Search in 2D Matrix in DSA Javascript - Time & Space Complexity

Time Complexity: Search in 2D Matrix

O(n * m)

Understanding Time Complexity

We want to understand how long it takes to find a number in a 2D matrix.

Specifically, how the search time grows as the matrix gets bigger.

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

function searchMatrix(matrix, target) {
  for (let i = 0; i < matrix.length; i++) {
    for (let j = 0; j < matrix[i].length; j++) {
      if (matrix[i][j] === target) {
        return true;
      }
    }
  }
  return false;
}
    

This code checks each element in the matrix one by one to find the target number.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Checking each element in the matrix.
• How many times: The outer loop runs for each row, and the inner loop runs for each column in that row.

How Execution Grows With Input

As the matrix size grows, the number of checks grows with the total number of elements.

Input Size (n x m)	Approx. Operations
10 x 10	100 checks
100 x 100	10,000 checks
1000 x 1000	1,000,000 checks

Pattern observation: The operations grow roughly by the total number of elements, so doubling rows and columns quadruples work.

Final Time Complexity

Time Complexity: O(n * m)

This means the time to search grows proportionally to the total number of elements in the matrix.

Common Mistake

[X] Wrong: "Since the matrix is sorted, we can find the target in constant time."

[OK] Correct: Even if sorted, this code checks every element one by one, so it still takes time proportional to all elements in the worst case.

Interview Connect

Understanding how nested loops affect time helps you explain and improve search methods in real problems.

Self-Check

"What if we used binary search on each row instead of checking every element? How would the time complexity change?"

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