Set Matrix Zeroes In Place in DSA Go - Cheat Sheet & Quick Revision

Recall & Review

beginner

What does the "Set Matrix Zeroes In Place" problem ask you to do?

It asks to modify a given matrix so that if any element is 0, its entire row and column are set to 0, without using extra space for another matrix.

intermediate

Why do we use the first row and first column as markers in the in-place solution?

Because we want to avoid extra space, we use the first row and first column to remember which rows and columns should be zeroed later.

intermediate

In the in-place approach, how do you handle the first row and first column separately?

We use two separate boolean flags to track if the first row or first column originally had any zero, so we can zero them at the end if needed.

beginner

What is the time complexity of the in-place solution for Set Matrix Zeroes?

The time complexity is O(m * n), where m is the number of rows and n is the number of columns, because we scan the matrix multiple times but each scan is linear.

beginner

What is the space complexity of the in-place solution?

The space complexity is O(1) because we only use a few extra variables and modify the matrix itself without extra storage.

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

A Find the maximum element in the matrix

B Transpose the matrix

C Sort the matrix rows

D Set entire rows and columns to zero if any element is zero

The problem requires setting entire rows and columns to zero if any element in them is zero.

Which part of the matrix is used to store markers in the in-place solution?

A Diagonal elements

B Last row and last column

C First row and first column

D Middle row and column

The first row and first column are used as markers to avoid extra space.

How do you know if the first row needs to be zeroed at the end?

A By checking a separate boolean flag set during the first scan

B By scanning the last row

C By checking the last column

D By counting zeros in the matrix

A boolean flag tracks if the first row originally had any zero.

What is the space complexity of the in-place solution?

A O(1)

B O(m + n)

C O(m * n)

D O(log(m * n))

The solution uses constant extra space by modifying the matrix in place.

What is the time complexity of the in-place solution?

A O(m + n)

B O(m * n)

C O(m^2 * n^2)

D O(log(m * n))

The matrix is scanned multiple times but each scan is linear, so total time is O(m * n).

Explain how to use the first row and first column as markers to set matrix zeroes in place.

Think about how to remember which rows and columns to zero without extra arrays.

Describe the steps to solve the Set Matrix Zeroes problem in place with O(1) extra space.

Break down the solution into scanning, marking, and updating phases.

Practice

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

easy

A. Modify the matrix so that if an element is 0, its entire row and column become 0

B. Find the sum of all elements in the matrix

C. Sort each row of the matrix in ascending order

D. Transpose the matrix without changing its elements

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: Identify the correct operation

   Only Modify the matrix so that if an element is 0, its entire row and column become 0 describes this exact operation, modifying rows and columns based on zero elements.

3. Final Answer:

   Modify the matrix so that if an element is 0, its entire row and column become 0 -> Option A

4. Quick Check:

   Set rows and columns to zero = Modify matrix [OK]

Hint: Remember: zero means zero out whole row and column [OK]

Common Mistakes:

• Confusing with matrix transpose
• Trying to sum elements instead of modifying
• Sorting rows instead of zeroing

2. Which of the following Go code snippets correctly declares a fixed-size 2D array of integers?

easy

A. matrix := [][]int{{1, 2}, {3, 4}}

B. matrix := [2][2]int{{1, 2}, {3, 4}}

C. matrix := [][]int{1, 2, 3, 4}

D. matrix := [2]int{1, 2, 3, 4}

Solution

1. Step 1: Understand Go 2D array declaration

   In Go, fixed-size 2D arrays use syntax like [rows][cols]type, so [2][2]int is a 2x2 matrix.

2. Step 2: Check each option

   matrix := [2][2]int{{1, 2}, {3, 4}} correctly declares a 2x2 integer array with values. matrix := [][]int{{1, 2}, {3, 4}} is a slice of slices (dynamic), not a fixed-size array. matrix := [][]int{1, 2, 3, 4} and D are invalid for 2D matrix.

