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Given a sorted array of integers, what can be the minimum worst case time complexity to find ceiling of a number x in given array? Ceiling of an element x is the smallest element present in array which is greater than or equal to x. Ceiling is not present if x is greater than the maximum element present in array. Group of answer choicesO(LogN*LogN)O(LogN)O(LogLogn)O(N)

Question

Given a sorted array of integers, what can be the minimum worst case time complexity to find ceiling of a number x in given array? Ceiling of an element x is the smallest element present in array which is greater than or equal to x. Ceiling is not present if x is greater than the maximum element present in array. Group of answer choicesO(LogN*LogN)O(LogN)O(LogLogn)O(N)

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Solution

The minimum worst case time complexity to find the ceiling of a number x in a given sorted array of integers is O(LogN).

Here's why:

  1. Since the array is sorted, we can use a binary search algorithm to find the ceiling of x.

  2. In binary search, we first compare x with the middle element of the array.

  3. If x matches with the middle element, we return the middle element.

  4. If x is greater than the middle element, then x can only lie in the right

This problem has been solved

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