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

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.

串联字符串的最大长度。

给定一个字符串数组 arr，字符串 s 是将 arr 某一子序列字符串连接所得的字符串，如果 s 中的每一个字符都只出现过一次，那么它就是一个可行解。

请返回所有可行解 s 中最长长度。

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/maximum-length-of-a-concatenated-string-with-unique-characters
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思路是DFS backtracking。这里我用一个 letters 数组判断当前拼凑出来的 path 里面是否有重复字母，其他部分基本是标准的DFS。

时间O(n*2^n)

空间O(n)

Java实现

class Solution {
    int res = 0;

    public int maxLength(List<String> arr) {
        // corner case
        if (arr == null || arr.size() == 0) {
            return 0;
        }
        dfs(arr, "", 0);
        return res;
    }

    private void dfs(List<String> arr, String path, int index) {
        boolean isUnique = helper(path);
        if (isUnique) {
            res = Math.max(res, path.length());
        }
        if (index == arr.size() || !isUnique) {
            return;
        }
        for (int i = index; i < arr.size(); i++) {
            dfs(arr, path + arr.get(i), i + 1);
        }
    }

    private boolean helper(String s) {
        boolean[] letters = new boolean[26];
        for (int i = 0; i < s.length(); i++) {
            if (letters[s.charAt(i) - 'a'] == true) {
                return false;
            }
            letters[s.charAt(i) - 'a'] = true;
        }
        return true;
    }
}

LeetCode 题目总结