How do you solve Make the XOR of All Segments Equal to Zero on LeetCode?

Make the XOR of All Segments Equal to Zero (LeetCode #1787) 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: For segments of size k, elements must be equal for their XOR to be zero.

What is the time complexity of Make the XOR of All Segments Equal to Zero?

The optimal solution for Make the XOR of All Segments Equal to Zero runs in O(n) time and uses O(n) space. This complexity arises because we traverse the array once for each group of k, leading to a linear time complexity overall.

What is the brute force solution for Make the XOR of All Segments Equal to Zero?

The brute force approach for Make the XOR of All Segments Equal to Zero is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach checks every possible segment of size k in the array and counts how many modifications are needed to make their XOR equal to zero. This involves recalculating the XOR for each segment, which can be inefficient. The optimal Optimal Solution approach improves this to O(n).

What are the interview tips for Make the XOR of All Segments Equal to Zero?

Always consider edge cases, such as when k equals 1 or when all elements are already the same. Explain your thought process clearly; it helps the interviewer understand your approach. Practice similar problems to recognize patterns in segment-related questions.

What are common mistakes when solving Make the XOR of All Segments Equal to Zero?

Not recognizing the need to group elements by their indices modulo k. Overcomplicating the counting process instead of using a simple frequency map.

#1787

Make the XOR of All Segments Equal to Zero

Hard

ArrayHash TableDynamic ProgrammingBit ManipulationCountingHash MapArray

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)

The optimal approach leverages the observation that for the XOR of all segments of size k to be zero, every k-th element must be equal. We can count how many changes are needed to make all elements in each group of k equal.

⚙️

Algorithm

3 steps

1Step 1: Create a frequency map to count occurrences of each number in each group of k.
2Step 2: For each group, determine the most common number and calculate how many changes are needed to make all elements in that group equal to this number.
3Step 3: Sum the changes across all groups.

solution.py16 lines

1# Full working Python code
2
3def min_changes_optimal(nums, k):
4    from collections import defaultdict
5    n = len(nums)
6    changes = 0
7    for i in range(k):
8        count = defaultdict(int)
9        for j in range(i, n, k):
10            count[nums[j]] += 1
11        max_freq = max(count.values(), default=0)
12        changes += (n // k) - max_freq
13    return changes
14
15# Example usage
16print(min_changes_optimal([1,2,0,3,0], 1))  # Output: 3

ℹ

Complexity note: This complexity arises because we traverse the array once for each group of k, leading to a linear time complexity overall.

1For segments of size k, elements must be equal for their XOR to be zero.
2Using frequency counts helps minimize changes efficiently.

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