Solution Explanation for 1252. Cells with Odd Values in a Matrix

This problem involves a matrix initialized with zeros and a series of operations that increment rows and columns based on given indices. The goal is to efficiently count the number of cells with odd values after all operations.

Core Idea: Instead of explicitly creating and modifying the matrix (which would be inefficient for large matrices), we can track the number of times each row and column is incremented. The final value of a cell (i, j) will be the sum of increments to row i and column j.

Three Approaches:

The problem can be solved in three ways, each with different time and space complexities:

1. Simulation:

• Approach: This is the most straightforward approach. We create the matrix and simulate the increment operations directly. After all increments, we iterate through the matrix, counting odd-valued cells.
• Time Complexity: O(k(m+n) + mn), where k is the length of indices, m is the number of rows, and n is the number of columns. The dominant factor is the nested loops for incrementing and counting.
• Space Complexity: O(mn) because we create and store the entire matrix.

2. Space Optimization:

• Approach: Instead of creating the entire matrix, we maintain two arrays: row and col. row[i] stores the number of times row i is incremented, and similarly for columns. The value of a cell (i, j) is then row[i] + col[j]. We iterate through all possible cell indices and count odd values.
• Time Complexity: O(k + mn). Creating row and col takes O(m+n), the loop for increments takes O(k), and the final counting loop takes O(mn).
• Space Complexity: O(m+n) since we only store the row and col arrays.

3. Mathematical Optimization:

• Approach: This is the most efficient approach. We observe that a cell (i, j) is odd only if exactly one of row[i] and col[j] is odd. We count odd rows (cnt1) and odd columns (cnt2). The number of odd cells is then cnt1 * (n - cnt2) + cnt2 * (m - cnt1).
• Time Complexity: O(k + m + n). Counting increments takes O(k), and counting odd rows and columns takes O(m+n).
• Space Complexity: O(m+n) to store row and col arrays.

Code Examples (Python):

Approach 1 (Simulation):

class Solution:
    def oddCells(self, m: int, n: int, indices: List[List[int]]) -> int:
        matrix = [[0] * n for _ in range(m)]
        for r, c in indices:
            for i in range(m):
                matrix[i][c] += 1
            for j in range(n):
                matrix[r][j] += 1
        count = 0
        for row in matrix:
            for cell in row:
                if cell % 2 != 0:
                    count += 1
        return count
 

Approach 2 (Space Optimization):

class Solution:
    def oddCells(self, m: int, n: int, indices: List[List[int]]) -> int:
        row_inc = [0] * m
        col_inc = [0] * n
        for r, c in indices:
            row_inc[r] += 1
            col_inc[c] += 1
        count = 0
        for i in range(m):
            for j in range(n):
                if (row_inc[i] + col_inc[j]) % 2 != 0:
                    count += 1
        return count
 

Approach 3 (Mathematical Optimization):

class Solution:
    def oddCells(self, m: int, n: int, indices: List[List[int]]) -> int:
        row_inc = [0] * m
        col_inc = [0] * n
        for r, c in indices:
            row_inc[r] += 1
            col_inc[c] += 1
        odd_rows = sum(1 for x in row_inc if x % 2 != 0)
        odd_cols = sum(1 for x in col_inc if x % 2 != 0)
        return odd_rows * (n - odd_cols) + odd_cols * (m - odd_rows)
 

The mathematical optimization (Approach 3) provides the best time complexity, making it the most efficient solution for large input sizes. The other approaches are provided for comparison and illustrative purposes. The space complexity is similar for approaches 2 and 3. Approach 1 is significantly less efficient in both time and space for large inputs.