Search in 2D Matrix in DSA Go - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to check if the matrix is empty.

DSA Go

if len(matrix) == [1] {
    return false
}

Drag options to blanks, or click blank then click option'

A1

B-1

C0

Dlen(matrix[0])

Attempts:

3 left

💡 Hint

Common Mistakes

Checking len(matrix[0]) without verifying matrix is not empty.

Using 1 instead of 0 for empty check.

We check if the number of rows is zero to know if the matrix is empty.

2fill in blank

medium

Complete the code to get the number of columns in the matrix.

DSA Go

cols := len(matrix[1])

Drag options to blanks, or click blank then click option'

A{0}

B0

C(0)

D[0]

Attempts:

3 left

💡 Hint

Common Mistakes

Using parentheses or braces instead of square brackets.

Using len(matrix) instead of len(matrix[0]) for columns.

We use matrix[0] to access the first row and get its length for columns.

3fill in blank

hard

Fix the error in the binary search condition to find the target in a row.

DSA Go

for left <= right {
    mid := left + (right - left) / 2
    if matrix[row][mid] == [1] {
        return true
    } else if matrix[row][mid] < target {
        left = mid + 1
    } else {
        right = mid - 1
    }
}

Drag options to blanks, or click blank then click option'

Amid

Btarget

Crow

Dleft

Attempts:

3 left

💡 Hint

Common Mistakes

Comparing with mid index instead of target value.

Using wrong variable names in condition.

The condition must compare the matrix element with the target value.

4fill in blank

hard

Fill both blanks to correctly perform binary search on the matrix rows.

DSA Go

for left, right := 0, [1] - 1; left <= right; {
    mid := left + (right - left) / 2
    if matrix[mid][0] == target {
        return true
    } else if matrix[mid][0] < [2] {
        left = mid + 1
    } else {
        right = mid - 1
    }
}

Drag options to blanks, or click blank then click option'

Alen(matrix)

Btarget

Clen(matrix[0])

Dmid

Attempts:

3 left

💡 Hint

Common Mistakes

Using len(matrix[0]) for rows count.

Comparing with mid index instead of target.

We set right to last row index (len(matrix)-1) and compare matrix[mid][0] with target.

5fill in blank

hard

Fill all three blanks to complete the function that searches target in the 2D matrix.

DSA Go

func searchMatrix(matrix [][]int, target int) bool {
    if len(matrix) == 0 || len(matrix[1]) == 0 {
        return false
    }
    rows, cols := len(matrix), len(matrix[0])
    left, right := 0, rows - 1
    for left <= right {
        mid := left + (right - left) / 2
        if matrix[mid][0] == [2] {
            return true
        } else if matrix[mid][0] < target {
            left = mid + 1
        } else {
            right = mid - 1
        }
    }
    row := right
    if row < 0 {
        return false
    }
    left, right = 0, cols - 1
    for left <= right {
        mid := left + (right - left) / 2
        if matrix[row][mid] == target {
            return true
        } else if matrix[row][mid] < [3] {
            left = mid + 1
        } else {
            right = mid - 1
        }
    }
    return false
}

Drag options to blanks, or click blank then click option'

A[0]

Btarget

Drow

Attempts:

3 left

💡 Hint

Common Mistakes

Using wrong index or variable in comparisons.

Not checking if matrix[0] exists before accessing.

Check if first row exists with matrix[0], compare matrix[mid][0] with target, and compare matrix[row][mid] with target.

Practice

1. What is the main idea behind searching a target value in a 2D matrix where each row is sorted and the first integer of each row is greater than the last integer of the previous row?

easy

A. Use a hash map to store all elements for quick lookup

B. Search each row linearly until the target is found

C. Treat the 2D matrix as a sorted 1D array and use binary search

D. Start from bottom-left corner and move diagonally

Solution

1. Step 1: Understand matrix properties

   The matrix rows are sorted and each row's first element is greater than the last element of the previous row, so the entire matrix can be seen as a sorted list.

2. Step 2: Apply binary search on the matrix

   By treating the matrix as a single sorted array, binary search can be used to efficiently find the target.

3. Final Answer:

   Treat the 2D matrix as a sorted 1D array and use binary search -> Option C

4. Quick Check:

   Sorted matrix + binary search = Treat the 2D matrix as a sorted 1D array and use binary search [OK]

Hint: Think of the matrix as one sorted list for fast search [OK]

Common Mistakes:

• Searching each row linearly wastes time
• Using hash map wastes extra space
• Starting from bottom-left works only for different matrix types

2. Which of the following Go code snippets correctly initializes a 2D matrix of integers with 3 rows and 4 columns?

easy

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

B. matrix := [][]int{{3,4},{1,2,3},{4,5,6,7}}

C. matrix := make([]int, 3*4)

D. matrix := make([][]int, 3); for i := range matrix { matrix[i] = make([]int, 4) }

Solution

1. Step 1: Understand Go 2D slice initialization

   In Go, a 2D slice is a slice of slices. We first create the outer slice with length 3, then initialize each inner slice with length 4.

