How do you solve Set Matrix Zeroes on LeetCode?

Set Matrix Zeroes (LeetCode #73) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(m * n) time complexity and O(1) space complexity. Key insight: Using the first row and column to mark zeros allows us to avoid using extra space.

What is the brute force solution for Set Matrix Zeroes?

The brute force approach for Set Matrix Zeroes is Brute Force, with O(m * n) time complexity and O(m * n) space complexity. The brute force approach involves scanning the entire matrix to find zeros and then modifying the rows and columns accordingly. This is straightforward but inefficient as it requires multiple passes over the matrix. The optimal Optimal Solution approach improves this to O(m * n).

What algorithm or data structure is used to solve Set Matrix Zeroes?

Set Matrix Zeroes uses the following concepts: Array, Hash Table, Matrix. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Set Matrix Zeroes?

Explain your thought process clearly; it shows your understanding. Consider edge cases, such as matrices with only one row or column. Practice explaining your code as you write it; this simulates the interview environment.

What are common mistakes when solving Set Matrix Zeroes?

Failing to handle the first row and column separately can lead to incorrect results. Not considering the implications of modifying the matrix while iterating through it.

#73

Set Matrix Zeroes

Medium

ArrayHash TableMatrixHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(m * n)	O(m * n)
Space	O(m * n)	O(1)

💡

Intuition

Time O(m * n)Space O(1)

The optimal approach uses the first row and first column of the matrix itself to mark which rows and columns need to be zeroed. This allows us to achieve the task in place without using extra space.

⚙️

Algorithm

4 steps

1Step 1: Use the first row and first column to mark which rows and columns should be zeroed.
2Step 2: Iterate through the matrix (excluding the first row and column) and mark the first row and column if a zero is found.
3Step 3: Zero out the marked rows and columns based on the markers in the first row and column.
4Step 4: Finally, zero out the first row and first column if they were marked.

solution.py19 lines

1def setZeroes(matrix):
2    m, n = len(matrix), len(matrix[0])
3    row_zero = any(matrix[0][j] == 0 for j in range(n))
4    col_zero = any(matrix[i][0] == 0 for i in range(m))
5    for i in range(1, m):
6        for j in range(1, n):
7            if matrix[i][j] == 0:
8                matrix[i][0] = 0
9                matrix[0][j] = 0
10    for i in range(1, m):
11        for j in range(1, n):
12            if matrix[i][0] == 0 or matrix[0][j] == 0:
13                matrix[i][j] = 0
14    if row_zero:
15        for j in range(n):
16            matrix[0][j] = 0
17    if col_zero:
18        for i in range(m):
19            matrix[i][0] = 0

ℹ

Complexity note: This complexity is due to the single pass through the matrix to mark rows and columns, and we use the matrix itself for marking, thus requiring constant space.

1Using the first row and column to mark zeros allows us to avoid using extra space.
2Always consider in-place modifications to save memory.

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