Check if array is sorted | Explained using Python Code

Learn a one-pass way to check if array is sorted in non-decreasing order. Intuition, commented Python code, dry run, and clear Big-O analysis included.

So let’s get started with the [Problem Link].

1. What does the problem ask?

Given an integer array arr, decide if it is sorted in non-decreasing order (each element is no larger than the next).
Return True if the whole array obeys this rule, otherwise return False.

2. Quick examples

3. Intuition & approach

1. Sliding window of size 2 – look at every neighbour pair arr[i] and arr[i + 1].
2. The array is sorted only if all pairs satisfy arr[i] ≤ arr[i+1].
3. The first pair that violates this instantly proves the array is not sorted, so we can stop early.

Think of walking from left to right; the moment you step down a hill, you know the landscape is not purely uphill.

4. Code

class Solution:
    def arraySortedOrNot(self, arr) -> bool:
        n = len(arr)
        # Scan pairs (i, i+1)
        for i in range(n - 1):           # last index is n-2
            if arr[i] > arr[i + 1]:      # found a drop
                return False
        return True                      # never dropped → sorted

5. Step-by-step explanation

1. Length check – no special case needed; the loop runs 0 times when n ≤ 1, returning True.
2. Loop from index 0 to n − 2.
   • Compare current element with the next.
   • If ever arr[i] > arr[i+1], immediately return False (unsorted).
3. If the loop finishes without early exit, every pair was in order → return True.

6. Dry run on [3, 5, 5, 2, 6]

The function returns False after meeting the first violation.

7. Complexity

8. Conclusion

To check if an array is sorted, you only need to compare each element with its neighbour once. This linear scan is:

• Simple – five short lines of code.
• Efficient – runs in O(n) time and O(1) space.
• Early-exit capable – stops the moment an out-of-order pair appears.

A perfect example where the straightforward approach is already optimal.