For an array, I'm using a left pointer (pointing to 0) and a right pointer (pointing to end).

For every iteration, if my search element is not found, I increment left and decrement right. This iteration continues until left <= right. Find the code snippet below

left = 0
right = length - 1

while left <= right:
    // search logic
    left += 1
    right -= 1

What will be worst the time complexity of this algorithm?

Also, is this a type of linear search?

  • $\begingroup$ what is the cost of "search logic"? $\endgroup$ – tstanisl Apr 27 at 21:12

$O(n)$, since you still will go through the entire array.

And yes, this is a linear search since $O(n)$ is linear.

  • $\begingroup$ The loop will run only n/2 times right? So wondering if this will be O(n) still. $\endgroup$ – Shylajhaa Apr 27 at 19:01
  • 1
    $\begingroup$ of course, constants do not make a difference in $O$ notation $\endgroup$ – nir shahar Apr 27 at 19:01

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