README_EN.md

1239. Maximum Length of a Concatenated String with Unique Characters

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Description

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.

Solutions

State compression uses a 32-bit number to record the appearance of letters, and masks stores the previously enumerated strings.

Python3

class Solution:
    def maxLength(self, arr: List[str]) -> int:
        def ones_count(x):
            c = 0
            while x:
                x &= x - 1
                c += 1
            return c

        ans = 0
        masks = [0]
        for s in arr:
            mask = 0
            for ch in s:
                ch = ord(ch) - ord('a')
                if (mask >> ch) & 1 == 1:
                    mask = 0
                    break
                mask |= 1 << ch
            if mask == 0:
                continue
            for m in masks:
                if m & mask == 0:
                    masks.append(m | mask)
                    ans = max(ans, ones_count(m | mask))

        return ans

Java

class Solution {
    public int maxLength(List<String> arr) {
        int ans = 0;
        List<Integer> masks = new ArrayList<>();
        masks.add(0);

    loop:
        for (String s : arr) {
            int mask = 0;
            for (char ch : s.toCharArray()) {
                ch -= 'a';
                if (((mask >> ch) & 1) == 1) {
                    continue loop;
                }
                mask |= 1 << ch;
            }
            int n = masks.size();
            for (int i = 0; i < n; i++) {
                int m = masks.get(i);
                if ((m & mask) == 0) {
                    masks.add(m | mask);
                    ans = Math.max(ans, Integer.bitCount(m | mask));
                }
            }
        }

        return ans;
    }
}

Go

func maxLength(arr []string) int {

	max := func(x, y int) int {
		if x > y {
			return x
		}
		return y
	}

	ans := 0
	masks := []int{0}

loop:
	for _, s := range arr {
		mask := 0
		for _, ch := range s {
			ch -= 'a'
			if (mask>>ch)&1 == 1 {
				continue loop
			}
			mask |= 1 << ch
		}
		for _, m := range masks {
			if m&mask == 0 {
				masks = append(masks, m|mask)
				ans = max(ans, bits.OnesCount(uint(m|mask)))
			}
		}
	}

	return ans
}

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