Search in 2D Matrix in DSA Typescript - Time & Space 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?
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 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.
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.
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).
[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.
Understanding this helps you explain how searching in a grid works and why sometimes it takes longer when the data grows.
"What if the matrix rows and columns were sorted, and we used a smarter search? How would the time complexity change?"