What is the time complexity of Maximum Length of a Concatenated String with Unique Characters?

The optimal solution for Maximum Length of a Concatenated String with Unique Characters runs in O(n * 2^n) time and uses O(1) space. The time complexity is O(n * 2^n) because we explore all subsets of the input array, and each subset can be checked for uniqueness in constant time using bit manipulation.

What is the brute force solution for Maximum Length of a Concatenated String with Unique Characters?

The brute force approach for Maximum Length of a Concatenated String with Unique Characters is Brute Force, with O(n²) time complexity and O(1) space complexity. We can generate all possible subsequences of the input array and check each one to see if it has unique characters. This approach is straightforward but inefficient for larger inputs. The optimal Optimal Solution approach improves this to O(n * 2^n).

What algorithm or data structure is used to solve Maximum Length of a Concatenated String with Unique Characters?

Maximum Length of a Concatenated String with Unique Characters uses the following concepts: Array, String, Backtracking, Bit Manipulation. The recommended patterns to study are: Backtracking, Bit Manipulation.

What are the interview tips for Maximum Length of a Concatenated String with Unique Characters?

Explain your thought process clearly while coding. Discuss edge cases, such as empty strings or strings with all unique characters. Practice writing clean, efficient code with proper variable names.

What are common mistakes when solving Maximum Length of a Concatenated String with Unique Characters?

Failing to check for uniqueness correctly. Not considering all combinations, especially when using recursion.

#1239

Maximum Length of a Concatenated String with Unique Characters

Medium

ArrayStringBacktrackingBit ManipulationBacktrackingBit Manipulation

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n * 2^n)
Space	O(1)	O(1)

💡

Intuition

Time O(n * 2^n)Space O(1)

We can use backtracking to explore all combinations of strings while maintaining a bitmask to track used characters. This avoids generating unnecessary combinations and checks for uniqueness efficiently.

⚙️

Algorithm

4 steps

1Step 1: Define a recursive function that takes the current index and the current bitmask of used characters.
2Step 2: For each string, check if adding it to the current bitmask maintains uniqueness.
3Step 3: If unique, update the maximum length and recurse with the next index.
4Step 4: Return the maximum length found.

solution.py18 lines

1# Full working Python code
2def maxLength(arr):
3    def backtrack(index, current_mask):
4        nonlocal max_len
5        max_len = max(max_len, bin(current_mask).count('1'))
6        for i in range(index, len(arr)):
7            new_mask = current_mask
8            for char in arr[i]:
9                new_mask |= (1 << (ord(char) - ord('a')))
10            if new_mask == current_mask + (1 << (ord(arr[i][0]) - ord('a'))):
11                backtrack(i + 1, new_mask)
12
13    max_len = 0
14    backtrack(0, 0)
15    return max_len
16
17# Example usage
18print(maxLength(['un', 'iq', 'ue']))  # Output: 4

ℹ

Complexity note: The time complexity is O(n * 2^n) because we explore all subsets of the input array, and each subset can be checked for uniqueness in constant time using bit manipulation.

1Using bit manipulation helps efficiently track used characters.
2Backtracking allows us to explore combinations without generating all explicitly.

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