#1252

Easy

LeetCode Problem Workspace

Cells with Odd Values in a Matrix

This problem involves updating a matrix based on given indices and counting cells with odd values afterward.

Problem Statement

You are given a matrix of size m x n, initialized to all zeroes. A 2D array of indices is provided, where each element [ri, ci] represents a row and column to apply an increment operation on. The task is to perform this operation on each specified index, incrementing all elements in the row and column of the specified index by 1.

After performing all the operations, return the number of odd-valued cells in the resulting matrix. Consider efficient ways to count odd values after applying the given changes based on the matrix dimensions and number of operations.

Examples

Example 1

Input: m = 2, n = 3, indices = [[0,1],[1,1]]

Output: 6

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

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

Solution Approach

Naive Approach

A straightforward solution is to directly apply the increment operation to the matrix for each index in the input. After applying all operations, iterate through the matrix and count the cells with odd values. This method is simple but can be inefficient when working with larger matrices.

Optimized Approach using Row and Column Tracking

Instead of directly modifying the matrix for each index, maintain a record of the number of increments applied to each row and column. Then, calculate the number of odd cells based on the parity of these row and column sums. This reduces the time complexity compared to directly modifying the matrix.

Efficient Simulation with Even and Odd Parity

Another optimization is to simulate the operations based on the parity of the increments. By tracking how many times each row and column is incremented, you can compute the parity of each cell based on whether its row and column sums are odd or even. This approach efficiently determines the final count of odd-valued cells.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

The time complexity depends on the approach. The naive approach takes O(m * n * k) where k is the number of operations. The optimized approach takes O(m + n + k), where m and n are the dimensions of the matrix and k is the number of operations. The efficient simulation method, using parity tracking, can further reduce the time complexity to O(m + n + k) while eliminating the need for matrix manipulation.

What Interviewers Usually Probe

Tests understanding of matrix manipulation and array operations.
Assesses ability to optimize solutions and recognize inefficiencies in naive approaches.
Evaluates problem-solving skills in simulating operations and managing matrix states efficiently.

Common Pitfalls or Variants

Common pitfalls

Misunderstanding the task and applying increments directly to the matrix rather than simulating row/column operations.
Not accounting for the fact that multiple increments can affect the same row and column multiple times.
Failing to optimize the approach, leading to unnecessary computations and higher time complexity.

Follow-up variants

What if the matrix is much larger, and performance becomes an issue?
How would the solution change if we had to also return the final matrix, not just the count of odd cells?
What if the number of operations is significantly higher, affecting performance?

FAQ

How does the problem 'Cells with Odd Values in a Matrix' work?

In this problem, you increment specific rows and columns in a matrix and count how many cells have odd values afterward.

What is the primary pattern used in solving 'Cells with Odd Values in a Matrix'?

This problem is solved using the 'Array plus Math' pattern, as it requires efficient row and column manipulations combined with math for counting odd cells.

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

Common mistakes include not optimizing the operations and performing inefficient matrix manipulations instead of simulating the changes.

How can I optimize my approach to solve 'Cells with Odd Values in a Matrix'?

Optimizing the solution involves tracking row and column operations instead of directly modifying the matrix. This helps reduce unnecessary computations.

What is the time complexity of solving 'Cells with Odd Values in a Matrix'?

The time complexity varies based on the approach. The optimized solution runs in O(m + n + k) time, where m and n are the matrix dimensions and k is the number of operations.

terminal

Solution

Solution 1: Simulation

We create a matrix $g$ to store the result of operations. For each pair $(r_i, c_i)$ in $\textit{indices}$, we add $1$ to all numbers in the $r_i$-th row of the matrix and add $1$ to all elements in the $c_i$-th column.

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

Solution 2: Space Optimization

We can use a row array $\textit{row}$ and a column array $\textit{col}$ to record the number of times each row and column is incremented. For each pair $(r_i, c_i)$ in $\textit{indices}$, we add $1$ to $\textit{row}[r_i]$ and $\textit{col}[c_i]$ respectively.

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

Solution 3: Mathematical Optimization

We notice that a number at position $(i, j)$ in the matrix will be odd only when exactly one of $\textit{row}[i]$ and $\textit{col}[j]$ is odd and the other is even.

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

Continue Practicing

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