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DSA Cprogramming~10 mins

Set Matrix Zeroes In Place in DSA C - Interactive Practice

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Practice - 5 Tasks
Answer the questions below
1fill in blank
easy

Complete the code to set the first element of the row to zero when a zero is found.

DSA C
if(matrix[i][j] == 0) {
    matrix[i][0] = [1];
}
Drag options to blanks, or click blank then click option'
A1
B-1
C0
Dmatrix[i][j]
Attempts:
3 left
💡 Hint
Common Mistakes
Setting matrix[i][0] to 1 instead of 0.
Using matrix[i][j] which is already zero, but not marking the row properly.
2fill in blank
medium

Complete the code to set the first element of the column to zero when a zero is found.

DSA C
if(matrix[i][j] == 0) {
    matrix[0][j] = [1];
}
Drag options to blanks, or click blank then click option'
A1
B0
Cmatrix[i][j]
D-1
Attempts:
3 left
💡 Hint
Common Mistakes
Setting matrix[0][j] to 1 instead of 0.
Using matrix[i][j] which is already zero, but not marking the column properly.
3fill in blank
hard

Fix the error in the loop condition to avoid overwriting the first row and column prematurely.

DSA C
for(int i = 1; i [1] rows; i++) {
    for(int j = 1; j < cols; j++) {
        if(matrix[i][0] == 0 || matrix[0][j] == 0) {
            matrix[i][j] = 0;
        }
    }
}
Drag options to blanks, or click blank then click option'
A<=
B>=
C>
D<
Attempts:
3 left
💡 Hint
Common Mistakes
Using '<=' causing index out of range or overwriting first row/column.
Using '>' or '>=' which are incorrect for forward iteration.
4fill in blank
hard

Fill both blanks to correctly zero out the first row and first column if needed.

DSA C
if(firstRowZero) {
    for(int j = 0; j [1] cols; j++) {
        matrix[0][j] = 0;
    }
}
if(firstColZero) {
    for(int i = 0; i [2] rows; i++) {
        matrix[i][0] = 0;
    }
}
Drag options to blanks, or click blank then click option'
A<
B<=
C>
D>=
Attempts:
3 left
💡 Hint
Common Mistakes
Using '<=' causing index out of range.
Using '>' or '>=' which are invalid for forward loops.
5fill in blank
hard

Fill all three blanks to correctly check and mark if the first row and first column contain zero.

DSA C
bool firstRowZero = false;
bool firstColZero = false;
for(int j = 0; j [1] cols; j++) {
    if(matrix[0][j] == [2]) {
        firstRowZero = true;
        break;
    }
}
for(int i = 0; i [3] rows; i++) {
    if(matrix[i][0] == 0) {
        firstColZero = true;
        break;
    }
}
Drag options to blanks, or click blank then click option'
A<
B0
C==
D<=
Attempts:
3 left
💡 Hint
Common Mistakes
Using '<=' causing out-of-bound errors.
Using '=' instead of '==' for comparison.
Checking for non-zero values instead of zero.

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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