Search in 2D Matrix in DSA Rust - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to check if the matrix is empty before searching.

DSA Rust

if matrix.is_empty() || matrix[0].is_empty() {
    [1];
}

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

Areturn true

Bbreak

Ccontinue

Dreturn false

Attempts:

3 left

💡 Hint

Common Mistakes

Returning true when matrix is empty

Not checking if matrix[0] is empty

If the matrix or its first row is empty, the target cannot be found, so return false.

2fill in blank

medium

Complete the code to calculate the middle index in binary search.

DSA Rust

let mid = left + (right - left) [1] 2;

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

A+

B/

C*

D-

Attempts:

3 left

💡 Hint

Common Mistakes

Using multiplication instead of division

Using addition instead of division

To find the middle index, we add left to half the distance between right and left using division by 2.

3fill in blank

hard

Fix the error in the comparison to move the search window.

DSA Rust

if matrix[row][col] [1] target {
    left = mid + 1;
} else {
    right = mid - 1;
}

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

A<

B!=

C>

D==

Attempts:

3 left

💡 Hint

Common Mistakes

Using '>' instead of '<'

Using '==' in condition instead of comparison

If the current element is less than the target, move the left boundary up to search higher values.

4fill in blank

hard

Fill both blanks to convert mid index to 2D row and column.

DSA Rust

let row = mid [1] n_cols;
let col = mid [2] n_cols;

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

A/

B*

C%

D+

Attempts:

3 left

💡 Hint

Common Mistakes

Swapping division and modulo

Using multiplication instead of division

Row is mid divided by number of columns; column is mid modulo number of columns.

5fill in blank

hard

Fill all three blanks to complete the binary search loop condition and update pointers.

DSA Rust

while [1] <= [2] {
    let mid = [3] + (right - left) / 2;
    // rest of the code
}

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

Aleft

Bright

Dmid

Attempts:

3 left

💡 Hint

Common Mistakes

Using mid in loop condition

Swapping left and right in condition

The loop runs while left is less than or equal to right; mid is calculated from left and right.

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 binary search treating the matrix as a flattened sorted list

B. Search each row linearly from left to right

C. Search each column linearly from top to bottom

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

Solution

1. Step 1: Understand matrix sorting property

   Each row is sorted and the first element of a row 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 flattened matrix

   We can treat the matrix as a single sorted list and use binary search to find the target efficiently.

3. Final Answer:

   Use binary search treating the matrix as a flattened sorted list -> Option A

4. Quick Check:

   Matrix sorted row-wise + binary search = Use binary search treating the matrix as a flattened sorted list [OK]

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

Common Mistakes:

• Searching each row or column linearly wastes time
• Using hash map wastes extra space
• Ignoring the matrix's sorted property

2. Which Rust code snippet correctly calculates the middle index for binary search on a 2D matrix treated as a 1D array with total length `len`?

easy

A. let mid = (start + end) / 2;

B. let mid = start + (end - start) / 2;

C. let mid = (start - end) / 2;

D. let mid = (start * end) / 2;

Solution

1. Step 1: Avoid overflow in mid calculation

   Using `(start + end) / 2` can overflow if start and end are large. The safer way is `start + (end - start) / 2`.

2. Step 2: Check other options

   let mid = (start - end) / 2; subtracts incorrectly, and D multiplies which is wrong for mid calculation.

3. Final Answer:

   let mid = start + (end - start) / 2; -> Option B

4. Quick Check:

   Safe mid calculation = let mid = start + (end - start) / 2; [OK]

Hint: Use start + (end - start) / 2 to avoid overflow [OK]

Common Mistakes:

• Using (start + end) / 2 causing overflow
• Using subtraction incorrectly for mid
• Multiplying indices instead of averaging

3. Given the following Rust code to search for target 5 in a 2D matrix, what will be the printed output?

let matrix = vec![
    vec![1, 3, 5, 7],
    vec![10, 11, 16, 20],
    vec![23, 30, 34, 60]
];
let target = 5;
let rows = matrix.len();
let cols = matrix[0].len();
let mut start = 0;
let mut end = rows * cols - 1;
let mut found = false;
while start <= end {
    let mid = start + (end - start) / 2;
    let r = mid / cols;
    let c = mid % cols;
    if matrix[r][c] == target {
        found = true;
        break;
    } else if matrix[r][c] < target {
        start = mid + 1;
    } else {
        end = mid - 1;
    }
}
println!("{}", found);

medium

A. Runtime panic

B. false

C. Compilation error

D. true

Solution

1. Step 1: Understand binary search on matrix

   The code treats the matrix as a sorted list and uses binary search to find target 5.

2. Step 2: Trace the search for target 5

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

3. Final Answer:

   true -> Option D

4. Quick Check:

   Target 5 found = true [OK]

Hint: Map mid to row and col, then compare target [OK]

Common Mistakes:

• Confusing row and column calculation
• Off-by-one errors in indices
• Not updating start or end correctly

4. Identify the bug in this Rust code snippet that searches for a target in a 2D matrix and suggest the fix:

let matrix = vec![
    vec![1, 4, 7, 11],
    vec![15, 20, 25, 30],
    vec![35, 40, 45, 50]
];
let target = 20;
let rows = matrix.len();
let cols = matrix[0].len();
let mut start = 0;
let mut end = rows * cols;
while start <= end {
    let mid = start + (end - start) / 2;
    let r = mid / cols;
    let c = mid % cols;
    if matrix[r][c] == target {
        println!("Found");
        break;
    } else if matrix[r][c] < target {
        start = mid + 1;
    } else {
        end = mid - 1;
    }
}

medium

A. end should be rows * cols - 1 instead of rows * cols

B. start should be 1 instead of 0

C. Use while start < end instead of while start <= end

D. Calculate r and c as mid % cols and mid / cols respectively

Solution

1. Step 1: Check the range of end index

   Matrix has rows * cols elements, so last index is rows * cols - 1, not rows * cols.

2. Step 2: Fix end initialization

   Set end = rows * cols - 1 to avoid out-of-bound access when mid equals rows * cols.

3. Final Answer:

   end should be rows * cols - 1 instead of rows * cols -> Option A

4. Quick Check:

   Index range off-by-one fixed = end should be rows * cols - 1 instead of rows * cols [OK]

Hint: End index = total elements - 1 to avoid overflow [OK]

Common Mistakes:

• Setting end to rows * cols causing panic
• Swapping row and column calculations
• Incorrect loop condition

5. You have 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 correctly searches for a target value efficiently?

hard

A. Use binary search on each row separately until target is found or all rows checked

B. Treat the matrix as a flattened sorted list and apply binary search once

C. Start from top-right corner and move left or down based on comparison

D. Use a hash set to store all elements and check membership

Solution

1. Step 1: Understand matrix property

   Rows are sorted but no guarantee that first element of a row is greater than last of previous row, so flattened binary search won't work.

2. Step 2: Use stair-step search from top-right

   Starting at top-right, move left if current > target, move down if current < target, efficiently narrowing search.

3. Final Answer:

   Start from top-right corner and move left or down based on comparison -> Option C

4. Quick Check:

   Stair-step search fits matrix property = Start from top-right corner and move left or down based on comparison [OK]

Hint: Start top-right, move left/down to find target fast [OK]

Common Mistakes:

• Assuming full matrix sorted for binary search
• Searching each row fully wasting time
• Using extra space unnecessarily