Introduction

In this blog post, we will explore how to solve the LeetCode problem "1252" 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

There is an m x n matrix that is initialized to all 0's. There is also a 2D array indices where each indices[i] = [ri, ci] represents a 0-indexed location to perform some increment operations on the matrix. For each location indices[i], do both of the following: Increment all the cells on row ri. Increment all the cells on column ci. Given m, n, and indices, return the number of odd-valued cells in the matrix after applying the increment to all locations in indices. Example 1: Input: m = 2, n = 3, indices = [[0,1],[1,1]] Output: 6 Explanation: Initial matrix = [[0,0,0],[0,0,0]]. After applying first increment it becomes [[1,2,1],[0,1,0]]. The final matrix is [[1,3,1],[1,3,1]], which contains 6 odd numbers. Example 2: Input: m = 2, n = 2, indices = [[1,1],[0,0]] Output: 0 Explanation: Final matrix = [[2,2],[2,2]]. There are no odd numbers in the final matrix. Constraints: 1 <= m, n <= 50 1 <= indices.length <= 100 0 <= ri < m 0 <= ci < n Follow up: Could you solve this in O(n + m + indices.length) time with only O(n + m) extra space?

Explanation

Here's an efficient solution to the problem, along with explanations:

• Key Idea: Instead of directly simulating the matrix and incrementing cells, we can keep track of the number of times each row and each column is incremented. Then, an element matrix[i][j] will be odd if row_increments[i] + col_increments[j] is odd.
• Efficient Counting: We can then efficiently count the number of odd cells by iterating through the row_increments and col_increments arrays and counting the combinations that result in an odd sum.

• Space Optimization: We only need to store the row and column increments, leading to O(m + n) space complexity.

• Runtime Complexity: O(m + n + indices.length)

• Storage Complexity: O(m + n)

Code

    def odd_cells(m: int, n: int, indices: list[list[int]]) -> int:
    """
    Given m, n, and indices, return the number of odd-valued cells in the matrix
    after applying the increment to all locations in indices.
    """

    row_increments = [0] * m
    col_increments = [0] * n

    for r, c in indices:
        row_increments[r] += 1
        col_increments[c] += 1

    odd_count = 0
    for i in range(m):
        for j in range(n):
            if (row_increments[i] + col_increments[j]) % 2 != 0:
                odd_count += 1

    return odd_count