LeetCode 1239. Maximum Length of a Concatenated String with Unique Characters

原题链接在这里：https://leetcode.com/problems/maximum-length-of-a-concatenated-string-with-unique-characters/

题目：

Given an array of strings arr. String s is a concatenation of a sub-sequence of arr which have unique characters.

Return the maximum possible length of s.

Example 1:

Input: arr = ["un","iq","ue"]
Output: 4
Explanation: All possible concatenations are "","un","iq","ue","uniq" and "ique".
Maximum length is 4.

Example 2:

Input: arr = ["cha","r","act","ers"]
Output: 6
Explanation: Possible solutions are "chaers" and "acters".

Example 3:

Input: arr = ["abcdefghijklmnopqrstuvwxyz"]
Output: 26

Constraints:

• 1 <= arr.length <= 16
• 1 <= arr[i].length <= 26
• arr[i] contains only lower case English letters.

题解：

In the arr, s could contain duplicate character. These strings with duplicate characters can't be used.

Have a list to maintain all previous bit masks. When checking a new string s, check all previous bit masks, if there is not intersection, then s could be used.

Update the maximum result and add the new bitmast into list.

Time Complexity: expontential. 

Space: expontential.

AC Java:

 1 class Solution {
 2     public int maxLength(List<String> arr) {
 3         int res = 0;
 4         List<Integer> candidates = new ArrayList<>();
 5         candidates.add(0);
 6         
 7         for(String s : arr){
 8             int dup = 0;
 9             int sBit = 0;
10             for(char c : s.toCharArray()){
11                 dup |= sBit & 1<<(c-'a');
12                 sBit |= 1<<(c-'a');
13             }
14             
15             if(dup > 0){
16                 continue;
17             }
18             
19             for(int i = 0; i<candidates.size(); i++){
20                 if((candidates.get(i) & sBit) > 0){
21                     continue;
22                 }
23                 
24                 candidates.add(candidates.get(i) | sBit);
25                 res = Math.max(res, Integer.bitCount(candidates.get(i) | sBit));
26             }
27         }
28         
29         return res;
30     }
31 }