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Leetcode 2917. Find the K

Problem Statement

You are given an integer array nums and an integer k. The k-or of an array is defined as the bitwise OR of the largest k elements in the array. Return the maximum k-or of any subarray of length at least k.

Clarifying Questions

1. What is the range of input values for nums and k?
   • The array nums can contain values which are integers, and k is a positive integer.
2. Can the array nums have duplicate elements?
   • Yes, duplicates are allowed in the array nums.
3. What should be returned if k is larger than the length of the array?
   • This case should be considered invalid, and you can either return 0 or raise an exception depending on the problem constraints.
4. Is there a specific order of complexity that needs to be met?
   • Optimizing for time complexity is ideal, but the specific target is not stated here.

Strategy

To solve this problem, we can:

1. Utilize a sliding window approach with a size of at least k to find all subarrays.
2. For each subarray of length at least k, calculate the k-or by taking the bitwise OR of the largest k elements.
3. Keep track of the maximum k-or found during this traversal.

Code

Here’s a potential implementation in Java:

import java.util.Arrays;
import java.util.Collections;
import java.util.PriorityQueue;

public class Solution {
    
    public int maxKOr(int[] nums, int k) {
        if (nums == null || nums.length < k) {
            return 0;  // Or throw an exception.
        }
        
        int maxOr = 0;
        
        // Sliding window approach, window size from k up to nums.length
        for (int windowSize = k; windowSize <= nums.length; windowSize++) {
            PriorityQueue<Integer> maxHeap = new PriorityQueue<>(Collections.reverseOrder());
            
            // Initial window
            for (int i = 0; i < windowSize; i++) {
                maxHeap.offer(nums[i]);
            }
            
            // Calculate k-or for the initial window
            maxOr = Math.max(maxOr, calculateKOr(maxHeap, k));
            
            // Slide the window across nums
            for (int i = windowSize; i < nums.length; i++) {
                maxHeap.offer(nums[i]);
                maxHeap.remove(nums[i - windowSize]);
                maxOr = Math.max(maxOr, calculateKOr(maxHeap, k));
            }
        }
        
        return maxOr;
    }
    
    private int calculateKOr(PriorityQueue<Integer> maxHeap, int k) {
        Integer[] largestK = maxHeap.toArray(new Integer[0]);
        Arrays.sort(largestK, Collections.reverseOrder());
        
        int kOr = 0;
        for (int i = 0; i < k; i++) {
            kOr |= largestK[i];
        }
        return kOr;
    }
}

Time Complexity

• The sliding window traversal is O(n * (n - k + 1)).
• For each window, maintaining a max-heap and calculating the k-or requires O(windowSize log windowSize + k log k) operations.

In the worst case, this could result in a relatively high time complexity, potentially O(n^2 log n). Optimizations might be necessary for very large inputs. For typical constraints, this approach should be efficient enough to handle the problem.

Additional Notes

• Further optimization may involve avoiding recalculating the entire k-or for overlapping windows.
• Depending on constraints, more efficient data structures and algorithms, such as segment trees or balanced binary search trees, might be necessary for handling larger inputs more efficiently.

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