2. Step 2: Check each option

   matrix := make([][]int, 3); for i := range matrix { matrix[i] = make([]int, 4) } correctly creates a 3x4 2D slice. matrix := [][]int{{3,4},{1,2,3},{4,5,6,7}} is irregular and not fixed size. matrix := make([]int, 3*4) creates a 1D slice. matrix := [3][4]int{} creates a fixed array, not a slice.

3. Final Answer:

   matrix := make([][]int, 3); for i := range matrix { matrix[i] = make([]int, 4) } -> Option D

4. Quick Check:

   2D slice init with make + loop = matrix := make([][]int, 3); for i := range matrix { matrix[i] = make([]int, 4) } [OK]

Hint: Use make for outer slice, then loop to make inner slices [OK]

Common Mistakes:

• Using make only once creates 1D slice
• Confusing arrays and slices
• Initializing inner slices incorrectly

3. What will be the output of the following Go code that searches for target 5 in the given matrix?

matrix := [][]int{
  {1, 3, 5, 7},
  {10, 11, 16, 20},
  {23, 30, 34, 50},
}
target := 5
found := false
rows := len(matrix)
cols := len(matrix[0])
left, right := 0, rows*cols-1
for left <= right {
  mid := (left + right) / 2
  val := matrix[mid/cols][mid%cols]
  if val == target {
    found = true
    break
  } else if val < target {
    left = mid + 1
  } else {
    right = mid - 1
  }
}
fmt.Println(found)

medium

A. true

B. false

C. Compilation error

D. Runtime panic

Solution

1. Step 1: Understand binary search on 2D matrix

   The code treats the matrix as a 1D sorted array and uses binary search with index mapping mid/cols and mid%cols to access elements.

2. Step 2: Trace the search for target 5

   Target 5 is at matrix[0][2]. The binary search will find it and set found to true.

3. Final Answer:

   true -> Option A

4. Quick Check:

   Binary search finds 5 = true [OK]

Hint: Map 1D index to 2D with row=mid/cols, col=mid%cols [OK]

Common Mistakes:

• Incorrect index mapping causing wrong element access
• Off-by-one errors in binary search boundaries
• Not breaking loop when found

4. The following Go code attempts to search for a target in a 2D matrix but has a bug. What is the bug?

func searchMatrix(matrix [][]int, target int) bool {
  rows := len(matrix)
  cols := len(matrix[0])
  left, right := 0, rows*cols
  for left <= right {
    mid := (left + right) / 2
    val := matrix[mid/cols][mid%cols]
    if val == target {
      return true
    } else if val < target {
      left = mid + 1
    } else {
      right = mid - 1
    }
  }
  return false
}

medium

A. right should be initialized to rows*cols - 1

B. mid calculation causes integer overflow

C. matrix indexing should be matrix[mid%rows][mid/cols]

D. Loop condition should be left < right

Solution

1. Step 1: Check initialization of right pointer

   Since indices go from 0 to rows*cols - 1, right should be rows*cols - 1, not rows*cols.

2. Step 2: Understand impact of wrong right value

   Setting right to rows*cols causes out-of-bounds access when mid equals rows*cols, leading to runtime error.

3. Final Answer:

   right should be initialized to rows*cols - 1 -> Option A

4. Quick Check:

   Right boundary off by one causes error = right should be initialized to rows*cols - 1 [OK]

Hint: Right index = total elements - 1, not total elements [OK]

Common Mistakes:

• Using rows*cols instead of rows*cols - 1 for right
• Swapping row and column indices
• Incorrect loop condition causing infinite loop

5. Given a 2D matrix where each row is sorted in ascending order but the first integer of each row may be smaller than the last integer of the previous row, which approach is best to search for a target value efficiently?

hard

A. Perform binary search on each row separately

B. Start from top-right corner and move left/down based on comparison

C. Use binary search treating the matrix as a sorted 1D array

D. Flatten the matrix into a 1D slice and sort it before searching

Solution

1. Step 1: Analyze matrix properties

   Since rows are sorted but rows are not strictly increasing overall, the matrix is not globally sorted.

2. Step 2: Choose search strategy

   Binary search on entire matrix won't work. Binary search on each row is O(m log n). Starting from top-right corner and moving left/down is O(m+n) and efficient.

3. Final Answer:

   Start from top-right corner and move left/down based on comparison -> Option B

4. Quick Check:

   Partial row sorting + top-right search = Start from top-right corner and move left/down based on comparison [OK]

Hint: Top-right corner search works when rows sorted but matrix not fully sorted [OK]

Common Mistakes:

• Assuming full matrix is sorted for binary search
• Flattening and sorting wastes time
• Ignoring matrix properties leads to wrong approach