Set Matrix Zeroes In Place in DSA Typescript - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to check if the matrix is empty.

DSA Typescript

if (matrix.length === [1]) return;

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

A1

B0

Cnull

Dundefined

Attempts:

3 left

💡 Hint

Common Mistakes

Using 1 instead of 0 causes the check to fail for empty matrices.

Using null or undefined is incorrect for array length comparison.

We check if the matrix has zero rows by comparing matrix.length to 0.

2fill in blank

medium

Complete the code to mark the first row and column if a zero is found.

DSA Typescript

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

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

A0

B1

C-1

Dnull

Attempts:

3 left

💡 Hint

Common Mistakes

Using 1 or -1 will not correctly mark the rows and columns.

Using null is invalid for number arrays.

We mark the first cell of the row and column as 0 to indicate that entire row and column should be zeroed later.

3fill in blank

hard

Fix the error in the loop condition to skip the first row and column.

DSA Typescript

for (let i = 1; i [1] matrix.length; i++) {
  for (let j = 1; j < matrix[0].length; j++) {
    if (matrix[i][0] === 0 || matrix[0][j] === 0) {
      matrix[i][j] = 0;
    }
  }
}

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

A>=

B<=

C<

D>

Attempts:

3 left

💡 Hint

Common Mistakes

Using <= causes runtime errors due to invalid index access.

Using > or >= will skip the loop entirely.

The loop should run while i is less than matrix.length to avoid going out of bounds.

4fill in blank

hard

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

DSA Typescript

if (firstRowHasZero) {
  for (let j = 0; j [1] matrix[0].length; j++) {
    matrix[0][j] = [2];
  }
}
if (firstColHasZero) {
  for (let i = 0; i [1] matrix.length; i++) {
    matrix[i][0] = 0;
  }
}

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

A<

B<=

C0

D1

Attempts:

3 left

💡 Hint

Common Mistakes

Using <= causes index errors.

Setting elements to 1 or other values does not zero them.

Use < in the loop condition to avoid out of bounds, and set elements to 0 to zero out the row and column.

5fill in blank

hard

Fill all three blanks to complete the function that sets matrix zeroes in place.

DSA Typescript

function setZeroes(matrix: number[][]): void {
  let firstRowHasZero = false;
  let firstColHasZero = false;

  for (let j = 0; j [1] matrix[0].length; j++) {
    if (matrix[0][j] === 0) firstRowHasZero = true;
  }
  for (let i = 0; i [2] matrix.length; i++) {
    if (matrix[i][0] === 0) firstColHasZero = true;
  }

  for (let i = 1; i < matrix.length; i++) {
    for (let j = 1; j < matrix[0].length; j++) {
      if (matrix[i][j] === 0) {
        matrix[i][0] = 0;
        matrix[0][j] = 0;
      }
    }
  }

  for (let i = 1; i < matrix.length; i++) {
    for (let j = 1; j < matrix[0].length; j++) {
      if (matrix[i][0] === 0 || matrix[0][j] === 0) {
        matrix[i][j] = [3];
      }
    }
  }

  if (firstRowHasZero) {
    for (let j = 0; j < matrix[0].length; j++) {
      matrix[0][j] = 0;
    }
  }

  if (firstColHasZero) {
    for (let i = 0; i < matrix.length; i++) {
      matrix[i][0] = 0;
    }
  }
}

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

A<

B<=

C0

D1

Attempts:

3 left

💡 Hint

Common Mistakes

Using <= causes index errors.

Setting elements to 1 or other values does not zero them.

Use < in loop conditions to avoid out of bounds, and set elements to 0 to zero them.

Practice

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

easy

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

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

C. Sort the matrix rows in ascending order.

D. Replace all non-zero elements with 1.

Solution

1. Step 1: Understand the problem requirement

   The problem requires changing the matrix so that any row or column containing a zero is set entirely to zero.

2. Step 2: Note the in-place constraint

   This must be done without using extra space for another matrix or large data structures.

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 A

4. Quick Check:

   In-place zeroing of rows and columns = Modify the matrix so that if an element is 0, its entire row and column become 0 without using extra space. [OK]

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

Common Mistakes:

• Using extra space like another matrix
• Only zeroing rows or only columns
• Not handling the in-place requirement

2. Which of the following TypeScript code snippets correctly declares a 2D matrix named matrix of numbers?

easy

A. const matrix: number[][] = [[1, 0], [3, 4]];

B. const matrix: number[] = [[1, 0], [3, 4]];

C. const matrix: number = [[1, 0], [3, 4]];

D. const matrix: any = 1, 0, 3, 4;

Solution

1. Step 1: Understand TypeScript 2D array syntax

   A 2D array of numbers is declared as number[][].

