How do you solve Cells with Odd Values in a Matrix on LeetCode?

Cells with Odd Values in a Matrix (LeetCode #1252) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(m + n + k) time complexity and O(m + n) space complexity. Key insight: Tracking increments separately for rows and columns allows for efficient calculation of odd cells.

What is the brute force solution for Cells with Odd Values in a Matrix?

The brute force approach for Cells with Odd Values in a Matrix is Brute Force, with O(n²) time complexity and O(1) space complexity. We can directly simulate the process by updating the matrix for each increment operation. This approach is straightforward but may be inefficient for larger inputs. The optimal Optimal Solution approach improves this to O(m + n + k).

What algorithm or data structure is used to solve Cells with Odd Values in a Matrix?

Cells with Odd Values in a Matrix uses the following concepts: Array, Math, Simulation. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Cells with Odd Values in a Matrix?

Always think about how you can optimize a brute-force solution. Clarify constraints and edge cases before diving into coding. Practice explaining your thought process clearly as you code.

What are common mistakes when solving Cells with Odd Values in a Matrix?

Forgetting to consider the effect of both row and column increments when counting odd cells. Not optimizing the solution to avoid unnecessary matrix construction.

#1252

Cells with Odd Values in a Matrix

Easy

ArrayMathSimulationHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(m + n + k)
Space	O(1)	O(m + n)

💡

Intuition

Time O(m + n + k)Space O(m + n)

Instead of simulating the entire matrix, we can keep track of how many times each row and column is incremented. This allows us to calculate the final odd counts without constructing the full matrix.

⚙️

Algorithm

3 steps

1Step 1: Create two arrays to count increments for rows and columns.
2Step 2: For each index in the indices array, increment the corresponding row and column counts.
3Step 3: Calculate the number of odd cells based on the counts from the two arrays.

solution.py14 lines

1# Full working Python code
2
3def oddCells(m, n, indices):
4    row_count = [0] * m
5    col_count = [0] * n
6    for r, c in indices:
7        row_count[r] += 1
8        col_count[c] += 1
9    odd_count = 0
10    for r in row_count:
11        for c in col_count:
12            if (r + c) % 2 != 0:
13                odd_count += 1
14    return odd_count

ℹ

Complexity note: The time complexity is O(m + n + k) where k is the number of indices since we only iterate through the row and column counts and the indices. The space complexity is O(m + n) due to the additional arrays used to store the counts.

1Tracking increments separately for rows and columns allows for efficient calculation of odd cells.
2The final odd count can be determined by the parity of row and column increments.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.