Sorted Array Check

Write a function that returns True if an array is sorted in ascending order, and False otherwise. For example, [1, 2, 3, 4, 5] returns True, while [1, 2, 3, 5, 4] returns False.

A sorted array has the property that every element is less than or equal to the next one. We only need to find a SINGLE violation to prove the array is unsorted. This is the "Early Exit" pattern — we return False immediately upon finding the first violation, without needing to check the rest of the array.

1. Iterate through the array from index 0 to len(arr) - 2. 2. At each position, compare arr[i] with arr[i+1]. 3. If arr[i] > arr[i+1], the array is not sorted — return False immediately. 4. If the loop completes without finding any violations, return True.

- Using range(len(arr)) instead of range(len(arr) - 1), causing an index-out-of-bounds error when accessing arr[i+1] - Not handling edge cases: an empty array [] or single-element array [5] should both return True - Checking for strict less-than (<) instead of less-than-or-equal (<=), incorrectly marking [1, 1, 2] as unsorted

The early exit pattern is a powerful optimization strategy. In the best case, you find the violation at the very first pair and return in O(1) time. This "fail fast" approach appears frequently in validation problems and is worth internalizing.