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.
private:
int hashs[256];
int hasht[256];
inline bool check()
{
for(int i=0;i<256;i++)
if(hashs[i]<hasht[i]) return false;
return true;
}
public:
string minWindow(string S, string T)
{
for(int i=0;i<256;i++)
{
hashs[i]=0;
hasht[i]=0;
}
for(int i=0;i<T.size();i++)
hasht[T[i]]++;
int minl=-1;
int minr=S.length();
for(int i=0;i<S.size();i++)
hashs[S[i]]++;
if(!check()) return "";
int l=0;
int r=0;
for(int i=0;i<256;i++) hashs[i]=0;
hashs[S[0]]=1;
while(true)
{
if(check())
{
if(r-l<minr-minl)
{
minl=l;
minr=r;
if(minr-minl+1==T.length()) break;
}
hashs[S[l]]--;
l++;
}
else
{
r++;
if(r==S.length()) break;
hashs[S[r]]++;
}
}
if(minl==-1) return "";
else return S.substr(minl,minr-minl+1);
}
};