Give you a string {s and a string} t. Returns the minimum substring of all characters of # t # in # s #. If there is no substring covering all characters of {t} in {s}, the empty string "" is returned.

be careful:

• For repeated characters in # t #, the number of characters in the substring we are looking for must not be less than the number of characters in # t #.
• If there is such a substring in ， s ， we guarantee that it is the only answer.

Example 1:

Input: S = "Adobe codebanc", t = "ABC"
Output: "BANC"

Example 2:

Input: s = "a", t = "a"
Output: "a"

Example 3:

Input: s = "a", t = "aa"
Output: ''
Explanation: both characters' a 'in t should be included in the substring of s,
Therefore, there is no qualified substring and an empty string is returned.

Tips:

• 1 <= s.length, t.length <= 105
• s ， and t ， are composed of English letters

class Solution {
    public String minWindow(String s, String t) {
        int[] tmap=new int[52];
        int[] smap=new int[52];
        int l=0,r=0;
        for (int i=0;i<t.length();i++){
            //26 lowercase in the front and capital in the back
            if ('a'<=t.charAt(i)) tmap[t.charAt(i)-'a'+26]++;
            else tmap[t.charAt(i)-'A']++;
        }
        int max=10000,index=-1;
        while (r<s.length()){
            if ('a'<=s.charAt(r)) smap[s.charAt(r)-'a'+26]++;
            else smap[s.charAt(r)-'A']++;
            r++;
　　　　　　　　//When the current conditions are met, try to reduce the left boundary as much as possible
            while (check(tmap,smap)){
　　　　　　　　　　//Judge whether the answer has been updated. If not, update the solution first
                if (index==-1) {
                    index=l;
                    max=r-l;
                }
                if ('a'<=s.charAt(l)) smap[s.charAt(l)-'a'+26]--;
                else smap[s.charAt(l)-'A']--;
                if (check(tmap,smap)) {
                    l++;
　　　　　　　　　　　　//The left boundary can be reduced to determine whether it is smaller than the current solution and update the solution
                    if (r-l<max){
                        index=l;
                        max=r-l;
                }
                }else{
                    if ('a'<=s.charAt(l)) smap[s.charAt(l)-'a'+26]++;
                    else smap[s.charAt(l)-'A']++;
                    break;
                }
            }
        }
        //If there is no updated solution, an empty string is returned
        return index!=-1?s.substring(index,index+max):"";
    }

    public boolean check(int[] a,int[] b){
        for (int i=0;i<a.length;i++){
            if (a[i]>b[i]) return false;
        }
        return true;
    }
}

Topic 2:

Input: s1 = "ab" s2 = "eidbaooo"
Output: true
 Explanation: s2 contains one of the permutations of s1 ("ba")

Input: s1= "ab" s2 = "eidboaoo"
Output: false

Input: S = "cbaebacd", P = "ABC"
Output: [0,6]
Explanation:
The substring with the starting index equal to 0 is "cba", which is an ectopic word of "abc".
The substring with the starting index equal to 6 is "bac", which is an ectopic word of "abc".

Input: s = "abab", p = "ab"
Output: [0,1,2]
Explanation:
The substring with the starting index equal to 0 is "ab", which is an ectopic word of "ab".
The substring with the starting index equal to 1 is "ba", which is an ectopic word of "ab".
The substring with the starting index equal to 2 is "ab", which is an ectopic word of "ab".

Input: s = ""
Output: 0