1239. Maximum Length of a Concatenated String with Unique Characters

Description

You are given an array of strings arr. A string s is formed by the concatenation of a subsequence of arr that has unique characters.

Return the maximum possible length of s.

A subsequence is an array that can be derived from another array by deleting some or no elements without changing the order of the remaining elements.

 

Example 1:

Input: arr = ["un","iq","ue"]
Output: 4
Explanation: All the valid concatenations are:
- ""
- "un"
- "iq"
- "ue"
- "uniq" ("un" + "iq")
- "ique" ("iq" + "ue")
Maximum length is 4.

Example 2:

Input: arr = ["cha","r","act","ers"]
Output: 6
Explanation: Possible longest valid concatenations are "chaers" ("cha" + "ers") and "acters" ("act" + "ers").

Example 3:

Input: arr = ["abcdefghijklmnopqrstuvwxyz"]
Output: 26
Explanation: The only string in arr has all 26 characters.

 

Constraints:

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

Solutions

Solution 1: Bit Manipulation + State Compression

State compression is used, with a 32-bit number recording the occurrence of letters, and masks storing the strings enumerated before.

The time complexity is $O(2^n + L)$, and the space complexity is $O(2^n)$. Where $n$ and $L$ are the length of the string array and the sum of the lengths of the strings in the array, respectively.

Python Code
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class Solution:
    def maxLength(self, arr: List[str]) -> int:
        ans = 0
        masks = [0]
        for s in arr:
            mask = 0
            for c in s:
                i = ord(c) - ord('a')
                if mask >> i & 1:
                    mask = 0
                    break
                mask |= 1 << i
            if mask == 0:
                continue
            for m in masks:
                if m & mask == 0:
                    masks.append(m | mask)
                    ans = max(ans, (m | mask).bit_count())
        return ans

Java Code
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class Solution {
    public int maxLength(List<String> arr) {
        int ans = 0;
        List<Integer> masks = new ArrayList<>();
        masks.add(0);
        for (var s : arr) {
            int mask = 0;
            for (int i = 0; i < s.length(); ++i) {
                int j = s.charAt(i) - 'a';
                if (((mask >> j) & 1) == 1) {
                    mask = 0;
                    break;
                }
                mask |= 1 << j;
            }
            if (mask == 0) {
                continue;
            }
            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;
    }
}

C++ Code
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class Solution {
public:
    int maxLength(vector<string>& arr) {
        int ans = 0;
        vector<int> masks = {0};
        for (auto& s : arr) {
            int mask = 0;
            for (auto& c : s) {
                int i = c - 'a';
                if (mask >> i & 1) {
                    mask = 0;
                    break;
                }
                mask |= 1 << i;
            }
            if (mask == 0) {
                continue;
            }
            int n = masks.size();
            for (int i = 0; i < n; ++i) {
                int m = masks[i];
                if ((m & mask) == 0) {
                    masks.push_back(m | mask);
                    ans = max(ans, __builtin_popcount(m | mask));
                }
            }
        }
        return ans;
    }
};

Go Code
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func maxLength(arr []string) (ans int) {
	masks := []int{0}
	for _, s := range arr {
		mask := 0
		for _, c := range s {
			i := int(c - 'a')
			if mask>>i&1 == 1 {
				mask = 0
				break
			}
			mask |= 1 << i
		}
		if mask == 0 {
			continue
		}
		n := len(masks)
		for _, m := range masks[:n] {
			if m&mask == 0 {
				masks = append(masks, m|mask)
				ans = max(ans, bits.OnesCount(uint(m|mask)))
			}
		}
	}
	return
}