Set Matrix Zeroes In Place in DSA Javascript - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

What is the output of the matrix after running this code?

Given the matrix below, what will be its state after the function setZeroes runs?

DSA Javascript

function setZeroes(matrix) {
  const m = matrix.length;
  const n = matrix[0].length;
  let firstRowZero = false;
  let firstColZero = false;

  for (let i = 0; i < m; i++) {
    if (matrix[i][0] === 0) firstColZero = true;
  }
  for (let j = 0; j < n; j++) {
    if (matrix[0][j] === 0) firstRowZero = true;
  }

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

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

  if (firstColZero) {
    for (let i = 0; i < m; i++) {
      matrix[i][0] = 0;
    }
  }

  if (firstRowZero) {
    for (let j = 0; j < n; j++) {
      matrix[0][j] = 0;
    }
  }
}

const matrix = [
  [1, 1, 1],
  [1, 0, 1],
  [1, 1, 1]
];
setZeroes(matrix);
console.log(matrix);

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

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

C [[1,0,1],[0,0,0],[1,0,1]]

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

💡 Hint

Check how the first row and first column are used to mark zeroes.

The code marks rows and columns to be zeroed using the first row and column as markers. Then it sets the cells to zero accordingly. The zero at position (1,1) causes row 1 and column 1 to be zeroed.

❓ Predict Output

intermediate

What is the output of the matrix after running this code?

Given the matrix below, what will be its state after the function setZeroes runs?

DSA Javascript

function setZeroes(matrix) {
  const m = matrix.length;
  const n = matrix[0].length;
  let firstRowZero = false;
  let firstColZero = false;

  for (let i = 0; i < m; i++) {
    if (matrix[i][0] === 0) firstColZero = true;
  }
  for (let j = 0; j < n; j++) {
    if (matrix[0][j] === 0) firstRowZero = true;
  }

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

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

  if (firstColZero) {
    for (let i = 0; i < m; i++) {
      matrix[i][0] = 0;
    }
  }

  if (firstRowZero) {
    for (let j = 0; j < n; j++) {
      matrix[0][j] = 0;
    }
  }
}

const matrix = [
  [0, 1, 2, 0],
  [3, 4, 5, 2],
  [1, 3, 1, 5]
];
setZeroes(matrix);
console.log(matrix);

A [[0,1,2,0],[3,4,5,2],[1,3,1,5]]

B [[0,0,0,0],[0,4,5,0],[0,3,1,0]]

C [[0,0,0,0],[3,4,5,2],[1,3,1,5]]

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

💡 Hint

Remember to zero the first row and column if needed after marking.

Zeros in first row and last column cause entire first row and last column to be zeroed. Also, zero in first column causes first column to be zeroed.

🔧 Debug

advanced

Why does this code cause an infinite loop?

Consider this code snippet for setting matrix zeroes. Why does it cause an infinite loop?

DSA Javascript

function setZeroes(matrix) {
  const m = matrix.length;
  const n = matrix[0].length;
  let firstRowZero = false;
  let firstColZero = false;

  for (let i = 0; i <= m; i++) {
    if (matrix[i][0] === 0) firstColZero = true;
  }

  for (let j = 0; j < n; j++) {
    if (matrix[0][j] === 0) firstRowZero = true;
  }

  // rest of code omitted
}

A The loop condition uses <= m which causes index out of bounds and infinite loop.

B The variable firstRowZero is never updated causing infinite loop.

C The matrix is modified inside the loop causing infinite loop.

D The loop variable i is not incremented causing infinite loop.

💡 Hint

Check the loop boundary conditions carefully.

The loop runs from i = 0 to i <= m, but valid indices are 0 to m-1. Accessing matrix[m][0] causes error or infinite behavior.

❓ Predict Output

advanced

What is the output of the matrix after running this code?

Given the matrix below, what will be its state after the function setZeroes runs?

DSA Javascript

function setZeroes(matrix) {
  const m = matrix.length;
  const n = matrix[0].length;
  let firstRowZero = false;
  let firstColZero = false;

  for (let i = 0; i < m; i++) {
    if (matrix[i][0] === 0) firstColZero = true;
  }
  for (let j = 0; j < n; j++) {
    if (matrix[0][j] === 0) firstRowZero = true;
  }

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

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

  if (firstColZero) {
    for (let i = 0; i < m; i++) {
      matrix[i][0] = 0;
    }
  }

  if (firstRowZero) {
    for (let j = 0; j < n; j++) {
      matrix[0][j] = 0;
    }
  }
}

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

A [[1,0,3],[4,0,6],[7,0,9]]

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

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

D [[1,0,3],[0,0,0],[7,0,9]]

💡 Hint

Focus on how the zero at (1,1) affects its row and column.

The zero at (1,1) causes row 1 and column 1 to be zeroed. The first row and first column are not zeroed because they have no zero initially.

❓ Predict Output

expert

What is the output of the matrix after running this code?

Given the matrix below, what will be its state after the function setZeroes runs?

DSA Javascript

function setZeroes(matrix) {
  const m = matrix.length;
  const n = matrix[0].length;
  let firstRowZero = false;
  let firstColZero = false;

  for (let i = 0; i < m; i++) {
    if (matrix[i][0] === 0) firstColZero = true;
  }
  for (let j = 0; j < n; j++) {
    if (matrix[0][j] === 0) firstRowZero = true;
  }

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

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

  if (firstColZero) {
    for (let i = 0; i < m; i++) {
      matrix[i][0] = 0;
    }
  }

  if (firstRowZero) {
    for (let j = 0; j < n; j++) {
      matrix[0][j] = 0;
    }
  }
}

const matrix = [
  [0, 1, 2, 0],
  [3, 4, 0, 2],
  [1, 3, 1, 5]
];
setZeroes(matrix);
console.log(matrix);

A [[0,0,0,0],[0,4,0,0],[0,3,0,0]]

B [[0,1,2,0],[3,4,0,2],[1,3,1,5]]

C [[0,0,0,0],[3,4,0,2],[1,3,1,5]]

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

💡 Hint

Multiple zeros in matrix affect rows and columns to zero.

