How do you solve Maximum Bitwise XOR After Rearrangement on LeetCode?

Maximum Bitwise XOR After Rearrangement (LeetCode #3849) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Maximize XOR by pairing opposite bits.

What is the time complexity of Maximum Bitwise XOR After Rearrangement?

The optimal solution for Maximum Bitwise XOR After Rearrangement runs in O(n) time and uses O(n) space. Counting bits is O(n) and constructing the result is also O(n), making it efficient.

What is the brute force solution for Maximum Bitwise XOR After Rearrangement?

The brute force approach for Maximum Bitwise XOR After Rearrangement is Brute Force, with O(n²) time complexity and O(1) space complexity. Generate all permutations of string t and compute the XOR with s for each. This is inefficient as it checks every possible arrangement. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Maximum Bitwise XOR After Rearrangement?

Maximum Bitwise XOR After Rearrangement uses the following concepts: String, Greedy, Bit Manipulation. The recommended patterns to study are: Bit Manipulation, Greedy.

What are the interview tips for Maximum Bitwise XOR After Rearrangement?

Explain your thought process clearly. Discuss edge cases. Practice similar problems.

What are common mistakes when solving Maximum Bitwise XOR After Rearrangement?

Not counting bits correctly. Assuming order of t matters.

#3849

Maximum Bitwise XOR After Rearrangement

Medium

StringGreedyBit ManipulationBit ManipulationGreedy

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

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Intuition

Time O(n)Space O(n)

To maximize the XOR, pair '1's in s with '0's in t and vice versa. Count the bits in t and construct the best arrangement.

⚙️

Algorithm

3 steps

1Step 1: Count the number of '0's and '1's in t.
2Step 2: For each bit in s, match '1' with '0' from t if available, otherwise use '1' from t.
3Step 3: Construct the result based on the matches.

solution.py12 lines

1def maxXOR(s, t):
2    count0 = t.count('0')
3    count1 = t.count('1')
4    result = []
5    for char in s:
6        if char == '1' and count0 > 0:
7            result.append('0')
8            count0 -= 1
9        else:
10            result.append('1')
11            count1 -= 1
12    return ''.join(result)

ℹ

Complexity note: Counting bits is O(n) and constructing the result is also O(n), making it efficient.

1Maximize XOR by pairing opposite bits.
2Count bits to determine optimal arrangement.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.