算法 - 未排序数组中删除的时间复杂度

Question

Suppose there is a unsorted array A, and it contains an element x (x is the pointer of the element), and every element has a satellite variable k. So, we can get the following time complexity (for worst cases):

If we want to Search for a particular K, then it costs O(n).

if we want to Insert an element, then it costs O(1), because A just adds the element to the end.

What if we know x, then Delete it from the array A?

We have to Search for xk first and get the index of x, then Delete x via its index in A, right?

So for Delete , it costs O(n) too, right?

thanks

Answer 1

Finding the element with a given value is linear.

Since the array isn't sorted anyway, you can do the deletion itself in constant time. First swap the element you want to delete to the end of the array, then reduce the array size by one element.

Answer 2

Yes, that's right. Also, if it's an array, deleting alone will take O(n) time because after you delete the element, you'll need to shift all the elements to the right of that element one place to the left. So, even if you know x (for example, you will only delete the first element), it will take O(n) time.

Answer 3

排序数组中删除操作的最坏情况时间复杂度为O（n） ，如果数组未排序，并且提到删除操作后不应更改数组的顺序，则时间复杂度将与O（n）相同）否则它将是O（1） 。

Answer 4

Unsorted Array

Ex:

Items Value     [3,   5,  1,  7,  4]
Items Address   [&1, &2, &3, &4, &5]
Deleting - Value 5

1. Deletion - Order to be preserved - O(n + n) - O(2n) ~> O(n)

i) O(n) - Finding the position of that element (Index = 1 for value 5)

ii) O(n) - After deleting that element, the rest of the items (1, 7, 4) needs to be reshuffled to hold the previous item's address. Like

Items Value     [3,   1,  7,  4]
Items Address   [&1, &2, &3, &4]

2. Deletion - Without preserving the Order - O(n + 1 + 1) - O(2 + n) ~> O(n)

i) O(n) - Finding the position of that element (Index = 1 for value 5)

ii) O(1) - Swape with the last element of the array.

Items Value     [3,   4,  1,  7,  5]
Items Address   [&1, &2, &3, &4, &5]

iii) O(1) - Delete the last element of the array.

Items Value     [3,   4,  1,  7]
Items Address   [&1, &2, &3, &4]

Answer 5

Yes. It takes O(n) time to find the element you want to delete. Then in order to delete it, you must shift all elements to the right of it one space to the left. This is also O(n) so the total complexity is linear.

Also, if you're talking about statically allocated arrays, insert takes O(n) as well. You have to resize the array in order to accommodate the extra element. There are ways to amortize this running time to O(1) though.