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,

  • 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.


#include <iostream>
#include <string>
#include <vector>
#include <cstring>
#include <limits.h>
#include <string.h>
#include <cstdio>
#include <cstdlib>

using namespace std;

bool is_all_non_neg(const vector<int> &v) 
    for (int i=0; i<v.size(); ++i)
        if (v[i]<0)
            return false;
    return true;

string minWindow(string S, string T) {
        if (S.size()==0 || T.size()==0) return "";
        int slow=0, fast=0;

        vector<bool> is_in_T(256,false);       
        for (int i=0; i<T.size(); ++i)  is_in_T[T[i]]=true;

        vector<int> dist(256, 0);
        for (int i=0; i<T.size(); ++i)  dist[T[i]]--;
        if (is_in_T[S[0]]) dist[S[0]]++;

        string min_w;   // the minimum window found so far
        int min_w_size = INT_MAX;   // the size of the minimum window
        bool have_found = false;    // if a window has been found
        while (slow<S.size() && fast<S.size()) {
            if (!is_in_T[S[slow]]) {    
            if (dist[S[slow]]>0) {
            } else {
                if (!have_found)
                    have_found = is_all_non_neg(dist);  
                if (have_found && min_w_size>fast-slow+1) { // once we have found a window, the window is always good.
                    min_w_size = fast-slow+1;
                    min_w = S.substr(slow, min_w_size);
                ++fast;     // the window must be extended given slow cannot move
                if (fast<S.size() && is_in_T[S[fast]]) ++dist[S[fast]];

        return min_w;

int main()
        string s, t;
            cout<<minWindow(s, t)<<endl;
        return 0;