Search in 2D Matrix in DSA Javascript - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of search in sorted 2D matrix

What is the output of the following JavaScript code that searches for the number 5 in a sorted 2D matrix?

DSA Javascript

const matrix = [
  [1, 3, 5],
  [7, 9, 11],
  [13, 15, 17]
];

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 [row, col];
    } else if (matrix[row][col] > target) {
      col--;
    } else {
      row++;
    }
  }
  return [-1, -1];
}

console.log(searchMatrix(matrix, 5));

A [0, 2]

B [2, 0]

C [-1, -1]

D [1, 0]

💡 Hint

Trace the search starting from the top-right corner of the matrix.

The search starts at the top-right element (0,2) which is 5. Since it matches the target, the function returns [0, 2].

❓ Predict Output

intermediate

Result of searching for a missing element

What does the following code print when searching for 8 in the given matrix?

DSA Javascript

const matrix = [
  [2, 4, 6],
  [8, 10, 12],
  [14, 16, 18]
];

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 [row, col];
    } else if (matrix[row][col] > target) {
      col--;
    } else {
      row++;
    }
  }
  return [-1, -1];
}

console.log(searchMatrix(matrix, 8));

A [-1, -1]

B [0, 2]

C [1, 0]

D [2, 1]

💡 Hint

Check the matrix elements row by row and column by column as per the search logic.

The search finds 8 at position (1, 0) in the matrix and returns [1, 0].

❓ Predict Output

advanced

Output of search in matrix with duplicates

What is the output of this code searching for 3 in a matrix containing duplicates?

DSA Javascript

const matrix = [
  [1, 2, 3],
  [3, 4, 5],
  [5, 6, 7]
];

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 [row, col];
    } else if (matrix[row][col] > target) {
      col--;
    } else {
      row++;
    }
  }
  return [-1, -1];
}

console.log(searchMatrix(matrix, 3));

A [0, 2]

B [1, 0]

C [-1, -1]

D [2, 0]

💡 Hint

The search stops at the first found occurrence from top-right.

The search starts at (0,2) which is 3, so it returns [0, 2] immediately.

🧠 Conceptual

advanced

Why start search from top-right corner?

Why is the top-right corner chosen as the starting point for searching a target in a sorted 2D matrix where each row and column is sorted ascending?

A Because starting from top-right guarantees the target is found in the first step.

B Because top-right is always the smallest element in the matrix.

C Because top-right corner is the middle element of the matrix.

D Because from top-right, moving left decreases values and moving down increases values, allowing efficient elimination.

💡 Hint

Think about how values change when moving left or down from the top-right.

Starting at top-right allows moving left to smaller values and down to larger values, efficiently narrowing the search space.

🔧 Debug

expert

Identify the error in this 2D matrix search code

What error will this code produce when searching for 10 in the matrix? Identify the exact error type.

DSA Javascript

const matrix = [
  [1, 4, 7],
  [8, 10, 12],
  [14, 16, 18]
];

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

console.log(searchMatrix(matrix, 10));

A RangeError: Maximum call stack size exceeded

B TypeError: Cannot read property 'undefined' of undefined

C ReferenceError: col is not defined

D TypeError: Cannot read property 'length' of undefined

💡 Hint

Check the initial value of 'col' and how it is used as an index.

Setting col = matrix[0].length sets col to 3, which is out of bounds for indices 0-2. Accessing matrix[row][3] is undefined, causing a TypeError when trying to compare or access properties.

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