Set Matrix Zeroes In Place in DSA Java - Time & Space Complexity

Time Complexity: Set Matrix Zeroes In Place

O(n * m)

Understanding Time Complexity

We want to know how the time needed to set matrix zeroes grows as the matrix size grows.

How does the number of steps change when the matrix gets bigger?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

public void setZeroes(int[][] matrix) {
    int rows = matrix.length;
    int cols = matrix[0].length;
    boolean firstColZero = false;

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

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

    if (matrix[0][0] == 0) {
        for (int j = 0; j < cols; j++) matrix[0][j] = 0;
    }

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

This code sets entire rows and columns to zero if any element in them is zero, using the matrix itself to store flags.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Nested loops scanning all matrix elements multiple times.
• How many times: Each element is checked or updated a few times, but loops run over all rows and columns.

How Execution Grows With Input

As the matrix size grows, the number of steps grows roughly with the number of elements.

Input Size (n x m)	Approx. Operations
10 x 10	About 300 steps
100 x 100	About 30,000 steps
1000 x 1000	About 3,000,000 steps

Pattern observation: Operations grow roughly proportional to the total number of elements (rows times columns).

Final Time Complexity

Time Complexity: O(n * m)

This means the time needed grows directly with the number of elements in the matrix.

Common Mistake

[X] Wrong: "Since we use nested loops, the time complexity must be quadratic like O(n²) always."

[OK] Correct: Here, n and m are different dimensions (rows and columns). The complexity depends on both, so it is O(n * m), not necessarily square if rows and columns differ.

Interview Connect

Understanding how nested loops over a matrix affect time helps you explain and optimize matrix problems clearly in interviews.

Self-Check

"What if we used extra space to store zero flags instead of the matrix itself? How would the time complexity change?"

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