Make the XOR of All Segments Equal to Zero
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Topics:
ArraysDynamic ProgrammingBit Manipulation
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.
Example 1:
Input: nums = [1,2,0,3,0], k = 1 Output: 3 Explanation: Modify the array from [1,2,0,3,0] to from [0,0,0,0,0].
Example 2:
Input: nums = [3,4,5,2,1,7,3,4,7], k = 3 Output: 3 Explanation: Modify the array from [3,4,5,2,1,7,3,4,7] to [3,4,7,3,4,7,3,4,7].
Example 3:
Input: nums = [1,2,4,1,2,5,1,2,6], k = 3 Output: 3 Explanation: Modify the array from [1,2,4,1,2,5,1,2,6] to [1,2,3,1,2,3,1,2,3].
Constraints:
1 <= k <= nums.length <= 20000 <= nums[i] < 210
Solution
Clarifying Questions
When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:
- What are the constraints on the size of the input array, and the range of values within the array?
- If it's impossible to make the XOR of all segments equal to zero, what should I return?
- Are there any restrictions on the number of segments I can create?
- Can you provide an example of an input and its expected output?
- Is the input array guaranteed to be non-empty?
Brute Force Solution
Approach
The brute force method for this problem is to explore every single way to divide the given sequence into segments. We check if the XOR of each segment in a division equals zero and keep track of the minimum changes needed among the valid divisions.
Here's how the algorithm would work step-by-step:
- Start by considering all possible ways to break the sequence into segments. For example, you could have one big segment, or many small segments.
- For each possible division into segments, calculate the XOR value for each segment separately.
- Check if the XOR value of every single segment in that particular division is equal to zero.
- If all segments XOR to zero, determine the number of changes needed to make that division possible. This is done by counting how many numbers in each segment are not zero.
- Keep track of the minimum number of changes needed across all the divisions where every segment XORs to zero.
- After checking all possible divisions, the minimum number of changes you recorded represents the answer.
Code Implementation
def make_xor_segments_zero_brute_force(sequence):
number_of_elements = len(sequence)
minimum_changes = float('inf')
for i in range(1 << (number_of_elements - 1)):
segments = []
current_segment = []
for j in range(number_of_elements):
current_segment.append(sequence[j])
if j < number_of_elements - 1 and (i >> j) & 1:
segments.append(current_segment)
current_segment = []
segments.append(current_segment)
all_segments_xor_to_zero = True
changes_needed = 0
for segment in segments:
xor_sum = 0
for number in segment:
xor_sum ^= number
if xor_sum != 0:
all_segments_xor_to_zero = False
break
if all_segments_xor_to_zero:
# If XOR is zero we count changes to determine the minimum
for segment in segments:
for number in segment:
if number != 0:
changes_needed += 1
minimum_changes = min(minimum_changes, changes_needed)
# Return the minimum changes needed to achieve XOR of zero across segments
if minimum_changes == float('inf'):
return -1
return minimum_changesBig(O) Analysis
Time Complexity
O(2^n * n) – The most computationally expensive part of this brute-force approach is generating all possible segmentations of the input array. An array of size n has 2^(n-1) possible segmentations. For each of these segmentations, we iterate through each segment to calculate its XOR value, which takes O(n) time in the worst-case segmentation where there's one large segment. Thus, the overall time complexity is O(2^(n-1) * n), which simplifies to O(2^n * n).
Space Complexity
O(N) – The brute force approach, as described, explores all possible segment divisions. The maximum depth of recursion for dividing an array of size N could be N in the worst-case scenario when each element forms its own segment. This recursive call stack consumes auxiliary space proportional to the depth of the recursion. Therefore, the space complexity is O(N), where N is the number of elements in the input sequence, primarily due to the recursion depth.
Optimal Solution
Approach
The core idea is to minimize changes needed in the given sequence. We achieve this by maximizing the length of the sub-sequence that remains unchanged when the entire sequence's XOR needs adjustment to meet the zero-sum segment requirement.
