Set Matrix Zeroes In Place in DSA Python - Time & Space Complexity

Time Complexity: Set Matrix Zeroes In Place

O(n x m)

Understanding Time Complexity

We want to know how the time to set zeroes in a matrix grows as the matrix size increases.

How does the number of steps change when the matrix gets bigger?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

def set_zeroes(matrix):
    rows, cols = len(matrix), len(matrix[0])
    col_zero = False

    for r in range(rows):
        if matrix[r][0] == 0:
            col_zero = True
        for c in range(1, cols):
            if matrix[r][c] == 0:
                matrix[r][0] = 0
                matrix[0][c] = 0

    for r in range(1, rows):
        for c in range(1, cols):
            if matrix[r][0] == 0 or matrix[0][c] == 0:
                matrix[r][c] = 0

    if matrix[0][0] == 0:
        for c in range(cols):
            matrix[0][c] = 0

    if col_zero:
        for r in range(rows):
            matrix[r][0] = 0

This code sets entire rows and columns to zero if any element in them is zero, using the first row and column as markers.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Nested loops scanning rows and columns of the matrix.
• How many times: Each element is checked or updated a few times, roughly proportional to the total number of elements (rows x columns).

How Execution Grows With Input

As the matrix size grows, the number of steps grows roughly with the number of elements.

Input Size (n x m)	Approx. Operations
10 x 10 = 100	About 300 to 400 steps
100 x 100 = 10,000	About 30,000 to 40,000 steps
1000 x 1000 = 1,000,000	About 3,000,000 to 4,000,000 steps

Pattern observation: The steps grow roughly in direct proportion to the number of elements, so doubling rows and columns quadruples the work.

Final Time Complexity

Time Complexity: O(n x m)

This means the time grows proportionally to the total number of elements in the matrix.

Common Mistake

[X] Wrong: "Since we only mark zeros in the first row and column, the time complexity is less than scanning the whole matrix twice."

[OK] Correct: We still scan almost every element multiple times to mark and then update, so the total steps scale with the entire matrix size.

Interview Connect

Understanding this helps you explain how to efficiently update large grids without extra space, a common real-world problem.

Self-Check

"What if we used extra space to store zero positions instead of marking in place? How would the time complexity change?"

Practice

1. What is the main goal of the Set Matrix Zeroes In Place problem?

easy

A. Find the sum of all elements in the matrix.

B. Modify the matrix so that if an element is 0, its entire row and column become 0.

C. Sort each row of the matrix in ascending order.

D. Transpose the matrix without changing its elements.

Solution

1. Step 1: Understand the problem requirement

   The problem asks to change the matrix so that if any element is zero, all elements in that element's row and column become zero.

2. Step 2: Identify the correct description

   Modify the matrix so that if an element is 0, its entire row and column become 0. exactly describes this transformation, while others describe unrelated operations.

3. Final Answer:

   Modify the matrix so that if an element is 0, its entire row and column become 0. -> Option B

4. Quick Check:

   Set matrix zeroes means zero rows and columns [OK]

Hint: Remember zero means zero entire row and column [OK]

Common Mistakes:

• Confusing with matrix transpose
• Thinking only the zero element changes
• Trying to sum or sort instead

2. Which of the following Python code snippets correctly initializes a 3x3 matrix with all zeros?

easy

A. matrix = [[0]*3]*3

B. matrix = [0,0,0]*3

C. matrix = [0]*3 for _ in range(3)

D. matrix = [[0 for _ in range(3)] for _ in range(3)]

Solution

1. Step 1: Analyze each initialization method

   matrix = [[0]*3]*3 creates shallow copies of the same row, causing changes in one row to affect others. matrix = [[0 for _ in range(3)] for _ in range(3)] creates independent rows correctly. matrix = [0]*3 for _ in range(3) is invalid syntax. matrix = [0,0,0]*3 creates a flat list, not a matrix.