2. Step 2: Check each option

   const matrix: number[][] = [[1, 0], [3, 4]]; uses number[][] and initializes correctly. Options B and C use incorrect types. const matrix: any = 1, 0, 3, 4; is invalid syntax.

3. Final Answer:

   const matrix: number[][] = [[1, 0], [3, 4]]; -> Option A

4. Quick Check:

   2D array type is number[][] = const matrix: number[][] = [[1, 0], [3, 4]]; [OK]

Hint: Use number[][] for 2D numeric arrays in TypeScript [OK]

Common Mistakes:

• Using number[] for 2D arrays
• Assigning arrays to number type
• Incorrect syntax for array declaration

3. Given the following TypeScript code, what will be the state of matrix after calling setZeroes(matrix)?

function setZeroes(matrix: number[][]): void {
  const rows = matrix.length;
  const cols = matrix[0].length;
  const rowZero = new Array(rows).fill(false);
  const colZero = new Array(cols).fill(false);

  for (let i = 0; i < rows; i++) {
    for (let j = 0; j < cols; j++) {
      if (matrix[i][j] === 0) {
        rowZero[i] = true;
        colZero[j] = true;
      }
    }
  }

  for (let i = 0; i < rows; i++) {
    for (let j = 0; j < cols; j++) {
      if (rowZero[i] || colZero[j]) {
        matrix[i][j] = 0;
      }
    }
  }
}

const matrix = [
  [1, 2, 3],
  [4, 0, 6],
  [7, 8, 9]
];
setZeroes(matrix);
console.log(matrix);

medium

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

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

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

D. [[1, 0, 3], [4, 0, 6], [7, 0, 9]]

Solution

1. Step 1: Identify zero positions

   Zero found at matrix[1][1]. So row 1 and column 1 must be zeroed.

2. Step 2: Apply zeroing to rows and columns

   Row 1 becomes [0,0,0]. Column 1 elements at rows 0 and 2 become 0. Resulting matrix is [[1,0,3],[0,0,0],[7,0,9]].

3. Final Answer:

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

4. Quick Check:

   Zero at (1,1) zeroes row 1 and column 1 = [[1, 0, 3], [0, 0, 0], [7, 0, 9]] [OK]

Hint: Mark rows and columns first, then zero them [OK]

Common Mistakes:

• Zeroing only rows or only columns
• Not zeroing all affected cells
• Modifying matrix during first scan

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

function setZeroes(matrix: number[][]): void {
  const rows = matrix.length;
  const cols = matrix[0].length;

  for (let i = 0; i < rows; i++) {
    for (let j = 0; j < cols; j++) {
      if (matrix[i][j] === 0) {
        for (let k = 0; k < cols; k++) {
          matrix[i][k] = 0;
        }
        for (let k = 0; k < rows; k++) {
          matrix[k][j] = 0;
        }
      }
    }
  }
}

medium

A. It only zeros the first row and column.

B. It modifies the matrix while scanning, causing incorrect zeroing.

C. It does not zero the columns correctly.

D. It uses extra space which is not allowed.

Solution

1. Step 1: Analyze matrix modification during iteration

   The code sets zeros immediately when found, changing matrix values during scanning.

2. Step 2: Understand impact of immediate modification

   This causes new zeros to be detected incorrectly, leading to extra rows and columns zeroed.

3. Final Answer:

   It modifies the matrix while scanning, causing incorrect zeroing. -> Option B

4. Quick Check:

   Immediate zeroing during scan causes errors = It modifies the matrix while scanning, causing incorrect zeroing. [OK]

Hint: Mark first, then zero to avoid premature changes [OK]

Common Mistakes:

• Zeroing matrix cells while scanning
• Not using markers to track zero rows/columns
• Assuming immediate zeroing is safe

5. You want to optimize the Set Matrix Zeroes In Place solution to use O(1) extra space by using the first row and first column as markers. Given the matrix:

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

What will be the matrix after applying this optimized approach?

hard

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

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

C. [[0, 0, 0], [0, 0, 0], [7, 0, 9]]

D. [[0, 0, 0], [0, 0, 0], [7, 0, 0]]

Solution

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

   Zeros at (0,2) and (1,1) mark first row and column accordingly.

2. Step 2: Zero rows and columns based on markers

   Row 0 and 1 become all zeros. Column 1 and 2 become zeros in all rows. Final matrix is [[0,0,0],[0,0,0],[7,0,0]].

3. Final Answer:

   [[0, 0, 0], [0, 0, 0], [7, 0, 0]] -> Option D

4. Quick Check:

   First row/column markers zero rows/columns = [[0, 0, 0], [0, 0, 0], [7, 0, 0]] [OK]

Hint: Use first row/column as flags to save space [OK]

Common Mistakes:

• Not handling first row and column separately
• Overwriting markers too early
• Missing zeros in last row or column