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] < 210This approach works by checking each segment of length k and ensuring that its XOR is zero. For each segment, calculate the XOR of the current elements. If the XOR is not zero, identify which element can be flipped to achieve the desired XOR, trying to minimize changes across overlapping segments. This greedy approach is quick but might not always be optimal due to the choice of elements flipped.
This solution iterates over the array, checking each position modulo k. It maintains a frequency dictionary for elements at positions that share the same modulo result. For each segment starting at position i % k, it calculates the XOR. If it isn't zero, it computes the changes needed to convert the XOR to zero by changing the least frequent element that brings the total XOR to zero.
Java
The time complexity is O(n), where n is the length of the array, due to the loop iterating through the array. The space complexity is O(n/k) because it stores frequency counts for elements in one segment of size n/k.
This approach uses dynamic programming to keep track of the minimum changes necessary to ensure all segments of an array of k have an XOR equal to zero. We calculate the possible values for all segments and use a DP table to store the minimum number of changes necessary for each prefix of the array.
The DP approach initializes a state array dp where each entry represents the minimum modifications required for each XOR possibility. For each position modulo k, the algorithm considers each possible XOR value and finds the number of changes needed to achieve this XOR value using the elements in the current segment. This allows efficient updates of XOR possibilities as we progress through the array.
Java
The time complexity is O((n/k) * (2^x)), with x being the bit length of elements in the array, due to managing XOR value possibilities for every segment. The space complexity is O(2^x) due to the DP array storing possible XOR states.
| Approach | Complexity |
|---|---|
| Greedy Approach | The time complexity is O(n), where n is the length of the array, due to the loop iterating through the array. The space complexity is O(n/k) because it stores frequency counts for elements in one segment of size n/k. |
| Dynamic Programming Approach | The time complexity is O((n/k) * (2^x)), with x being the bit length of elements in the array, due to managing XOR value possibilities for every segment. The space complexity is O(2^x) due to the DP array storing possible XOR states. |
Sum of All Subsets XOR Total - Leetcode 1863 - Python • NeetCodeIO • 14,573 views views
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