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Set Matrix Zeroes

DESCRIPTION (credit Leetcode.com)

Write a function to modify an m x n integer matrix matrix directly, such that if any element in the matrix is 0, its entire corresponding row and column are set to 0. This transformation should be done in place, without using any additional data structures for storage.

EXAMPLES

Input:


matrix = [
    [0,2,3],
    [4,5,6],
    [7,8,9]
]

Output:


[
    [0,0,0],
    [0,5,6],
    [0,8,9]
]

Explanation: Since the element at the first row and first column is 0, set all elements in the first row and first column to 0.

Solution


def setZeroes(matrix):
  rows, cols = len(matrix), len(matrix[0])
  zero_rows, zero_cols = set(), set()

  for i in range(rows):
    for j in range(cols):
      if matrix[i][j] == 0:
        zero_rows.add(i)
        zero_cols.add(j)

  for row in zero_rows:
    for col in range(cols):
      matrix[row][col] = 0

  for col in zero_cols:
    for row in range(rows):
      matrix[row][col] = 0

  return matrix

set matrix zeroes

0 / 14

Explanation

This solution solves the problem in three steps:

Step 1: Mark the rows and columns to be zeroed

The first step is to iterate over each cell in the matrix, and find the indexes of the rows and columns that need to be zeroed. We can use store those indexes in two sets rows and cols - each time we find a zero in the matrix, we add the row and column indexes to the respective sets.


def setZeroes(matrix):
  rows, cols = len(matrix), len(matrix[0])
  zero_rows, zero_cols = set(), set()

  for i in range(rows):
    for j in range(cols):
      if matrix[i][j] == 0:
        zero_rows.add(i)
        zero_cols.add(j)

  for row in zero_rows:
    for col in range(cols):
      matrix[row][col] = 0

  for col in zero_cols:
    for row in range(rows):
      matrix[row][col] = 0

  return matrix

i = 1, j = 1

0 / 1

Processing zeros by adding to zero_rows and zero_cols

Step 2: Zero the rows

The second step is to zero the rows. We iterate through the rows in the zero_rows set, and set all the elements in that row to zero by iterating through each column in the matrix, and setting those cells to zero.


def setZeroes(matrix):
  rows, cols = len(matrix), len(matrix[0])
  zero_rows, zero_cols = set(), set()

  for i in range(rows):
    for j in range(cols):
      if matrix[i][j] == 0:
        zero_rows.add(i)
        zero_cols.add(j)

  for row in zero_rows:
    for col in range(cols):
      matrix[row][col] = 0

  for col in zero_cols:
    for row in range(rows):
      matrix[row][col] = 0

  return matrix

i = 2, j = 2

0 / 1

Step 3: Zero the columns

The final step is to zero the columns. We iterate through the columns in the zero_cols set, and set all the elements in that column to zero by iterating through each row in the matrix, and setting those cells to zero.


def setZeroes(matrix):
  rows, cols = len(matrix), len(matrix[0])
  zero_rows, zero_cols = set(), set()

  for i in range(rows):
    for j in range(cols):
      if matrix[i][j] == 0:
        zero_rows.add(i)
        zero_cols.add(j)

  for row in zero_rows:
    for col in range(cols):
      matrix[row][col] = 0

  for col in zero_cols:
    for row in range(rows):
      matrix[row][col] = 0

  return matrix

set row 1 to 0

0 / 1

Complexity Analysis

Time complexity: O(m * n): We iterate through the entire matrix once.
Space complexity: O(1): We use only a constant amount of space.

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Set Matrix Zeroes

Rotate Image

Set Matrix Zeroes

DESCRIPTION (credit Leetcode.com)

EXAMPLES

Input:


matrix = [
    [0,2,3],
    [4,5,6],
    [7,8,9]
]

Output:


[
    [0,0,0],
    [0,5,6],
    [0,8,9]
]

Explanation: Since the element at the first row and first column is 0, set all elements in the first row and first column to 0.

Solution


def setZeroes(matrix):
  rows, cols = len(matrix), len(matrix[0])
  zero_rows, zero_cols = set(), set()

  for i in range(rows):
    for j in range(cols):
      if matrix[i][j] == 0:
        zero_rows.add(i)
        zero_cols.add(j)

  for row in zero_rows:
    for col in range(cols):
      matrix[row][col] = 0

  for col in zero_cols:
    for row in range(rows):
      matrix[row][col] = 0

  return matrix

set matrix zeroes

0 / 14

Explanation

This solution solves the problem in three steps:

Step 1: Mark the rows and columns to be zeroed


def setZeroes(matrix):
  rows, cols = len(matrix), len(matrix[0])
  zero_rows, zero_cols = set(), set()

  for i in range(rows):
    for j in range(cols):
      if matrix[i][j] == 0:
        zero_rows.add(i)
        zero_cols.add(j)

  for row in zero_rows:
    for col in range(cols):
      matrix[row][col] = 0

  for col in zero_cols:
    for row in range(rows):
      matrix[row][col] = 0

  return matrix

i = 1, j = 1

0 / 1

Processing zeros by adding to zero_rows and zero_cols

Step 2: Zero the rows


def setZeroes(matrix):
  rows, cols = len(matrix), len(matrix[0])
  zero_rows, zero_cols = set(), set()

  for i in range(rows):
    for j in range(cols):
      if matrix[i][j] == 0:
        zero_rows.add(i)
        zero_cols.add(j)

  for row in zero_rows:
    for col in range(cols):
      matrix[row][col] = 0

  for col in zero_cols:
    for row in range(rows):
      matrix[row][col] = 0

  return matrix

i = 2, j = 2

0 / 1

Step 3: Zero the columns


def setZeroes(matrix):
  rows, cols = len(matrix), len(matrix[0])
  zero_rows, zero_cols = set(), set()

  for i in range(rows):
    for j in range(cols):
      if matrix[i][j] == 0:
        zero_rows.add(i)
        zero_cols.add(j)

  for row in zero_rows:
    for col in range(cols):
      matrix[row][col] = 0

  for col in zero_cols:
    for row in range(rows):
      matrix[row][col] = 0

  return matrix

set row 1 to 0

0 / 1

Complexity Analysis

Time complexity: O(m * n): We iterate through the entire matrix once.
Space complexity: O(1): We use only a constant amount of space.

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