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

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of Set Matrix Zeroes In Place - Simple Case

What is the state of the matrix after running the setZeroes method on the given matrix?

DSA Java

int[][] matrix = {
  {1, 1, 1},
  {1, 0, 1},
  {1, 1, 1}
};

// After calling setZeroes(matrix), what is matrix?

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

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

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

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

💡 Hint

Remember that if an element is zero, its entire row and column become zero.

The zero at position (1,1) causes the entire row 1 and column 1 to be set to zero. So the middle row and middle column become zero.

❓ Predict Output

intermediate

Output of Set Matrix Zeroes In Place - Multiple Zeroes

What is the matrix after applying setZeroes on this matrix?

DSA Java

int[][] matrix = {
  {0, 1, 2, 0},
  {3, 4, 5, 2},
  {1, 3, 1, 5}
};

// After setZeroes(matrix), what is matrix?

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

B [[0, 1, 2, 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

Check all zero positions and mark their rows and columns.

Zeroes at (0,0) and (0,3) cause row 0 and columns 0 and 3 to be zeroed. Rows 1 and 2 have columns 0 and 3 zeroed.

🔧 Debug

advanced

Identify the Bug in This Set Matrix Zeroes Code

What error or problem will this code cause when run on a matrix with zeroes?

DSA Java

public void setZeroes(int[][] matrix) {
  int rows = matrix.length;
  int cols = matrix[0].length;
  boolean[] rowZero = new boolean[rows];
  boolean[] colZero = new boolean[cols];

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

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

A No bug; code works correctly for all matrices.

B Does not set zeroes in place; uses extra space incorrectly.

C Fails to set zeroes for the first row and column correctly.

D Throws ArrayIndexOutOfBoundsException if matrix is empty (0 rows).

💡 Hint

Check what happens if matrix has zero rows or zero columns.

If matrix.length == 0, matrix[0] access causes ArrayIndexOutOfBoundsException. Also if matrix[0].length == 0, cols is zero but code still runs loops.

❓ Predict Output

advanced

Output of Optimized In-Place Set Matrix Zeroes

Given this optimized code using first row and column as markers, what is the matrix after running setZeroes?

DSA Java

int[][] matrix = {
  {1, 2, 3},
  {4, 0, 6},
  {7, 8, 9}
};

// After setZeroes(matrix), what is matrix?

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

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

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

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

💡 Hint

Remember the first row and column mark which rows and columns to zero.

Zero at (1,1) marks row 1 and column 1. So row 1 and column 1 become zero. First row and column are used as markers.

🧠 Conceptual

expert

Space Complexity of In-Place Set Matrix Zeroes Algorithm

What is the space complexity of the optimized in-place setZeroes algorithm that uses the first row and first column as markers?

A O(1) - Constant extra space regardless of matrix size

B O(m + n) - Extra space proportional to rows plus columns

C O(m * n) - Extra space proportional to total elements

D O(log(m + n)) - Extra space grows logarithmically with matrix size

💡 Hint

The algorithm only uses a few variables and modifies the matrix itself.

Using the first row and column as markers avoids extra arrays, so only constant extra space is used.

Practice

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

easy

A. Count the number of 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. Sort each row of the matrix in ascending order.

D. Find the sum of all elements in the matrix.

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.

2. Step 2: Recognize the constraint of in-place modification

   We must do this 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 B

4. Quick Check:

   Goal = Set rows and columns to zero in-place [OK]

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

Common Mistakes:

• Using extra space like another matrix
• Only zeroing rows or columns, not both
• Not modifying the matrix in-place

2. Which of the following Java code snippets correctly declares a 2D integer matrix named matrix with 3 rows and 4 columns?

easy

A. int[][] matrix = new int[3][4];

B. int matrix[][] = new int[4][3];

C. int matrix[3][4];

D. int matrix = new int[3][4];

Solution

1. Step 1: Recall Java 2D array declaration syntax

   In Java, 2D arrays are declared as int[][] name = new int[rows][columns];.

2. Step 2: Match the correct dimensions

   int[][] matrix = new int[3][4]; uses 3 rows and 4 columns correctly. int matrix[][] = new int[4][3]; reverses dimensions, C is invalid syntax, D misses array brackets.

