How do you solve Find Maximum Balanced XOR Subarray Length on LeetCode?

Find Maximum Balanced XOR Subarray Length (LeetCode #3755) 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: Prefix XOR helps track cumulative XOR efficiently.

What is the time complexity of Find Maximum Balanced XOR Subarray Length?

The optimal solution for Find Maximum Balanced XOR Subarray Length runs in O(n) time and uses O(n) space. Single pass through the array with a hash map for state tracking leads to linear time complexity.

What is the brute force solution for Find Maximum Balanced XOR Subarray Length?

The brute force approach for Find Maximum Balanced XOR Subarray Length is Brute Force, with O(n²) time complexity and O(1) space complexity. Check every possible subarray to see if it has an XOR of zero and equal even and odd counts. This is straightforward but inefficient. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Find Maximum Balanced XOR Subarray Length?

Find Maximum Balanced XOR Subarray Length uses the following concepts: Array, Hash Table, Bit Manipulation, Prefix Sum. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Find Maximum Balanced XOR Subarray Length?

Explain your thought process clearly. Discuss edge cases and constraints. Practice similar problems to build confidence.

What are common mistakes when solving Find Maximum Balanced XOR Subarray Length?

Not accounting for the case where the entire array is valid. Forgetting to update the hash map with the first occurrence.

#3755

Find Maximum Balanced XOR Subarray Length

Medium

ArrayHash TableBit ManipulationPrefix SumHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(n)Space O(n)

Use prefix XOR and a hash map to track the first occurrence of each (pxor, diff) state, allowing for efficient length calculation.

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Algorithm

3 steps

1Step 1: Initialize a hash map to store the first occurrence of (pxor, diff) and set initial values.
2Step 2: Iterate through the array, updating the prefix XOR and the difference between even and odd counts.
3Step 3: Check if the current (pxor, diff) exists in the map; if it does, calculate the length and update max length.

solution.py13 lines

1def maxBalancedXOR(nums):
2    prefix_map = {(0, 0): -1}
3    pxor, evens, odds, max_len = 0, 0, 0, 0
4    for i, num in enumerate(nums):
5        pxor ^= num
6        if num % 2 == 0: evens += 1
7        else: odds += 1
8        diff = evens - odds
9        if (pxor, diff) in prefix_map:
10            max_len = max(max_len, i - prefix_map[(pxor, diff)])
11        else:
12            prefix_map[(pxor, diff)] = i
13    return max_len

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Complexity note: Single pass through the array with a hash map for state tracking leads to linear time complexity.

1Prefix XOR helps track cumulative XOR efficiently.
2Using a hash map allows for quick lookups of previously seen states.

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