2. Step 2: Choose the correct matrix initialization

   Only matrix = [[0 for _ in range(3)] for _ in range(3)] correctly creates a 3x3 matrix with independent rows filled with zeros.

3. Final Answer:

   matrix = [[0 for _ in range(3)] for _ in range(3)] -> Option D

4. Quick Check:

   Use nested list comprehension for independent rows [OK]

Hint: Use nested list comprehension for 2D matrix [OK]

Common Mistakes:

• Using [[0]*3]*3 causes shared references
• Incorrect syntax in option C
• Flat list instead of 2D matrix

3. What will be the matrix after applying the Set Matrix Zeroes In Place logic on this matrix?

[[1, 2, 3],
 [4, 0, 6],
 [7, 8, 9]]

medium

A. [[1, 0, 3], [0, 0, 0], [7, 0, 9]]

B. [[1, 2, 3], [4, 0, 6], [7, 8, 9]]

C. [[1, 2, 3], [0, 0, 0], [7, 8, 9]]

D. [[0, 0, 0], [0, 0, 0], [0, 0, 0]]

Solution

1. Step 1: Identify zero positions

   The zero is at position (1,1) (second row, second column).

2. Step 2: Set entire row 1 and column 1 to zero

   Row 1 becomes [0,0,0]. Column 1 elements at rows 0 and 2 become zero, changing matrix to [[1,0,3],[0,0,0],[7,0,9]].

3. Final Answer:

   [[1, 0, 3], [0, 0, 0], [7, 0, 9]] -> Option A

4. Quick Check:

   Zero at (1,1) zeroes row 1 and column 1 [OK]

Hint: Zero row and column for each zero element [OK]

Common Mistakes:

• Not zeroing the entire column
• Changing elements not in zero row/column
• Confusing row and column indices

4. Given this code snippet to set matrix zeroes in place, what is the main bug?

def setZeroes(matrix):
    rows = len(matrix)
    cols = len(matrix[0])
    zero_rows = set()
    zero_cols = 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 i in zero_rows:
        for j in range(cols):
            matrix[i][j] = 0
    for j in zero_cols:
        for i in range(rows):
            matrix[i][j] = 0

medium

A. It misses zeroing the diagonal elements.

B. It modifies the matrix before identifying all zeroes.

C. It uses extra space, not in-place modification.

D. It does not handle empty matrices.

Solution

1. Step 1: Analyze space usage

   The code uses sets zero_rows and zero_cols to store indices, which uses extra space proportional to matrix size.

2. Step 2: Understand in-place requirement

   The problem requires modifying the matrix without extra space (in-place). Using sets violates this.

3. Final Answer:

   It uses extra space, not in-place modification. -> Option C

4. Quick Check:

   Extra sets mean not in-place [OK]

Hint: In-place means no extra sets or arrays [OK]

Common Mistakes:

• Ignoring extra space used by sets
• Thinking diagonal needs special handling
• Assuming empty matrix causes error here

5. How can you modify the matrix [[1,2,3],[4,0,6],[7,8,9]] in place to set zeroes without using extra space for rows and columns? Which approach is correct?

hard

A. Use first row and first column as markers to track zero rows and columns, then update matrix accordingly.

B. Create two separate lists to store zero rows and columns, then update matrix.

C. Copy the matrix, modify the copy, then replace original matrix.

D. Only zero the rows where zero is found, ignore columns.

Solution

1. Step 1: Use first row and column as markers

   Mark zeros in first row and column to remember which rows and columns to zero later, avoiding extra space.

2. Step 2: Update matrix based on markers

   After marking, iterate matrix to set zeros in rows and columns indicated by markers, then handle first row and column separately.

3. Final Answer:

   Use first row and first column as markers to track zero rows and columns, then update matrix accordingly. -> Option A

4. Quick Check:

   First row/column markers enable in-place zeroing [OK]

Hint: Use matrix edges as markers to avoid extra space [OK]

Common Mistakes:

• Using extra lists violates in-place rule
• Copying matrix wastes space
• Ignoring zero columns leads to wrong result