## Algorithm for finding the first bigger integer with o-small(n) complexity

I need to write an algorithm, which will find the first integer, which is bigger than x in a sorted array, where integers may repeat. The algorithm should have the complexity of o(n), where o is small. What is the difference between the algorithms with O(n) an o(n) difficulty?

You can using binary search method to find the first biggest integer of `x`

. It would in `O(log(n)) = small_o(n)`

.

**Analysis of Algorithms,** First, suppose that we invent a mathematical function Iaw(N) that computes the This is called big O notation, and we use it to specify the complexity class of an In this example, the O(N) algorithm takes more time for small N. Of course, This is one of the fundamental sorting algorithms, and most of us started learning about sorting using this algorithm. The time complexity of this is O(n ^2) which makes it unusable for a large

Here is a very good and in depth explanation of big-o vs little-o notation: Difference between Big-O and Little-O Notation

The most simple yet efficient algorithm is Binary Search as @OmG mentioned, and here is the link to start: https://en.wikipedia.org/wiki/Binary_search_algorithm

The idea is pretty simple: you just compare the number you search with the middle element of the array, if the middle is less then your number it's obvious that you need to find on the right half, ... otherwise left half. You stop when there is only one element.

**Complexity,** T(n) = O(n2). says that an algorithm has a quadratic time complexity. Definition of "big Oh". For any monotonic functions f(n) So I found this tasks in one book I am practicing from where it says: "Find a divide-and-conquer algorithm for finding square roots for large integers and along this, find its asymptotic time complexity". Results can be in integer range. I am not sure which algorithm should I use, but I started with recursive method where I count the medium between two borders and go on until i find the

As people have pointed out before me, you want to do a binary search. This post includes example code:

In python this is easiest with the bisect function: https://docs.python.org/2/library/bisect.html

from bisect import bisect_right def find_gt(a, x): 'Find leftmost value greater than x' i = bisect_right(a, x) if i != len(a): return a[i] raise ValueError

**Big O Notation and Algorithm Analysis with Python Examples,** We will see how Big-O notation can be used to find algorithm complexity with the Notice that the algorithm simply takes an integer as an argument. the algorithm performs only 2 steps: Finding the square of the first element and printing the For example, one might say that a sorting algorithm has O(n * lg(n)) complexity, where n is the number of items to sort. Big O notation is used in Computer Science to describe the performance or

**Analysis of algorithms,** Big-Ο is used as a tight upper-bound on the growth of an algorithm's effort (this effort is Definition : Let f(n) and g(n) be functions that map positive integers to positive real numbers. a little more imagination, but we'll be able to find an n0. for your first answer the complexity √i-1 is for finding if number is prime or not. but to find all number till n i will take n√n as n times the loop will go n times and each time it will take √i-1 to check wheather that number is prime or not so overall will be n√n. i hope i answered your doubt – Kumar Saurabh Aug 28 '15 at 12:14

**Analysis of Algorithms,** We can safely say that the time complexity of Insertion sort is O(n^2). Note that O(n^2) also covers linear time. The Big-O A Comparison-Free Sorting Algorithm Sorting Binary Numbers in Hardware - A Novel Algorithm and its Implementation. Laser Domino Sorting Algorithm - a thought experiment by me based on Counting Sort with an intention to achieve O(n) time complexity over Counting Sort's O(n + k).

**Binary search algorithm,** Worst-case space complexity · O(1). In computer science, binary search, also known as half-interval search, is Big O notation, and log {\displaystyle \log } \log is the However, the array must be sorted first to be able to apply binary search. It is possible to find the k smallest of n elements in O(n) time (by which I mean true O(n) time, not O(n + some function of k)). Refer to the Wikipedia article "Selection algorithm" , especially the subsections on "unordered partial sorting" and "median selection as pivot strategy", and also to the article "Median of medians" for the essential piece that makes this O(n) .

##### Comments

- o(n) means that lim f(n) /n=0, so you can have sqrt(n) algorithm or log(n) they both are o(n)
- The binary search algorithm is for searching the exact number, as far as I understand. Do I understand right, that after finding the exact x element in the array, I need to check every next element in this array (iterate) to find the next bigger one? Wouldn't the whole complexity be worse than log(n) in this case?
- you do not find the exact element, you find the first bigger element than your target.
- the next element is obviously the next bigger, the next next is 3rd bigger and so on.
- I think not necessarily, if the elements may repeat, or miss.
- well yes if the elements repeat then yes.