R – the O time in determining if a value is in a sorted array

arraysbig-obinary-searchcomplexity-theory

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 ...]

Best Solution

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).

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