What is the O time in determining if a value is in a sorted array?

Question

I have a sorted array of 5000 integers. How fast can I tell if a random integer is a member of the array? An answer in general, C and Ruby would be nice.

The array values are of the form

c * c + 1

where c can be any integer from 1 to 5000.

For example:

[2, 5, 10, 17, 26, 37, 50 ...]

Answer 1

log(n) for binary search on c

Answer 2

I would say it's O(const)! :)

Given a random number r, it's trivial to check whether it's a number that could be represented in the form (n*n+1). Just check whether the sqrt(r-1) is an integer or not!

(Well, it might be a little more complicated than that since your programming language can introduce some complexity into dealing with integers vs floating point numbers, but still: you do not need to search the array at all: just check whether the number is in this particular form.)

Answer 3

Binary search, as others have mentioned, is O(log2N), and can be coded either recursively:

   BinarySearch(A[0..N-1], value, low, high) {
       if (high < low)
           return -1 // not found
       mid = (low + high) / 2
       if (A[mid] > value)
           return BinarySearch(A, value, low, mid-1)
       else if (A[mid] < value)
           return BinarySearch(A, value, mid+1, high)
       else
           return mid // found
   }

or iteratively:

   BinarySearch(A[0..N-1], value) {
       low = 0
       high = N - 1
       while (low <= high) {
           mid = (low + high) / 2
           if (A[mid] > value)
               high = mid - 1
           else if (A[mid] < value)
               low = mid + 1
           else
               return mid // found
       }
       return -1 // not found
   }

However, if you're looking for the fastest possible way, you can set up a look up table based on the sqrt(N-1) of your numbers. With just 5,000 words of memory you can achieve O(1) lookups this way.

Explanation:

Since all your numbers are of the form N^2 + 1 for an integer N from 1 to N, you can create a table of N elements. The element at position i will specify if i^2 + 1 is in your array or not. The table can be implemented with a simple array of length N. It will take O(N) to build, and N words of space. But once you have the table, all lookups are O(1).

Example:

Here's sample code in Python, which reads like pseudocode, as always :-)

import math

N = 5000
ar = [17, 26, 37, 50, 10001, 40001]

lookup_table = [0] * N

for val in ar:
    idx = int(math.sqrt(val - 1))
    lookup_table[idx] = 1

def val_exists(val):
    return lookup_table[int(math.sqrt(val - 1))] == 1

print val_exists(37)
print val_exists(65)
print val_exists(40001)
print val_exists(90001)

Building the table takes up O(N) at most, and lookups are O(1).

Answer 4

Technically, the complexity of finding an element in a fixed-size array is constant, since log₂ 5000 isn't going to change.

Answer 5

Binary Search is O(log n)

WikiPedia

Answer 6

O(log n) if the array has n elements

Answer 7

Just to expand on that: it's lg n tests, that is log₂ n. That makes it O(log n). Why? because each trial of a binary search divides the array in half; thus it takes lg n trials.

Answer 8

Using a binary search, it's Log(N) search time.

bool ContainsBinarySearch(int[] array, int value) {
  return Array.BinarySearch(arrray, value) >= 0;
}

Answer 9

In Perl:

I would load the values into a static hash and then it would be O(1).

Build the Lookup hash

lookup_hash{$_} = 1 foreach (@original_array);

Lookup syntax

($lookup_hash{$lookup_value}) && print "Found it in O(1) - no loop here\n";