Search in 2D Matrix in DSA Typescript - Build from Scratch

Search in 2D Matrix

📖 Scenario: You are working with a grid-like map that stores numbers in rows and columns. You want to find if a specific number exists in this map.

🎯 Goal: Build a program that searches for a given number in a 2D matrix and tells if it is found or not.

📋 What You'll Learn

Create a 2D array called matrix with exact numbers

Create a variable called target with the number to search

Use nested for loops with variables row and col to check each element

Print "Found" if the target is in the matrix, otherwise print "Not Found"

💡 Why This Matters

🌍 Real World

Searching in 2D grids is common in games, maps, and image processing where data is arranged in rows and columns.

💼 Career

Understanding how to search in 2D arrays is important for software developers working with matrices, tables, or grid-based data.

1

Create the 2D matrix

Create a 2D array called matrix with these exact rows: [1, 3, 5], [7, 9, 11], [13, 15, 17]

DSA Typescript

// Your code here

Hint

Use square brackets to create arrays inside an array.

2

Set the target number

Create a variable called target and set it to 9

DSA Typescript

const matrix: number[][] = [
  [1, 3, 5],
  [7, 9, 11],
  [13, 15, 17]
];
// Your code here

Hint

Use const target: number = 9; to create the variable.

3

Search for the target in the matrix

Use nested for loops with variables row and col to check each element in matrix. Create a boolean variable found set to false before loops. If matrix[row][col] equals target, set found to true and break both loops.

DSA Typescript

const matrix: number[][] = [
  [1, 3, 5],
  [7, 9, 11],
  [13, 15, 17]
];
const target: number = 9;
// Your code here

Hint

Use two for loops, one inside the other, to check each element.

4

Print the search result

Print "Found" if found is true, otherwise print "Not Found"

DSA Typescript

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

let found: boolean = false;

for (let row = 0; row < matrix.length; row++) {
  for (let col = 0; col < matrix[row].length; col++) {
    if (matrix[row][col] === target) {
      found = true;
      break;
    }
  }
  if (found) {
    break;
  }
}
// Your code here

Hint

Use console.log to print the result based on the found variable.

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