Search in 2D Matrix in DSA Typescript - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to return false if the matrix is empty.

DSA Typescript

if (matrix.length === [1]) return false;

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

A1

B0

Cnull

D-1

Attempts:

3 left

💡 Hint

Common Mistakes

Checking matrix[0].length instead of matrix.length

Returning true instead of false when empty

If the matrix has zero rows, it means it's empty, so return false.

2fill in blank

medium

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

DSA Typescript

const cols = matrix[0].[1];

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Alength

Bsize

Ccount

Dwidth

Attempts:

3 left

💡 Hint

Common Mistakes

Using matrix.length instead of matrix[0].length

Using a non-existent property like size

The length property gives the number of columns in the first row.

3fill in blank

hard

Fix the error in the while loop condition to search correctly.

DSA Typescript

while (row < rows && col >= [1]) {

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

A-1

Bcols

C0

Drows

Attempts:

3 left

💡 Hint

Common Mistakes

Using cols instead of 0, which causes out-of-bound errors

Using -1 which is invalid index

Column index should be greater or equal to 0 to stay inside matrix bounds.

4fill in blank

hard

Fill both blanks to move the search correctly in the matrix.

DSA Typescript

if (matrix[row][col] === target) return true;
else if (matrix[row][col] < target) row[1];
else col[2];

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A++

B--

C+=

D-=

Attempts:

3 left

💡 Hint

Common Mistakes

Using -- for row increment or ++ for column decrement

Using += or -= which are valid but not the simplest here

If current value is less than target, move down (row++). Else move left (col--).

5fill in blank

hard

Fill all three blanks to complete the function that searches the matrix.

DSA Typescript

function searchMatrix(matrix: number[][], target: number): boolean {
  if (matrix.length === 0) return false;
  const rows = matrix.length;
  const cols = matrix[0].[1];
  let row = 0;
  let col = cols [2] 1;
  while (row < rows && col >= [3]) {
    if (matrix[row][col] === target) return true;
    else if (matrix[row][col] < target) row++;
    else col--;
  }
  return false;
}

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

Alength

B-

C0

D+

Attempts:

3 left

💡 Hint

Common Mistakes

Using cols instead of cols - 1 for starting column

Using 1 or -1 incorrectly in loop condition

Using + instead of - for col initialization

cols is matrix[0].length; col starts at cols - 1; col must be >= 0 in loop.

Practice

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

easy

A. Start from the bottom-left corner and move right or up based on comparison

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

C. Search each element row by row until the target is found

D. Use binary search on each row independently

Solution

1. Step 1: Understand matrix properties

   Since each row and column is sorted ascending, the top-right element is the largest in its row and smallest in its column.

2. Step 2: Use top-right corner to decide movement

   If the target is less than the current element, move left; if greater, move down. This efficiently narrows search space.

3. Final Answer:

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

4. Quick Check:

   Top-right corner search = Start from the top-right corner and move left or down based on comparison [OK]

Hint: Start at top-right, move left if bigger, down if smaller [OK]

Common Mistakes:

• Starting from bottom-left without adjusting moves correctly
• Searching row by row wastes time
• Using binary search on rows ignores column sorting

2. Which of the following TypeScript code snippets correctly initializes a 2D matrix of numbers?

easy

A. const matrix: number[][] = [[1,2,3],[4,5],[6,7,8,9]];

B. const matrix: number[] = [1,2,3,4,5,6,7,8,9];

C. const matrix: number[][] = [[1,2,3],[4,5,6],[7,8,9]];

D. const matrix: number[][] = [1,2,3,4,5,6,7,8,9];

Solution

1. Step 1: Check matrix type declaration

   TypeScript 2D array is declared as number[][]. const matrix: number[][] = [[1,2,3],[4,5,6],[7,8,9]];.

