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: $a_i \leq a_{i+1}$ for all $i$. We only need to find a single violation to prove the array is unsorted. This is the "early exit" pattern — return False immediately upon finding the first violation.

1. Iterate through indices $0$ to $N-2$ (using range(len(arr) - 1)).
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 strict less-than (<) instead of less-than-or-equal (<=), incorrectly marking [1, 1, 2] as unsorted

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