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DSA Cprogramming~5 mins

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

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Time Complexity: Set Matrix Zeroes In Place
O(m x n)
Understanding Time Complexity

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

Specifically, how many steps does it take to update the matrix in place?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.


void setZeroes(int** matrix, int matrixSize, int* matrixColSize) {
    int rows = matrixSize;
    int cols = *matrixColSize;
    int firstRowZero = 0;
    int firstColZero = 0;

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

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

This code sets entire rows and columns to zero if any element in them is zero, modifying the matrix in place.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

  • Primary operation: Nested loops scanning the matrix elements twice.
  • How many times: Each nested loop runs roughly rows x cols times.
How Execution Grows With Input

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

Input Size (rows x cols)Approx. Operations
10 x 10 = 100About 400 (4 passes over matrix)
100 x 100 = 10,000About 40,000
1000 x 1000 = 1,000,000About 4,000,000

Pattern observation: Operations grow roughly proportional to the total number of elements, multiplied by a small constant.

Final Time Complexity

Time Complexity: O(m x n)

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

Common Mistake

[X] Wrong: "The time complexity is O((m x n)²) because of nested loops inside nested loops."

[OK] Correct: The nested loops run sequentially, not nested inside each other repeatedly; each element is visited a fixed number of times, not multiplied repeatedly.

Interview Connect

Understanding how to analyze matrix operations helps you solve many real problems efficiently and shows you can reason about code performance clearly.

Self-Check

"What if we used extra space to store zero positions instead of modifying the matrix in place? How would the time complexity change?"

Practice

(1/5)
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. Find the sum of all elements in the matrix.
C. Modify the matrix so that if an element is 0, its entire row and column become 0 without using extra space.
D. Sort each row of the matrix in ascending order.

Solution

  1. Step 1: Understand the problem requirement

    The problem asks to change the matrix so that any row or column containing a zero should be set entirely to zero.
  2. Step 2: Note the constraint of in-place modification

    The matrix must be modified 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 C
  4. Quick Check:

    Set matrix zeroes in place = Modify matrix rows and columns with zeros [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 modifying the matrix in place
2. Which of the following C code snippets correctly declares a 3x3 matrix to be used for the Set Matrix Zeroes problem?
easy
A. int matrix[3][3] = {{1,2,3},{4,5,6},{7,8,9}};
B. int matrix[3] = {1,2,3,4,5,6,7,8,9};
C. int matrix[][] = {{1,2,3},{4,5,6},{7,8,9}};
D. int matrix[3][3] = {1,2,3,4,5,6,7,8};

Solution

  1. Step 1: Check correct 2D array declaration syntax in C

    In C, a 2D array must have fixed sizes for both dimensions, and initialization uses nested braces.
  2. Step 2: Validate each option

    int matrix[3][3] = {{1,2,3},{4,5,6},{7,8,9}}; correctly declares and initializes a 3x3 matrix. int matrix[3] = {1,2,3,4,5,6,7,8,9}; is a 1D array with too many elements. int matrix[][] = {{1,2,3},{4,5,6},{7,8,9}}; misses size for second dimension. int matrix[3][3] = {1,2,3,4,5,6,7,8}; has too few initializers (matrix[2][2] uninitialized).
  3. Final Answer:

    int matrix[3][3] = {{1,2,3},{4,5,6},{7,8,9}}; -> Option A
  4. Quick Check:

    Correct 2D array syntax = int matrix[3][3] = {{1,2,3},{4,5,6},{7,8,9}}; [OK]
Hint: Use nested braces for 2D array initialization [OK]
Common Mistakes:
  • Missing size for second dimension
  • Using 1D array for 2D data
  • Incorrect initializer braces
3. What will be the matrix after running the in-place zeroing algorithm on this 3x3 matrix?
int matrix[3][3] = {
  {1, 0, 3},
  {4, 5, 6},
  {7, 8, 9}
};
medium
A. [[1, 0, 3], [4, 5, 6], [7, 8, 9]]
B. [[0, 0, 0], [4, 0, 6], [7, 0, 9]]
C. [[0, 0, 0], [0, 0, 0], [7, 0, 9]]
D. [[0, 0, 0], [4, 5, 6], [0, 0, 0]]

Solution

  1. Step 1: Identify zero positions in original matrix

    Zero is at position (0,1). So row 0 and column 1 must be zeroed.
  2. Step 2: Set entire row 0 and column 1 to zero

