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LeetCode 1738: Find Kth Largest XOR Coordinate Value Solution

Master LeetCode problem 1738 (Find Kth Largest XOR Coordinate Value), a medium challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

Problem Explanation

Explanation

To solve this problem, we can create a prefix XOR matrix based on the given matrix. Then, we can iterate over all possible coordinates and store the XOR values in a min heap of size k. Finally, the top element of the heap will be the kth largest XOR coordinate value.

  1. Create a prefix XOR matrix by iterating over the given matrix and calculating the XOR values.
  2. Iterate over all possible coordinates (a, b) and calculate the XOR value using the prefix XOR matrix.
  3. Maintain a min heap of size k to store the k largest XOR values.
  4. Return the top element of the heap as the kth largest XOR coordinate value.

Time Complexity: O(m * n * log k)
Space Complexity: O(m * n + k)

Solution Code

import java.util.PriorityQueue;

class Solution {
    public int kthLargestValue(int[][] matrix, int k) {
        int m = matrix.length;
        int n = matrix[0].length;
        
        int[][] prefixXOR = new int[m][n];
        PriorityQueue<Integer> minHeap = new PriorityQueue<>();
        
        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                int left = j > 0 ? prefixXOR[i][j - 1] : 0;
                int top = i > 0 ? prefixXOR[i - 1][j] : 0;
                int topLeft = (i > 0 && j > 0) ? prefixXOR[i - 1][j - 1] : 0;
                
                prefixXOR[i][j] = matrix[i][j] ^ left ^ top ^ topLeft;
                minHeap.offer(prefixXOR[i][j]);
                
                if (minHeap.size() > k) {
                    minHeap.poll();
                }
            }
        }
        
        return minHeap.peek();
    }
}

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