Zeros at (0,0), (0,3), and (1,2) cause first row, first column, third column, and second row to be zeroed accordingly.

Practice

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

easy

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

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

C. Count how many zeroes are in the matrix.

D. Sort the matrix rows and columns after setting zeroes.

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 tasks

   Options B, C, and D describe different tasks like summing, counting, or sorting, which are not the goal here.

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 B

4. Quick Check:

   Goal = Modify matrix in place [OK]

Hint: Focus on in-place zeroing of rows and columns [OK]

Common Mistakes:

• Thinking the problem is about counting zeroes
• Assuming extra space is allowed
• Confusing with sorting or summing tasks

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

easy

A. const matrix = [[0,0,0],[0,0],[0,0,0]];

B. const matrix = Array(3).fill(Array(3).fill(0));

C. const matrix = Array.from({ length: 3 }, () => Array(3).fill(0));

D. const matrix = new Array(3).map(new Array(3).fill(0));

Solution

1. Step 1: Check each initialization method

   const matrix = Array(3).fill(Array(3).fill(0)); uses fill with the same inner array reference, causing all rows to point to the same array, which is incorrect.

2. Step 2: Identify correct independent row creation

   const matrix = Array.from({ length: 3 }, () => Array(3).fill(0)); uses Array.from with a function to create new arrays for each row, correctly initializing independent rows filled with zeros.

3. Final Answer:

   const matrix = Array.from({ length: 3 }, () => Array(3).fill(0)); -> Option C

4. Quick Check:

   Independent rows = const matrix = Array.from({ length: 3 }, () => Array(3).fill(0)); [OK]

Hint: Use Array.from with a function for independent rows [OK]

Common Mistakes:

• Using fill with array directly causing shared references
• Incorrect row lengths in nested arrays
• Using map on uninitialized arrays

3. What will be the matrix after running the following code?

function setZeroes(matrix) {
  const rows = matrix.length;
  const cols = matrix[0].length;
  let firstRowZero = false;
  let firstColZero = false;

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

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

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

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

let mat = [
  [1,1,1],
  [1,0,1],
  [1,1,1]
];
setZeroes(mat);
console.log(mat);

medium

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

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

C. [[1,1,1],[1,0,1],[1,1,1]]

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

Solution

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

   The zero is at position (1,1). The first row and column do not have zero initially. The code marks matrix[1][0] and matrix[0][1] as zero.

2. Step 2: Set zeroes based on marks and update first row/column if needed

   Using marks, the code sets matrix[1][1], matrix[1][2], matrix[0][1], and matrix[2][1] to zero. Since firstRowZero and firstColZero are false, no changes to first row or column.

3. Final Answer:

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

4. Quick Check:

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

Hint: Mark first row/col to track zero rows/columns [OK]

Common Mistakes:

• Not handling first row and column separately
• Overwriting marks too early
• Missing zeroing of entire row or column

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

function setZeroes(matrix) {
  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 modifies the matrix while iterating, causing incorrect zeroing.

B. It uses extra space which is not allowed.

C. It does not check for zero elements correctly.

D. It only sets zeroes in the first row and column.

Solution

1. Step 1: Analyze the nested loops and zero setting

   The code sets zeroes immediately when it finds a zero, modifying the matrix during iteration.

2. Step 2: Understand the effect of modifying during iteration

   Modifying the matrix while scanning causes new zeroes to appear, leading to extra rows and columns being zeroed incorrectly.

3. Final Answer:

   It modifies the matrix while iterating, causing incorrect zeroing. -> Option A

4. Quick Check:

   In-place modification during iteration causes errors [OK]

Hint: Avoid changing matrix while scanning for zeroes [OK]

Common Mistakes:

• Setting zeroes immediately instead of marking first
• Not separating scanning and updating phases
• Assuming immediate zeroing is safe

5. Given a 4x4 matrix:

[
  [0,1,2,0],
  [3,4,5,2],
  [1,3,1,5],
  [1,0,1,7]
]

After applying the in-place zeroing algorithm, what is the final matrix?

hard

A. [[0,0,0,0],[0,0,5,0],[0,0,1,0],[0,0,0,0]]

B. [[0,0,0,0],[3,4,5,2],[1,3,1,5],[1,0,1,7]]

C. [[0,1,2,0],[3,4,5,2],[1,3,1,5],[1,0,1,7]]

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

Solution

1. Step 1: Identify zero positions in the matrix

   Zeroes are at positions (0,0), (0,3), and (3,1). These mark rows 0 and 3, and columns 0, 1, and 3 for zeroing.

2. Step 2: Set entire rows and columns to zero based on marks

   Rows 0 and 3 become all zero. Columns 0, 1, and 3 become zero in all rows. The resulting matrix is [[0,0,0,0],[0,0,5,0],[0,0,1,0],[0,0,0,0]].

3. Final Answer:

   [[0,0,0,0],[0,0,5,0],[0,0,1,0],[0,0,0,0]] -> Option A

4. Quick Check:

   Zero rows and columns set correctly [OK]

Hint: Mark zero rows and columns, then update matrix once [OK]

Common Mistakes:

• Missing zeroing some columns or rows
• Overwriting matrix too early
• Not handling multiple zeroes correctly