    Row 0 becomes [0,0,0]. Column 1 elements at rows 1 and 2 become 0, so matrix[1][1]=0 and matrix[2][1]=0.
  3. Final Answer:

    [[0, 0, 0], [4, 0, 6], [7, 0, 9]] -> Option B
  4. Quick Check:

    Zero at (0,1) zeroes row 0 and col 1 = [[0, 0, 0], [4, 0, 6], [7, 0, 9]] [OK]
Hint: Zero row and column of each zero element [OK]
Common Mistakes:
  • Zeroing entire matrix
  • Missing zeroing column elements
  • Not zeroing row elements
4. Consider this code snippet for setting matrix zeroes in place. What is the main bug?
void setZeroes(int matrix[3][3]) {
  int i, j;
  for(i=0; i<3; i++) {
    for(j=0; j<3; j++) {
      if(matrix[i][j] == 0) {
        for(int k=0; k<3; k++) {
          matrix[i][k] = 0;
          matrix[k][j] = 0;
        }
      }
    }
  }
}
medium
A. It uses wrong loop bounds causing out-of-bounds access.
B. It does not check for zeros at all.
C. It uses extra space which is not allowed.
D. It modifies zeros immediately causing cascading zeroing and incorrect results.

Solution

  1. Step 1: Analyze immediate zeroing inside nested loops

    The code sets zeros immediately when it finds a zero, which changes matrix during iteration.
  2. Step 2: Understand cascading zeroing problem

    New zeros created cause further zeroing in the same pass, leading to incorrect results.
  3. Final Answer:

    It modifies zeros immediately causing cascading zeroing and incorrect results. -> Option D
  4. Quick Check:

    Immediate zeroing causes cascading errors = It modifies zeros immediately causing cascading zeroing and incorrect results. [OK]
Hint: Mark zeros first, then update matrix after [OK]
Common Mistakes:
  • Zeroing matrix while scanning
  • Not separating marking and updating steps
  • Assuming immediate zeroing is safe
5. Given a 4x4 matrix:
int matrix[4][4] = {
  {1, 2, 3, 4},
  {5, 0, 7, 8},
  {9, 10, 11, 12},
  {0, 14, 15, 16}
};
Which is the correct matrix after applying the optimal in-place zeroing algorithm using first row and column as markers?
hard
A. [[0, 0, 3, 4], [0, 0, 0, 0], [0, 0, 11, 12], [0, 0, 0, 0]]
B. [[1, 0, 3, 4], [0, 0, 0, 0], [9, 0, 11, 12], [0, 0, 0, 0]]
C. [[0, 0, 0, 0], [0, 0, 0, 0], [0, 0, 0, 0], [0, 0, 0, 0]]
D. [[1, 2, 3, 4], [5, 0, 7, 8], [9, 10, 11, 12], [0, 14, 15, 16]]

Solution

  1. Step 1: Identify zero positions

    Zeros at (1,1) and (3,0).
  2. Step 2: Determine first row and first column zero flags

    First row: no zeros (firstRowZero = false). First column: zero at (3,0) (firstColZero = true).
  3. Step 3: Mark zeros using first row and column (inner matrix)

    Scan i=1 to 3, j=1 to 3: (1,1)=0 sets matrix[1][0]=0, matrix[0][1]=0.
  4. Step 4: Zero inner matrix using marks

    For i=1 to 3, j=1 to 3: if matrix[i][0]==0 or matrix[0][j]==0, set matrix[i][j]=0. Zeros row 1 (mark at [1][0]), col 1 (mark at [0][1]), row 3 (orig [3][0]=0).
  5. Step 5: Apply flags to borders

    First row unchanged (false). First col all zeroed (true): matrix[0][0]=0, matrix[2][0]=0 (among others).
  6. Final Answer:

    [[0, 0, 3, 4], [0, 0, 0, 0], [0, 0, 11, 12], [0, 0, 0, 0]] -> Option A
  7. Quick Check:

    Zeros at (1,1),(3,0) zero rows 1&3, cols 0&1 = [[0, 0, 3, 4], [0, 0, 0, 0], [0, 0, 11, 12], [0, 0, 0, 0]] [OK]
Hint: Use first row/column as zero markers to save space [OK]
Common Mistakes:
  • Zeroing entire first row or column incorrectly
  • Not marking zeros properly
  • Overwriting markers too early
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