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

Time Complexity: Set Matrix Zeroes In Place

O(m x n)

Understanding Time Complexity

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

Specifically, how does the number of steps change when we mark rows and columns to zero in place?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

void setZeroes(std::vector<std::vector<int>>& matrix) {
    int rows = matrix.size();
    int cols = matrix[0].size();
    bool 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 first row and column as markers.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Nested loops traversing the matrix rows and columns.
• How many times: Each nested loop runs approximately rows x cols times.

How Execution Grows With Input

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

Input Size (rows x cols)	Approx. Operations
10 x 10 = 100	About 300 steps (3 passes over matrix)
100 x 100 = 10,000	About 30,000 steps
1000 x 1000 = 1,000,000	About 3,000,000 steps

Pattern observation: Operations grow linearly with the number of elements in the matrix.

Final Time Complexity

Time Complexity: O(m x n)

This means the time needed grows proportionally to the total number of elements in the matrix.

Common Mistake

[X] Wrong: "Since we use the first row and column as markers, the time complexity is less than scanning the whole matrix twice."

[OK] Correct: Even with markers, we still scan almost every element multiple times, so the time grows with the total elements.

Interview Connect

Understanding this time complexity helps you explain how your solution scales with matrix size, a key skill in interviews.

Self-Check

"What if we used extra space to store zero rows and columns instead of marking in place? How would the time complexity change?"

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. Find the sum of all elements in the matrix.

C. Count the number of zero elements in the matrix.

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

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

   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 D

4. Quick Check:

   In-place zeroing = Modify matrix rows and columns [OK]

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

Common Mistakes:

• Using extra space like another matrix
• Only zeroing the element itself, not row and column
• Not handling the in-place requirement

2. Which of the following C++ code snippets correctly declares a 2D vector to represent a matrix for the Set Matrix Zeroes In Place problem?

easy

A. std::vector matrix(cols, std::vector<int>(rows));

B. std::vector<int> matrix(rows, cols);

C. std::vector matrix(rows, std::vector<int>(cols));

D. std::vector<int> matrix[rows][cols];

Solution

1. Step 1: Understand 2D vector declaration

   To create a matrix with 'rows' rows and 'cols' columns, use vector of vectors with inner vector size 'cols'.

2. Step 2: Check each option

   std::vector matrix(rows, std::vector<int>(cols)); correctly creates a vector of vectors with 'rows' outer vectors and 'cols' inner vectors each.

3. Final Answer:

   std::vector matrix(rows, std::vector<int>(cols)); -> Option C

4. Quick Check:

   Correct 2D vector syntax = std::vector matrix(rows, std::vector<int>(cols)); [OK]

Hint: Use vector of vectors with inner vector size as columns [OK]

Common Mistakes:

• Using single vector with two sizes incorrectly
• Swapping rows and columns in inner vector
• Using array syntax instead of vector

3. What will be the state of the matrix after running the following code snippet for Set Matrix Zeroes In Place?

std::vector matrix = {
  {1, 2, 3},
  {4, 0, 6},
  {7, 8, 9}
};
// Assume correct in-place zeroing function is called here
setZeroes(matrix);
// Print matrix

medium

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

Solution

1. Step 1: Identify zero positions

   The zero is at position (1,1) (second row, second column).

2. Step 2: Set entire row and column to zero

   Row 1 becomes [0,0,0], column 1 becomes zero in rows 0 and 2, so matrix becomes [[1,0,3],[0,0,0],[7,0,9]].

3. Final Answer:

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

4. Quick Check:

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

Hint: Zero row and column where zero found [OK]

Common Mistakes:

• Zeroing entire matrix instead of row and column
• Not zeroing column correctly
• Leaving original zero row unchanged

4. Consider this buggy code snippet for Set Matrix Zeroes In Place. What is the main error?

void setZeroes(std::vector& matrix) {
  int rows = matrix.size();
  int cols = matrix[0].size();
  for (int i = 0; i < rows; i++) {
    for (int j = 0; j < cols; j++) {
      if (matrix[i][j] == 0) {
        for (int k = 0; k < cols; k++) matrix[i][k] = 0;
        for (int k = 0; k < rows; k++) matrix[k][j] = 0;
      }
    }
  }
}

medium

A. Modifying matrix inside loops causes multiple zeroing and incorrect results.

B. The loops do not cover all rows and columns.

C. The function does not return the modified matrix.

D. The matrix size is not checked for empty input.

Solution

1. Step 1: Analyze modification during iteration

   The code sets zeros immediately when a zero is found, changing matrix during iteration.

2. Step 2: Understand side effects

   This causes new zeros to appear and cause more zeroing, leading to incorrect results.

3. Final Answer:

   Modifying matrix inside loops causes multiple zeroing and incorrect results. -> Option A

4. Quick Check:

   Immediate modification causes cascading zeroing [OK]

Hint: Mark zeros first, then modify matrix after loops [OK]

Common Mistakes:

• Changing matrix while scanning causes wrong zeroing
• Not using markers or flags to delay zeroing
• Ignoring empty matrix edge cases

5. Given a matrix with multiple zeros, how can you efficiently set matrix zeroes in place without extra space? Choose the best approach.

hard

A. Use first row and first column as markers to store zero flags, then update matrix accordingly.

B. Create a copy of the matrix and update zeros in the copy.

C. Use a hash map to store zero positions and then update matrix.

D. Iterate multiple times over the matrix zeroing rows and columns immediately when zero found.

Solution

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

   Mark zeros found in first row and column to remember which rows and columns to zero later.

2. Step 2: Update matrix based on markers

   After marking, zero out rows and columns except first row and column, then handle first row and column separately.

3. Final Answer:

   Use first row and first column as markers to store zero flags, then update matrix accordingly. -> Option A

4. Quick Check:

   In-place marking with first row/column = Use first row and first column as markers to store zero flags, then update matrix accordingly. [OK]

Hint: Use matrix edges as flags to avoid extra space [OK]

Common Mistakes:

• Using extra space violating in-place rule
• Zeroing immediately causing cascading errors
• Not handling first row and column separately