2. Step 2: Validate matrix structure

   const matrix: number[][] = [[1,2,3],[4,5,6],[7,8,9]]; has uniform rows with 3 elements each, valid for a matrix. const matrix: number[][] = [[1,2,3],[4,5],[6,7,8,9]]; has rows of different lengths, which is allowed but less common for matrices.

3. Final Answer:

   const matrix: number[][] = [[1,2,3],[4,5,6],[7,8,9]]; -> Option C

4. Quick Check:

   Proper 2D array syntax = const matrix: number[][] = [[1,2,3],[4,5,6],[7,8,9]]; [OK]

Hint: Use number[][] with nested arrays for 2D matrix [OK]

Common Mistakes:

• Using single array for 2D matrix
• Incorrect type declaration like number[] for 2D
• Mixing array elements without nested arrays

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

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

function searchMatrix(matrix: number[][], target: number): boolean {
  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, 5));

medium

A. Runtime error

B. false

C. undefined

D. true

Solution

1. Step 1: Start search from top-right corner

   Initial position is row=0, col=3 (value 11). Since 11 > 5, move left to col=2 (value 7).

2. Step 2: Compare and move accordingly

   7 > 5, move left to col=1 (value 4). 4 < 5, move down to row=1, col=1 (value 5). Found target 5, return true.

3. Final Answer:

   true -> Option D

4. Quick Check:

   Search finds 5 = true [OK]

Hint: Start top-right, move left if bigger, down if smaller [OK]

Common Mistakes:

• Not updating row and col correctly
• Returning false too early
• Confusing row and column indices

4. Identify the bug in this TypeScript function that searches a target in a sorted 2D matrix:

function searchMatrix(matrix: number[][], target: number): boolean {
  let row = 0;
  let col = matrix.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;
}

medium

A. col should be initialized to matrix[0].length - 1, not matrix.length - 1

B. row should start from matrix.length - 1

C. The while loop condition should use col < matrix[0].length

D. The function should return true at the end instead of false

Solution

1. Step 1: Check initialization of col variable

   col is set to matrix.length - 1, which is number of rows minus one, but columns count is matrix[0].length.

2. Step 2: Correct col initialization

   col should be matrix[0].length - 1 to point to the last column index in the first row.

3. Final Answer:

   col should be initialized to matrix[0].length - 1, not matrix.length - 1 -> Option A

4. Quick Check:

   col init must match columns count = col should be initialized to matrix[0].length - 1, not matrix.length - 1 [OK]

Hint: col index = last column, use matrix[0].length - 1 [OK]

Common Mistakes:

• Using rows count for column index
• Wrong loop conditions causing out-of-bounds
• Returning true instead of false when not found

5. Given a 2D matrix where each row and column is sorted ascending, how would you modify the search function to return the position [row, col] of the target if found, or [-1, -1] if not found? Choose the correct code snippet.

function searchMatrix(matrix: number[][], target: number): number[] {
  let row = 0;
  let col = matrix[0].length - 1;
  while (row < matrix.length && col >= 0) {
    if (matrix[row][col] === target) {
      // What to return here?
    } else if (matrix[row][col] > target) {
      col--;
    } else {
      row++;
    }
  }
  // What to return if not found?
}

hard

A. Return [row, col] inside if, and [-1, -1] after loop

B. Return [col, row] inside if, and [0, 0] after loop

C. Return true inside if, and false after loop

D. Return matrix[row][col] inside if, and null after loop

Solution

1. Step 1: Return position when target found

   Inside the if condition, return the current indices as [row, col] to indicate target position.

2. Step 2: Return [-1, -1] if target not found

   After the loop ends without finding target, return [-1, -1] to indicate absence.

3. Final Answer:

   Return [row, col] inside if, and [-1, -1] after loop -> Option A

4. Quick Check:

   Return position or [-1, -1] if missing = Return [row, col] inside if, and [-1, -1] after loop [OK]

Hint: Return indices when found, [-1, -1] if not found [OK]

Common Mistakes:

• Swapping row and col in return
• Returning boolean instead of position
• Returning wrong default value when not found