Search in 2D Matrix in DSA Javascript - 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 values

Create a variable called target with the number to search

Use nested for loops with variables i and j to search the matrix

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

💡 Why This Matters

🌍 Real World

Searching in a 2D matrix is like looking for a seat in a theater or a book in a shelf arranged in rows and columns.

💼 Career

Understanding how to search in 2D data structures is important for jobs involving data analysis, game development, and software engineering.

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 Javascript

// Your code here

Hint

Use square brackets to create arrays inside an array.

2

Set the target number to search

Create a variable called target and set it to 9

DSA Javascript

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

Hint

Use const to create the variable and assign the number 9.

3

Search for the target in the matrix

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

DSA Javascript

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

Hint

Use two loops: outer for rows (i), inner for columns (j). Use break to stop loops when found.

4

Print the search result

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

DSA Javascript

const matrix = [
  [1, 3, 5],
  [7, 9, 11],
  [13, 15, 17]
];
const target = 9;
let found = false;
for (let i = 0; i < matrix.length; i++) {
  for (let j = 0; j < matrix[i].length; j++) {
    if (matrix[i][j] === 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 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