# Solving Leetcode Interviews in Seconds with AI: Cells with Odd Values in a Matrix


	# 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](https://www.chatmagic.app), 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
	```python
	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
	```