Introduction

In this blog post, we will explore how to solve the LeetCode problem "1738" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

You are given a 2D matrix of size m x n, consisting of non-negative integers. You are also given an integer k. The value of coordinate (a, b) of the matrix is the XOR of all matrix[i][j] where 0 <= i <= a < m and 0 <= j <= b < n (0-indexed). Find the kth largest value (1-indexed) of all the coordinates of matrix. Example 1: Input: matrix = [[5,2],[1,6]], k = 1 Output: 7 Explanation: The value of coordinate (0,1) is 5 XOR 2 = 7, which is the largest value. Example 2: Input: matrix = [[5,2],[1,6]], k = 2 Output: 5 Explanation: The value of coordinate (0,0) is 5 = 5, which is the 2nd largest value. Example 3: Input: matrix = [[5,2],[1,6]], k = 3 Output: 4 Explanation: The value of coordinate (1,0) is 5 XOR 1 = 4, which is the 3rd largest value. Constraints: m == matrix.length n == matrix[i].length 1 <= m, n <= 1000 0 <= matrix[i][j] <= 106 1 <= k <= m * n

Explanation

• Dynamic Programming with XOR: Calculate the XOR value for each coordinate (i, j) using the XOR values of its neighbors (i-1, j), (i, j-1), and (i-1, j-1) to avoid redundant calculations.
  • Min-Heap for Kth Largest: Maintain a min-heap of size k to store the k largest XOR values encountered so far. For each new XOR value, compare it with the smallest element in the heap. If it's larger, replace the smallest element with the new value.
  • Final Result: The root of the min-heap after processing all coordinates will be the kth largest XOR value.

• Runtime Complexity: O(m n log(k)), where m and n are the dimensions of the matrix.
• Storage Complexity: O(k)

Code

    import heapq

def kthLargestValue(matrix, k):
    m, n = len(matrix), len(matrix[0])
    xor_matrix = [[0] * n for _ in range(m)]
    min_heap = []

    for i in range(m):
        for j in range(n):
            if i == 0 and j == 0:
                xor_matrix[i][j] = matrix[i][j]
            elif i == 0:
                xor_matrix[i][j] = xor_matrix[i][j - 1] ^ matrix[i][j]
            elif j == 0:
                xor_matrix[i][j] = xor_matrix[i - 1][j] ^ matrix[i][j]
            else:
                xor_matrix[i][j] = xor_matrix[i - 1][j] ^ xor_matrix[i][j - 1] ^ xor_matrix[i - 1][j - 1] ^ matrix[i][j]

            if len(min_heap) < k:
                heapq.heappush(min_heap, xor_matrix[i][j])
            elif xor_matrix[i][j] > min_heap[0]:
                heapq.heapreplace(min_heap, xor_matrix[i][j])

    return min_heap[0]