---
title: '73. Set Matrix Zeroes'
description: Given an m x n integer matrix matrix, if an element is 0, set its entire row and column to 0's
icon: dot
topics:
  - { name: "Array", slug: "array" }
  - { name: "Hash Table", slug: "hash-table" }
  - { name: "Matrix", slug: "matrix" }
video: "T41rL0L3Pnw"
issue: "https://github.com/prdlk/leetcode/issues/93"
companies:
  - { name: "Bloomberg", period: "Three Months", frequency: 54 }
  - { name: "Amazon", period: "Three Months", frequency: 48.7 }
  - { name: "Meta", period: "Three Months", frequency: 43.3 }
  - { name: "Microsoft", period: "Three Months", frequency: 33.4 }
  - { name: "Google", period: "Three Months", frequency: 32.7 }
  - { name: "tcs", period: "Six Months", frequency: 43.5 }
sidebar:
  label: 'Set Matrix Zeroes'
  badge: 'Medium'
---

::::warning
You must do it in place.
::::

### Example 1:
- Input: `matrix = [[1,1,1],[1,0,1],[1,1,1]]`
- Output: `[[1,0,1],[0,0,0],[1,0,1]]`

### Example 2:
- Input: `matrix = [[0,1,2,0],[3,4,5,2],[1,3,1,5]]`
- Output: `[[0,0,0,0],[0,4,5,0],[0,3,1,0]]`

### Constraints:

- `m == matrix.length`
- `n == matrix[0].length`
- `1 <= m, n <= 200`
- `-2^31 <= matrix[i][j] <= 2^31 - 1`
- A straightforward solution using O(mn) space is probably a bad idea.
- A simple improvement uses O(m + n) space, but still not the best solution.
- Could you devise a constant space solution?

## Approach

```mermaid
flowchart TD
  S(["setZeroes(matrix)"]) --> I["m = len(matrix), n = len(matrix[0])"]
  I --> Z["row0_zero = any(matrix[0][c] == 0), col0_zero = any(matrix[r][0] == 0) — remember before they become markers"]
  Z --> F{"more r in range(1, m)?"}
  F -- yes --> G{"more c in range(1, n)?"}
  G -- yes --> Q{"matrix[r][c] == 0?"}
  Q -- yes --> M["matrix[r][0] = 0, matrix[0][c] = 0 — row 0 and column 0 are the marker lists"]
  M --> G
  Q -- no --> G
  G -- no --> F
  F -- no --> H{"more r in range(1, m)?"}
  H -- yes --> J{"more c in range(1, n)?"}
  J -- yes --> K{"matrix[r][0] == 0 or matrix[0][c] == 0?"}
  K -- yes --> W["matrix[r][c] = 0"]
  W --> J
  K -- no --> J
  J -- no --> H
  H -- no --> A{"row0_zero?"}
  A -- yes --> B["matrix[0][c] = 0 for c in range(n)"]
  B --> C{"col0_zero?"}
  A -- no --> C
  C -- yes --> D["matrix[r][0] = 0 for r in range(m)"]
  D --> E(["return None — zeroed in place"])
  C -- no --> E
```

## Solution

```py
class Solution:
    def setZeroes(self, matrix: List[List[int]]) -> None:
        """
        Do not return anything, modify matrix in-place instead.
        """
        m, n = len(matrix), len(matrix[0])

        # 1. remember if row 0 or column 0 have their own zeros
        row0_zero = any(matrix[0][c] == 0 for c in range(n))
        col0_zero = any(matrix[r][0] == 0 for r in range(m))

        # 2. use row 0 and column 0 as marker lists
        for r in range(1, m):
            for c in range(1, n):
                if matrix[r][c] == 0:
                    matrix[r][0] = 0
                    matrix[0][c] = 0

        # 3. zero the inner cells from the markers
        for r in range(1, m):
            for c in range(1, n):
                if matrix[r][0] == 0 or matrix[0][c] == 0:
                    matrix[r][c] = 0

        # 4. handle row 0 and column 0 last
        if row0_zero:
            for c in range(n):
                matrix[0][c] = 0
        if col0_zero:
            for r in range(m):
                matrix[r][0] = 0
```