Here's how the algorithm would work step-by-step:
- Realize that a sub-sequence where all XORs are zero does not need any changes.
- Identify such sub-sequences within the original sequence.
- Focus on the remainder of the sequence, outside of the zero-XOR segments.
- The minimum changes are equal to the total number of segments minus the length of the longest zero-XOR sub-sequence.
Code Implementation
def min_changes_to_xor_zero(sequence):
segment_count = len(sequence)
max_zero_xor_subsequence_length = 0
for i in range(1 << segment_count):
subsequence = []
for j in range(segment_count):
if (i >> j) & 1:
subsequence.append(sequence[j])
# Only consider subsequences where XOR is zero
if len(subsequence) > 0:
xor_sum = 0
for element in subsequence:
xor_sum ^= element
if xor_sum == 0:
# Keep track of the longest such sequence
max_zero_xor_subsequence_length = max(max_zero_xor_subsequence_length, len(subsequence))
else:
max_zero_xor_subsequence_length = max(max_zero_xor_subsequence_length, len(subsequence))
# Calculate the minimum changes needed
# This subtraction isolates segments needing changes
minimum_changes = segment_count - max_zero_xor_subsequence_length
return minimum_changesBig(O) Analysis
Time Complexity
O(n²) – The dominant operation is finding the longest zero-XOR subsequence. To achieve this, for each index i in the input array of size n, we iterate from i to n, computing the XOR sum of the segment. This involves a nested loop structure. In the worst-case scenario, we are checking every possible segment's XOR value. Therefore, the number of operations approximates to n * n/2, which simplifies to O(n²).
Space Complexity
O(N) – The algorithm implicitly requires storing XOR values to identify zero-XOR sub-sequences. This could be implemented using a hash map (or a similar data structure) to store prefix XORs and their counts, where the size of the hash map can grow up to N in the worst case if all prefix XORs are unique where N is the length of the input sequence. Therefore, the auxiliary space complexity is O(N).
Edge Cases
| Case | How to Handle |
|---|---|
| Empty input array | Return 0, as the XOR of an empty segment is defined as 0, thus satisfying the condition. |
| Array with a single element | Return 0, because a single-element array can be considered as a segment whose XOR sum equals 0. |
| All elements in the array are 0 | No operations needed, return 0 because XOR of all segments is already zero |
| Array with a length of 2 where both numbers are same | Return 0 as the array's XOR sum is already zero |
| Array where the XOR sum of the entire array is non-zero and no single element is 0 | It might not be possible to split array such that all segments equal zero, consider all element XOR sum as one segment and proceed. |
| Array of all identical non-zero values, and length isn't a multiple of k (segment size) | If k = 1 the solution is impossible, but if the length is a multiple of k, the solution is simple (empty array). |
| Large array size leading to potential performance issues (e.g. exceeding time limit). | The solution should have a time complexity no worse than O(n) if possible. |
| Integer overflow if XORing large numbers | Use appropriate data types (e.g., long) to handle potentially large XOR sums. |
Empty input array
How to Handle:
Return 0, as the XOR of an empty segment is defined as 0, thus satisfying the condition.
Array with a single element
How to Handle:
Return 0, because a single-element array can be considered as a segment whose XOR sum equals 0.
All elements in the array are 0
How to Handle:
No operations needed, return 0 because XOR of all segments is already zero
Array with a length of 2 where both numbers are same
How to Handle:
Return 0 as the array's XOR sum is already zero
Array where the XOR sum of the entire array is non-zero and no single element is 0
How to Handle:
It might not be possible to split array such that all segments equal zero, consider all element XOR sum as one segment and proceed.
Array of all identical non-zero values, and length isn't a multiple of k (segment size)
How to Handle:
If k = 1 the solution is impossible, but if the length is a multiple of k, the solution is simple (empty array).
Large array size leading to potential performance issues (e.g. exceeding time limit).
How to Handle:
The solution should have a time complexity no worse than O(n) if possible.
Integer overflow if XORing large numbers
How to Handle:
Use appropriate data types (e.g., long) to handle potentially large XOR sums.
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