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

Time Complexity: Search in 2D Matrix

O(m x n)

Understanding Time Complexity

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

How does the search time grow when the matrix gets bigger?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

function searchMatrix(matrix: number[][], target: number): boolean {
  for (let row = 0; row < matrix.length; row++) {
    for (let col = 0; col < matrix[row].length; col++) {
      if (matrix[row][col] === target) {
        return true;
      }
    }
  }
  return false;
}
    

This code checks every element in the matrix until it finds the target or finishes all elements.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Checking each element in the matrix.
• How many times: For each row, it checks every column element, so total checks = number of rows x number of columns.

How Execution Grows With Input

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

Input Size (rows x cols)	Approx. Operations
10 x 10 = 100	About 100 checks
100 x 100 = 10,000	About 10,000 checks
1000 x 1000 = 1,000,000	About 1,000,000 checks

Pattern observation: The operations grow proportionally to the total number of elements in the matrix.

Final Time Complexity

Time Complexity: O(m x n)

This means the time to search grows directly with the number of rows (m) times the number of columns (n).

Common Mistake

[X] Wrong: "The search will always be fast because we stop when we find the target."

[OK] Correct: In the worst case, the target might be at the very end or not present, so the code checks every element, making the time grow with the whole matrix size.

Interview Connect

Understanding this helps you explain how searching in a grid works and why sometimes it takes longer when the data grows.

Self-Check

"What if the matrix rows and columns were sorted, and we used a smarter search? How would the time complexity change?"

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