3. Final Answer:

   matrix := [2][2]int{{1, 2}, {3, 4}} -> Option B

4. Quick Check:

   Fixed 2D array syntax = [2][2]int [OK]

Hint: Use [rows][cols]type for fixed 2D arrays in Go [OK]

Common Mistakes:

• Using slice syntax when fixed array needed
• Incorrect nesting of braces
• Declaring 1D array instead of 2D

3. What will be the matrix after running the following Go code snippet?

matrix := [][]int{
  {1, 2, 3},
  {4, 0, 6},
  {7, 8, 9},
}
// After applying Set Matrix Zeroes In Place logic
// (zero out rows and columns containing zero)

medium

A. [[1, 0, 3], [0, 0, 0], [0, 0, 0]]

B. [[1, 2, 3], [0, 0, 0], [7, 0, 9]]

C. [[1, 0, 3], [0, 0, 0], [7, 0, 9]]

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

Solution

1. Step 1: Identify zero positions

   Zero is at matrix[1][1]. So row 1 and column 1 must be zeroed.

2. Step 2: Apply zeroing

   Row 1 becomes [0,0,0]. Column 1 elements at rows 0 and 2 become 0. So matrix[0][1] and matrix[2][1] are zero.

3. Final Answer:

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

4. Quick Check:

   Zero row and column applied correctly = [[1,0,3],[0,0,0],[7,0,9]] [OK]

Hint: Zero entire row and column for each zero found [OK]

Common Mistakes:

• Zeroing entire matrix
• Missing zero in column 1 for rows 0 and 2
• Zeroing wrong rows or columns

4. Consider this Go function snippet for setting matrix zeroes:

func setZeroes(matrix [][]int) {
  for i := 0; i < len(matrix); i++ {
    for j := 0; j < len(matrix[0]); j++ {
      if matrix[i][j] == 0 {
        for k := 0; k < len(matrix); k++ {
          matrix[k][j] = 0
        }
        for l := 0; l < len(matrix[0]); l++ {
          matrix[i][l] = 0
        }
      }
    }
  }
}

What is the main problem with this code?

medium

A. It only zeros rows but not columns

B. It uses wrong loop ranges for rows and columns

C. It does not check for zero elements at all

D. It modifies the matrix while iterating, causing incorrect zeroing

Solution

1. Step 1: Analyze matrix modification during iteration

   The code sets zeros immediately when it finds a zero, changing matrix elements during iteration.

2. Step 2: Understand side effects

   This causes new zeros to appear and triggers more zeroing incorrectly, leading to wrong final matrix.

3. Final Answer:

   It modifies the matrix while iterating, causing incorrect zeroing -> Option D

4. Quick Check:

   In-place modification during scan = incorrect zeroing [OK]

Hint: Mark zeros first, then update matrix to avoid side effects [OK]

Common Mistakes:

• Changing matrix while scanning
• Not using extra storage or flags
• Assuming immediate zeroing is safe

5. You want to set matrix zeroes in place with O(1) extra space. Which approach below correctly achieves this in Go?

hard

A. Use first row and first column as markers to track zero rows and columns, then update matrix accordingly

B. Create a separate boolean matrix to mark zero rows and columns, then update original matrix

C. Copy matrix to a new matrix and update zeros there to avoid modifying original during iteration

D. Use recursion to zero rows and columns when zero found

Solution

1. Step 1: Understand space constraints

   O(1) extra space means no additional matrix or large data structures allowed.

2. Step 2: Use first row and column as markers

   Mark zeros in first row and column to remember which rows and columns to zero later, then update matrix in place.

3. Step 3: Avoid extra space and recursion

   Options A and B use extra space, D uses recursion unnecessarily.

4. Final Answer:

   Use first row and first column as markers to track zero rows and columns, then update matrix accordingly -> Option A

5. Quick Check:

   In-place markers = O(1) space zeroing [OK]

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

Common Mistakes:

• Using extra arrays violating space constraint
• Modifying matrix during iteration without markers
• Overcomplicating with recursion