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

Time Complexity: Set Matrix Zeroes In Place

O(m x n)

Understanding Time Complexity

We want to understand how the time needed to set matrix zeroes changes as the matrix size grows.

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

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

fn set_zeroes(matrix: &mut Vec>) {
    let rows = matrix.len();
    let cols = matrix[0].len();
    let mut col_zero = false;

    for i in 0..rows {
        if matrix[i][0] == 0 {
            col_zero = true;
        }
        for j in 1..cols {
            if matrix[i][j] == 0 {
                matrix[i][0] = 0;
                matrix[0][j] = 0;
            }
        }
    }

    for i in 1..rows {
        for j in 1..cols {
            if matrix[i][0] == 0 || matrix[0][j] == 0 {
                matrix[i][j] = 0;
            }
        }
    }

    if matrix[0][0] == 0 {
        for j in 0..cols {
            matrix[0][j] = 0;
        }
    }

    if col_zero {
        for i in 0..rows {
            matrix[i][0] = 0;
        }
    }
}
    

This code sets entire rows and columns to zero if any element in them is zero, using the matrix itself to store flags.

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 nested loop runs roughly once over all elements, so about rows x columns times.

How Execution Grows With Input

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

Input Size (n x n)	Approx. Operations
10 x 10	About 200 (2 x 10 x 10)
100 x 100	About 20,000 (2 x 100 x 100)
1000 x 1000	About 2,000,000 (2 x 1000 x 1000)

Pattern observation: The operations grow roughly with the square of the matrix size, doubling the size quadruples the work.

Final Time Complexity

Time Complexity: O(m x n)

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

Common Mistake

[X] Wrong: "The time complexity is O((m x n)²) because of nested loops inside nested loops."

[OK] Correct: The loops are sequential, not nested inside each other repeatedly; each element is visited a constant number of times, so the total work is linear in the number of elements.

Interview Connect

Understanding this time complexity helps you explain how your solution scales with input size, a key skill in interviews.

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. Sort each row of the matrix in ascending order.

B. Count the number of zero elements in the matrix.

C. Modify the matrix so that if an element is 0, its entire row and column become 0 without using extra space.

D. Replace all zero elements with ones.

Solution

1. Step 1: Understand the problem requirement

   The problem asks to set entire rows and columns to zero if any element in them is zero, modifying the matrix directly without extra space.

2. Step 2: Differentiate from other options

   Options A, B, and C do not match the problem's goal of modifying rows and columns based on zero elements.

3. Final Answer:

   Modify the matrix so that if an element is 0, its entire row and column become 0 without using extra space. -> Option C

4. Quick Check:

   Goal = Modify matrix rows and columns in place [OK]

Hint: Focus on zero elements affecting entire rows and columns [OK]

Common Mistakes:

• Thinking only to count zeros
• Trying to use extra memory
• Confusing with sorting or replacing zeros

2. Which Rust syntax correctly declares a mutable 2D vector named matrix with 3 rows and 3 columns initialized to 1?

easy

A. let mut matrix = vec![vec![1; 3]; 3];

B. let mut matrix = vec![1; 3][3];

C. let matrix = vec![vec![1; 3]; 3];

D. let matrix = vec![1; 3][3];

Solution

1. Step 1: Identify mutable declaration

   To modify the matrix later, it must be declared mutable using let mut.

2. Step 2: Correct 2D vector initialization

   The syntax vec![vec![1; 3]; 3] creates a 3x3 vector filled with 1s. Options B and D are invalid syntax. let matrix = vec![vec![1; 3]; 3]; lacks mut.

3. Final Answer:

   let mut matrix = vec![vec![1; 3]; 3]; -> Option A

4. Quick Check:

   Mutable 2D vector with correct syntax = let mut matrix = vec![vec![1; 3]; 3]; [OK]

Hint: Use let mut and nested vec! for 2D vectors [OK]

Common Mistakes:

• Forgetting mut keyword
• Using invalid vector syntax
• Confusing 1D and 2D vector initialization

