最佳情况时间复杂度检查数组是否未排序

Question

I am trying to find out the best case time complexity while checking if the given array is unsorted.

I think, this is the fastest way to check if array is sorted or unsorted and the time complexity should be O(n) for this one.

for (i = 0; i < a.length-1; i++) { 
  if (a[i] < a[i + 1]) {
    return true;
  } else {
    return false;
  }
}

Or am I wrong?

Answer 1

Yes this takes O(n) , and it is as fast as it gets .

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Moreover, you need to change this:

for (i = 0; i < a.length; i++)

to this:

for (i = 0; i < a.length - 1; i++)

since you access a[i + 1 and you do not want to go out of bounds. Furthermore, your return statements should be swapped.

Answer 2

The average and worst-case complexity for your algorithm to check if an array is unsorted is O(N) . But the best-case complexity is O(1) .

The best case occurs when the first two elements of the array are out of order. This is detected in the first loop iteration, and takes a constant time irrespective of N .