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MinimumWindowSubstring.cpp
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59 lines (52 loc) · 1.74 KB
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//============================================================================
// 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
// empty string "".
//
// If there are multiple such windows, you are guaranteed that there will
// always be only one unique minimum window in S.
//============================================================================
#include <algorithm>
#include <iostream>
#include <climits>
using namespace std;
class Solution {
public:
string minWindow(string S, string T) {
int needToFind[256] = {0};
int hasFound[256] = {0};
size_t count = 0;
for (size_t i = 0; i < T.size(); i++)
needToFind[(int)T[i]]++;
string res = "";
int min = INT_MAX;
for (size_t begin = 0, end = 0; end < S.size(); end++) {
if (needToFind[(int)S[end]] == 0) continue;
hasFound[(int)S[end]]++;
if (hasFound[(int)S[end]] <= needToFind[(int)S[end]]) count++;
if (count == T.size()) {
while (needToFind[(int)S[begin]] == 0 || hasFound[(int)S[begin]] > needToFind[(int)S[begin]]) {
if (hasFound[(int)S[begin]] > needToFind[(int)S[begin]]) hasFound[(int)S[begin]]--;
begin++;
}
int len = end - begin + 1;
if (len < min) {
min = len;
res = S.substr(begin, len);
}
}
}
return res;
}
};
int main() {
return 0;
}