3. Final Answer:

   int[][] matrix = new int[3][4]; -> Option A

4. Quick Check:

   Java 2D array syntax = int[][] matrix = new int[3][4]; [OK]

Hint: Use int[][] for 2D arrays with new int[rows][cols] [OK]

Common Mistakes:

• Swapping rows and columns in new int[][]
• Using C-style array declaration
• Missing brackets in declaration

3. What will be the state of the matrix after running the following Java code snippet for setting matrix zeroes in place?

int[][] matrix = {
  {1, 2, 0},
  {4, 5, 6},
  {7, 0, 9}
};

// After applying setZeroes(matrix) method
// (method sets entire row and column to zero if element is zero)

medium

A. [[0, 0, 0], [4, 0, 0], [0, 0, 0]]

B. [[1, 0, 0], [4, 0, 0], [0, 0, 0]]

C. [[0, 0, 0], [4, 5, 0], [0, 0, 0]]

D. [[0, 2, 0], [4, 0, 6], [0, 0, 0]]

Solution

1. Step 1: Identify zero elements and their rows and columns

   Zeros are at positions (0,2) and (2,1). Rows 0 and 2, columns 1 and 2 must be zeroed.

2. Step 2: Set entire rows and columns to zero

   Row 0 becomes [0,0,0], row 2 becomes [0,0,0]. Column 1 and 2 in row 1 become 0, so row 1 becomes [4,0,0].

3. Final Answer:

   [[0, 0, 0], [4, 0, 0], [0, 0, 0]] -> Option A

4. Quick Check:

   Zero rows and columns set correctly = [[0, 0, 0], [4, 0, 0], [0, 0, 0]] [OK]

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

Common Mistakes:

• Only zeroing rows or columns, not both
• Not updating all affected cells
• Overwriting zeros too early

4. Consider this buggy Java code snippet for setting matrix zeroes in place:

public void setZeroes(int[][] matrix) {
  int rows = matrix.length;
  int cols = matrix[0].length;
  boolean[] rowZero = new boolean[rows];
  boolean[] colZero = new boolean[cols];

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

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

What is the main issue with this code regarding the problem constraints?

medium

A. It modifies the matrix before checking all zeros, causing incorrect results.

B. It does not correctly identify zero elements in the matrix.

C. It causes an ArrayIndexOutOfBoundsException.

D. It uses extra space proportional to rows and columns, violating in-place requirement.

Solution

1. Step 1: Analyze space usage of the code

   The code uses two boolean arrays, rowZero and colZero, sized by rows and columns, which is extra space.

2. Step 2: Compare with problem constraints

   The problem requires in-place modification without extra space, so using these arrays violates the constraint.

3. Final Answer:

   It uses extra space proportional to rows and columns, violating in-place requirement. -> Option D

4. Quick Check:

   Extra space used = violates in-place [OK]

Hint: Avoid extra arrays; use matrix first row/column as markers [OK]

Common Mistakes:

• Ignoring space constraints
• Using extra arrays for marking
• Not handling first row/column carefully

5. You have a 4x4 matrix:

1 2 3 4
5 0 7 8
9 10 11 12
0 14 15 16

Using the in-place method for setting matrix zeroes, what will be the matrix after the operation?

hard

A. [[0, 0, 3, 4], [0, 0, 0, 0], [0, 0, 11, 12], [0, 0, 15, 16]]

B. [[0, 0, 0, 4], [0, 0, 0, 0], [0, 0, 0, 12], [0, 0, 0, 0]]

C. [[0, 0, 3, 4], [0, 0, 0, 0], [0, 0, 11, 12], [0, 0, 0, 0]]

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

Solution

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

   Zeros at (1,1) and (3,0). Rows 1 and 3, columns 0 and 1 must be zeroed.

2. Step 2: Set zeros for affected rows and columns

   Rows 1 and 3 become entirely zeros. Columns 0 and 1 become zeros across all rows, resulting in row 0: [0,0,3,4]; row 2: [0,0,11,12].

3. Final Answer:

   [[0, 0, 3, 4], [0, 0, 0, 0], [0, 0, 11, 12], [0, 0, 0, 0]] -> Option C

4. Quick Check:

   Zero rows and columns updated correctly = [[0, 0, 3, 4], [0, 0, 0, 0], [0, 0, 11, 12], [0, 0, 0, 0]] [OK]

Hint: Mark first row/column, then update matrix carefully [OK]

Common Mistakes:

• Zeroing entire rows or columns unnecessarily
• Missing zeros in some cells
• Not handling first row and column markers properly