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Make the XOR of All Segments Equal to Zero

Difficulty: Hard

Problem Description

You are given an array nums and an integer k. The XOR of a segment [left, right] where left <= right is the XOR of all the elements with indices between left and right, inclusive: nums[left] XOR nums[left+1] XOR ... XOR nums[right]. Return the minimum number of elements to change in the array such that the XOR of all segments of size k is equal to zero.

Key Insights

  • The XOR of a segment results in zero if the segment contains an even number of identical elements or if the elements cancel each other out.
  • We need to ensure that each segment of size k has an XOR of zero.
  • The problem can be approached as a dynamic programming or greedy problem where we track the number of changes required for each segment.

Space and Time Complexity

Time Complexity: O(n)
Space Complexity: O(1)

Solution

To solve the problem, we can iterate through the array while maintaining a frequency count of the elements in the current segment of size k. If the XOR of the current segment is not zero, we will need to calculate how many changes are necessary. The strategy involves:

  1. Using a frequency array to count occurrences of each number in the current window of size k.
  2. Calculating the XOR of the current segment.
  3. If the XOR is not zero, we determine how many changes are required to make the XOR zero by adjusting the elements in the segment.

This approach is efficient because it only requires a single pass through the array, making it linear in complexity.

Code Solutions

def minChanges(nums, k):
    n = len(nums)
    changes = 0

    for i in range(0, n, k):
        segment_xor = 0
        count = {}
        
        # Process the current segment
        for j in range(i, min(i + k, n)):
            segment_xor ^= nums[j]
            if nums[j] in count:
                count[nums[j]] += 1
            else:
                count[nums[j]] = 1
        
        # If the segment XOR is not zero, we need to change elements
        if segment_xor != 0:
            changes += 1  # At least one change is required
            
            # To make the XOR zero, we can change one of the elements
            # We can choose to change it to a value that makes the segment XOR zero
            # This is simplified since we are only counting necessary changes
        
    return changes
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