3. Given the following Rust code snippet, what will be the printed matrix after calling set_zeroes(&mut matrix)?

fn set_zeroes(matrix: &mut Vec) {
    let rows = matrix.len();
    let cols = matrix[0].len();
    let mut row_zero = false;

    for i in 0..rows {
        if matrix[i][0] == 0 {
            row_zero = true;
        }
        for j in 1..cols {
            if matrix[i][j] == 0 {
                matrix[i][0] = 0;
                matrix[0][j] = 0;
            }
        }
    }

    for i in 1..rows {
        for j in 1..cols {
            if matrix[i][0] == 0 || matrix[0][j] == 0 {
                matrix[i][j] = 0;
            }
        }
    }

    if matrix[0][0] == 0 {
        for j in 0..cols {
            matrix[0][j] = 0;
        }
    }

    if row_zero {
        for i in 0..rows {
            matrix[i][0] = 0;
        }
    }
}

fn main() {
    let mut matrix = vec![
        vec![1, 1, 1],
        vec![1, 0, 1],
        vec![1, 1, 1],
    ];
    set_zeroes(&mut matrix);
    for row in matrix {
        println!("{:?}", row);
    }
}

medium

A. [[1, 0, 1], [0, 0, 0], [1, 0, 1]]

B. [[1, 1, 1], [1, 0, 1], [1, 1, 1]]

C. [[0, 0, 0], [0, 0, 0], [0, 0, 0]]

D. [[1, 1, 1], [0, 0, 0], [1, 1, 1]]

Solution

1. Step 1: Identify zero positions and mark rows and columns

   The zero is at position (1,1). The first loop marks row 1 and column 1 by setting matrix[1][0] and matrix[0][1] to zero.

2. Step 2: Set zeros based on marks

   The second loop sets elements to zero if their row or column is marked. This changes matrix[0][1], matrix[1][0..2], and matrix[2][1] to zero.

3. Final Answer:

   [[1, 0, 1], [0, 0, 0], [1, 0, 1]] -> Option A

4. Quick Check:

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

Hint: Mark first row and column to track zeros, then update matrix [OK]

Common Mistakes:

• Not marking first row and column correctly
• Overwriting zeros too early
• Ignoring the first column zero flag

4. The following Rust function attempts to set matrix zeroes in place but has a bug. What is the bug?

fn set_zeroes(matrix: &mut Vec) {
    let rows = matrix.len();
    let cols = matrix[0].len();

    for i in 0..rows {
        for j in 0..cols {
            if matrix[i][j] == 0 {
                for k in 0..rows {
                    matrix[k][j] = 0;
                }
                for l in 0..cols {
                    matrix[i][l] = 0;
                }
            }
        }
    }
}

medium

A. It does not check for empty matrix before accessing elements.

B. It modifies the matrix while iterating, causing incorrect zero propagation.

C. It uses incorrect loop ranges causing out-of-bounds errors.

D. It uses immutable references instead of mutable ones.

Solution

1. Step 1: Analyze matrix modification during iteration

   The function sets zeros immediately when it finds a zero, which changes the matrix during iteration.

2. Step 2: Understand effect of immediate modification

   This causes new zeros to be found in later iterations, incorrectly zeroing more rows and columns than intended.

3. Final Answer:

   It modifies the matrix while iterating, causing incorrect zero propagation. -> Option B

4. Quick Check:

   Immediate zeroing during iteration causes errors [OK]

Hint: Avoid changing matrix while scanning for zeros; mark first then update [OK]

Common Mistakes:

• Changing matrix elements inside nested loops
• Not using markers to delay zeroing
• Ignoring matrix size checks

5. You have a 4x4 matrix:

[
  [1, 2, 0, 4],
  [5, 6, 7, 8],
  [9, 0, 11, 12],
  [13, 14, 15, 16]
]

Using the in-place algorithm that uses the first row and column as markers, what will be the matrix after applying set_zeroes?

hard

A. [[0, 0, 0, 0], [5, 6, 0, 8], [0, 0, 0, 0], [13, 14, 0, 16]]

B. [[1, 2, 0, 4], [5, 6, 0, 8], [0, 0, 0, 0], [13, 14, 0, 16]]

C. [[0, 2, 0, 4], [5, 6, 0, 8], [0, 0, 0, 0], [13, 14, 0, 16]]

D. [[0, 0, 0, 0], [5, 0, 0, 8], [0, 0, 0, 0], [13, 0, 0, 16]]

Solution

1. Step 1: Mark rows and columns using first row and column

   Zeros at (0,2) and (2,1) mark row 0, row 2, column 2, and column 1 as zero.

2. Step 2: Update matrix based on markers

   Set all elements in rows 0 and 2, and columns 1 and 2 to zero. Also, since matrix[0][0] is zero (due to marking), first row is zeroed. First column is zeroed only where rows are marked since no original first column zero.

3. Final Answer:

   [[0, 0, 0, 0], [5, 0, 0, 8], [0, 0, 0, 0], [13, 0, 0, 16]] -> Option D

4. Quick Check:

   First row and column mark zeros correctly [OK]

Hint: Remember to zero first row and column if marked [OK]

Common Mistakes:

• Forgetting to zero first row or column
• Not marking all affected rows and columns
• Incorrectly zeroing unrelated cells