Set Matrix Zeroes In Place in DSA Rust - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to iterate over each row in the matrix.

DSA Rust

for [1] in 0..matrix.len() {
    // process each row
}

Drag options to blanks, or click blank then click option'

Aindex

Brow

Ccol

Di

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'col' instead of a row index variable.

Using a variable name that conflicts with Rust keywords.

We use 'i' as the loop variable to iterate over row indices from 0 to matrix length.

2fill in blank

medium

Complete the code to mark the first column zero flag when a zero is found in the matrix.

DSA Rust

if matrix[i][0] == 0 {
    [1] = true;
}

Drag options to blanks, or click blank then click option'

Afirst_col_zero

Brow_zero

Ccol_zero

Dzero_found

Attempts:

3 left

💡 Hint

Common Mistakes

Using a variable that tracks rows instead of the first column.

Forgetting to set the flag when zero is found.

We set 'first_col_zero' to true to remember that the first column needs to be zeroed later.

3fill in blank

hard

Fix the error in the code to correctly mark zeroes in the first row and column.

DSA Rust

if matrix[i][j] == 0 {
    matrix[i][0] = 0;
    matrix[[1]][j] = 0;
}

Drag options to blanks, or click blank then click option'

A1

B0

Cj

Di

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'i' instead of 0 for the first row index.

Using 'j' which is the column index, not row.

We mark the first row at column j by setting matrix[0][j] = 0 to track zero columns.

4fill in blank

hard

Fill both blanks to correctly zero out cells based on markers in first row and column.

DSA Rust

for i in (1..matrix.len()).rev() {
    for j in (1..matrix[0].len()).rev() {
        if matrix[i][0] == [1] || matrix[0][j] == [2] {
            matrix[i][j] = 0;
        }
    }
}

Drag options to blanks, or click blank then click option'

A0

B1

Ctrue

Dfalse

Attempts:

3 left

💡 Hint

Common Mistakes

Using boolean values instead of integer zero.

Using 1 which is not the marker for zero.

We check if the first column or first row has zero marker (0) to set matrix[i][j] to zero.

5fill in blank

hard

Fill all three blanks to zero out the first row and first column if needed.

DSA Rust

if first_col_zero {
    for i in 0..matrix.len() {
        matrix[i][[1]] = 0;
    }
}

if matrix.len() > 0 && matrix[0].len() > 0 {
    for j in 0..matrix[0].len() {
        matrix[[2]][j] = 0;
    }
}

let first_row_zero = matrix[[3]].iter().any(|&x| x == 0);

Drag options to blanks, or click blank then click option'

A0

D1

Attempts:

3 left

💡 Hint

Common Mistakes

Using 1 instead of 0 for first row or column.

Confusing row and column indices.

We zero out the first column (index 0), the first row (index 0), and check if the first row (index 0) has any zero.

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