1738. Find Kth Largest XOR Coordinate Value

Problem Description

Given an m x n matrix of non-negative integers and an integer k, find the kth largest value among all coordinate values. The value of coordinate (a, b) is the XOR of all matrix[i][j] where 0 <= i <= a < m and 0 <= j <= b < n (0-indexed).

Solution Approach: 2D Prefix XOR and Sorting/Quickselect

This problem can be efficiently solved using a two-dimensional prefix XOR array and either sorting or quickselect.

1. 2D Prefix XOR Array: We create a prefix XOR array s of size (m+1) x (n+1). s[i][j] stores the XOR of all elements from matrix[0][0] up to and including matrix[i-1][j-1]. This array allows us to quickly calculate the XOR value for any given coordinate (a, b) using the formula:

   value(a, b) = s[a+1][b+1]

2. Calculating the Prefix XOR Array: We iteratively build the s array. The formula for calculating s[i][j] is:

   s[i][j] = s[i-1][j] ^ s[i][j-1] ^ s[i-1][j-1] ^ matrix[i-1][j-1]

   This cleverly uses the XOR property that a ^ a = 0. It ensures that only the elements within the rectangle defined by (0,0) and (i-1,j-1) are included in the XOR.

3. Finding the Kth Largest Value: Once the s array is constructed, we collect all the coordinate values (which are simply the elements of s excluding the first row and column) into a list. Then, we can either:

   • Sort: Sort the list and return the kth largest element (which is at index len(list) - k since we're using 1-based indexing for k). This method has a time complexity of O(mn log(mn)).
   • Quickselect: Use a quickselect algorithm (like the one used in finding the median of an unsorted array), which typically has an average time complexity of O(mn). This would be more efficient for larger inputs.

Code Implementation (Python3)

This implementation uses sorting for simplicity:

from heapq import nlargest
 
class Solution:
    def kthLargestValue(self, matrix: List[List[int]], k: int) -> int:
        m, n = len(matrix), len(matrix[0])
        s = [[0] * (n + 1) for _ in range(m + 1)]
        ans = []
        for i in range(m):
            for j in range(n):
                s[i + 1][j + 1] = s[i + 1][j] ^ s[i][j + 1] ^ s[i][j] ^ matrix[i][j]
                ans.append(s[i + 1][j + 1])
        return nlargest(k, ans)[-1]

This uses the nlargest function from the heapq module, which is an efficient way to find the k largest elements.

Time and Space Complexity Analysis

• Time Complexity: O(mn log(mn)) if using sorting, O(mn) on average if using quickselect. The prefix XOR calculation takes O(mn) time, and sorting takes O(mn log(mn)). Quickselect has a average-case complexity of O(mn).

• Space Complexity: O(mn) to store the prefix XOR array s.

Other Language Implementations

The solution can be adapted to other languages like Java, C++, Go, etc., with similar logic and complexity. The main difference will be in syntax and library functions used for sorting or quickselect. The core algorithm remains the same.