resize(52,0);
for(int i=0;i<T.length();i++){
int ind=sub(T[i]);
if(ind!=-1)count[ind]++;
}
int dis=0;
for(int i=0;i<52;i++){
dis+=count[i];
}
int mins=-1,mine;
int start=0,end=0;
int ind=sub(S[end]);
if(ind!=-1){
count2[ind]++;
if(count2[ind]<=count[ind])dis--;
}
while(true){
if(dis!=0){
end++;
if(end==S.length())break;
int ind=sub(S[end]);
if(ind!=-1){
count2[ind]++;
if(count2[ind]<=count[ind])dis--;
}
}
else{
if(mins==-1||end-start<mine-mins){
mins=start;
mine=end;
}
int ind=sub(S[start]);
if(ind!=-1){
count2[ind]--;
if(count2[ind]<count[ind]){
dis++;
}
}
start++;
if(start>end)break;
}
}
if(mins==-1)return "";
return S.substr(mins,mine-mins+1);
}
};

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