Why Set Matrix Zeroes In Place in DSA Rust?

The Big Idea

What if you could fix an entire grid's zeros without extra memory or mistakes?

The Scenario

Imagine you have a big spreadsheet with numbers. If any cell is zero, you want to make its entire row and column zero. Doing this by hand means checking every cell, then going back and changing rows and columns one by one.

The Problem

Doing this manually or with simple code means you might change cells too early, causing wrong zeros. Also, it takes a lot of time and memory if you try to remember all zero positions separately.

The Solution

This method cleverly uses the first row and first column of the matrix itself to remember which rows and columns need to be zeroed. This way, it changes the matrix in place without extra memory and avoids early changes that cause errors.

Before vs After

✗ Before

let mut zero_positions = Vec::new();
for i in 0..rows {
    for j in 0..cols {
        if matrix[i][j] == 0 {
            zero_positions.push((i, j));
        }
    }
}
// Then zero rows and columns separately

✓ After

for i in 0..rows {
    for j in 0..cols {
        if matrix[i][j] == 0 {
            matrix[0][j] = 0;
            matrix[i][0] = 0;
        }
    }
}
// Use first row and column to zero others

What It Enables

You can efficiently update large grids or tables in place, saving memory and avoiding mistakes.

Real Life Example

In image processing, if a pixel is corrupted (zero), you might want to clear its entire row and column to avoid errors in the final image.

Key Takeaways

Manual tracking of zeros wastes memory and risks errors.

Using the matrix's first row and column as markers solves this elegantly.

This method updates the matrix in place with minimal extra space.

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