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

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).