Search in 2D Matrix in DSA Javascript - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to start searching from the top-right corner of the matrix.

DSA Javascript

let row = 0;
let col = matrix[0].length [1] 1;

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

A+

B/

C*

D-

Attempts:

3 left

💡 Hint

Common Mistakes

Using + instead of - causes index out of range.

Using * or / is not correct for index calculation.

We start from the last column index, so we subtract 1 from the length.

2fill in blank

medium

Complete the code to move left when the current element is greater than the target.

DSA Javascript

if (matrix[row][col] > target) {
  col [1] 1;
}

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

A++

B+=

C-=

D--

Attempts:

3 left

💡 Hint

Common Mistakes

Using += increases the column index, moving right instead of left.

Using ++ or -- alone changes by 1 but does not assign.

We decrease the column index by 1 to move left.

3fill in blank

hard

Fix the error in the loop condition to prevent out-of-bound errors.

DSA Javascript

while (row < matrix.length && col [1] 0) {
  // search logic
}

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

A<

B>=

C<=

D>

Attempts:

3 left

💡 Hint

Common Mistakes

Using > 0 skips checking column 0.

Using < or <= causes logic errors in loop.

Column index must be greater than or equal to 0 to stay within bounds and check column 0.

4fill in blank

hard

Fill both blanks to correctly move down or left based on comparison.

DSA Javascript

if (matrix[row][col] < target) {
  row [1] 1;
} else {
  col [2] 1;
}

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

A+=

B-=

C++

D--

Attempts:

3 left

💡 Hint

Common Mistakes

Using -= to move down or += to move left is incorrect.

Using ++ or -- without assignment causes errors.

Increase row to move down, decrease col to move left.

5fill in blank

hard

Fill all three blanks to complete the search function returning true if target found, else false.

DSA Javascript

function searchMatrix(matrix, target) {
  let row = 0;
  let col = matrix[0].length [1] 1;
  while (row < matrix.length && col [2] 0) {
    if (matrix[row][col] === target) {
      return [3];
    } else if (matrix[row][col] < target) {
      row += 1;
    } else {
      col -= 1;
    }
  }
  return false;
}

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

A+

B>=

Ctrue

D-

E<

Ffalse

Attempts:

3 left

💡 Hint

Common Mistakes

Using + instead of - for last column index.

Using < or <= in loop condition causing errors.

Returning false instead of true when found.

Start from last column (length - 1), loop while col >= 0, return true if found.

Practice

1. What is the main idea behind searching a target value in a 2D matrix where each row and column is sorted in ascending order?

easy

A. Use binary search on each row independently

B. Start from bottom-left corner and move right or down based on comparison

C. Search each element one by one from top-left to bottom-right

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

Solution

1. Step 1: Understand matrix properties

   Each row and column is sorted ascending, so top-right element is a good start to decide direction.

2. Step 2: Use top-right corner to guide search

   If target is less than current element, move left; if greater, move down.

3. Final Answer:

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

4. Quick Check:

   Top-right start = efficient search [OK]

Hint: Start at top-right to decide move direction quickly [OK]

Common Mistakes:

• Searching all elements one by one wastes time
• Starting from bottom-left without clear logic
• Using binary search on each row ignores column sorting

2. Which of the following is the correct syntax to check if a target exists in a 2D matrix named matrix in JavaScript?

easy

A. for(let i=0; i

B. for(let i=0; i

C. for(let i=0; i

D. for(let i=0; i

Solution

1. Step 1: Identify correct loop ranges

   Outer loop runs over rows (matrix.length), inner loop over columns (matrix[0].length).

2. Step 2: Check element equality with strict equality

   Use === for exact match and return true if found.

3. Final Answer:

   for(let i=0; i -> Option B

4. Quick Check:

   Correct loop ranges and strict equality [OK]

Hint: Outer loop rows, inner loop columns, use === for comparison [OK]

Common Mistakes:

• Swapping row and column lengths in loops
• Using == instead of === causing type coercion
• Looping inner loop with wrong length causing errors

3. What will be the output of the following code snippet?

const matrix = [
  [1, 4, 7, 11],
  [2, 5, 8, 12],
  [3, 6, 9, 16],
  [10, 13, 14, 17]
];
const target = 5;

function searchMatrix(matrix, target) {
  let row = 0;
  let col = matrix[0].length - 1;
  while(row < matrix.length && col >= 0) {
    if(matrix[row][col] === target) return true;
    else if(matrix[row][col] > target) col--;
    else row++;
  }
  return false;
}

console.log(searchMatrix(matrix, target));

medium

A. Error

B. false

C. true

D. undefined

Solution

1. Step 1: Start from top-right element

   Initial position is matrix[0][3] = 11, compare with target 5.

2. Step 2: Move left since 11 > 5, now at matrix[0][2] = 7

   7 > 5, move left again to matrix[0][1] = 4.

3. Step 3: 4 < 5, move down to matrix[1][1] = 5

   Found target 5, return true.

4. Final Answer:

   true -> Option C

5. Quick Check:

   Top-right search finds 5 = true [OK]

Hint: Move left if bigger, down if smaller from top-right [OK]

Common Mistakes:

• Starting from wrong corner causing wrong moves
• Not updating row or column correctly
• Returning false too early without full search

4. Find the bug in the following code that searches for a target in a 2D matrix where rows and columns are sorted.

function searchMatrix(matrix, target) {
  let row = matrix.length - 1;
  let col = 0;
  while(row >= 0 && col < matrix[0].length) {
    if(matrix[row][col] === target) return true;
    else if(matrix[row][col] < target) row--;
    else col++;
  }
  return false;
}

medium

A. The directions to move row and col are swapped in conditions

B. The initial row and col values are incorrect

C. The while loop condition should use '||' instead of '&&'

D. The function should return false inside the loop

Solution

1. Step 1: Analyze movement logic

   Starting at bottom-left, if current < target, should move right (col++), else move up (row--).

2. Step 2: Check conditions in code

   Code moves row-- when current < target, which is wrong; it should move col++ instead.

3. Final Answer:

   The directions to move row and col are swapped in conditions -> Option A

4. Quick Check:

   Correct move directions fix search [OK]

Hint: From bottom-left, move right if smaller, up if bigger [OK]

Common Mistakes:

• Swapping row and column moves
• Wrong initial position for search
• Incorrect loop condition causing infinite loop

5. Given a 2D matrix where each row and column is sorted ascending, write a function to count how many times a target appears in the matrix efficiently. Which approach is best?

hard

A. Start from top-right and move left/down counting matches (O(m+n))

B. Traverse entire matrix counting matches (O(m*n))

C. Use binary search on each row and sum counts (O(m*log n))

D. Flatten matrix to 1D array and use binary search once (O(m*n))

Solution

1. Step 1: Understand time complexities

   Full traversal is slow (O(m*n)), binary search per row is better but still costly.

2. Step 2: Use top-right start to count efficiently

   Moving left/down lets you skip rows or columns, counting matches in O(m+n) time.

3. Final Answer:

   Start from top-right and move left/down counting matches (O(m+n)) -> Option A

4. Quick Check:

   Top-right traversal counts efficiently [OK]

Hint: Top-right start counts matches in linear time [OK]

Common Mistakes:

• Using full traversal wasting time
• Flattening matrix losing sorted structure
• Binary search per row misses column info