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.

基本解题思路：

sliding window: start from a window that is tight.

Try moving the start forward and then see if it is possible to tighten the window.

public class Solution {    public String minWindow(String S, String T) {        // solve this problem by sliding window        int tlen = T.length(), slen = S.length();        int[] cnt = count(T);        int[] covered = new int[256];                int min_len = slen+1, start = 0;        String ans = "";        int cover = 0;        for(int i=0; i<slen; i++){            char ch = S.charAt(i);            if(cnt[ch] > 0){                if(covered[ch] < cnt[ch])       cover++;                covered[ch] ++;            }            if(cover == tlen){                // can be the end of the minimum substring                while(start < i){                    char chs = S.charAt(start);                    if(cnt[chs] > 0 && cnt[chs] == covered[chs])        break;                    else if(cnt[chs] > 0){     covered[chs]--;      }                    start++;                }                int len = i-start+1;                if(len < min_len){                    min_len = len;                    ans = S.substring(start, i+1);                }            }                    }        return ans;    }                private int[] count(String str){        int[] cnt = new int[256];        int len = str.length();        for(int i=0; i<len; i++){            char ch = str.charAt(i);            cnt[ch] ++;        }                return cnt;    }}

Space Complexity: O(1) though the count array/hashmap could take 256, but it is still constant

Time complexity: O(max(S.len, T.len)), need to scan T once and scan S twice.