The binary search is an excellent algorithm to find an item in a sorted list, being O(Log n) it is very efficient. This is an implementation in Python:

def binary_search(items, required_item):
    low = 0
    high = len(items) - 1

    while low <= high:
        index = (low + high) // 2
        item = items[index]

        if item == required_item:
            return index

        if item > required_item:
            high = index - 1
        else:
            low = index + 1

Essentially it works like this:

• start with low set to the first item, high to the last, index to the middle
• if the item at the index is a match, we are done, return the index
• if the item is greater, halve the list by ignoring the item and all items after it
• otherwise, halve the list by ignoring the item and all items before it
• repeat until there are no more items to check

For example, consider this :

animals = ["cat", "dog", "elephant", "lion", "sparrow", "tiger", "wombat"]

print('index of tiger : ', binary_search(animals, "tiger")  or "not found")
print('index of dog   : ', binary_search(animals, "dog")    or "not found")
print('index of turtle: ', binary_search(animals, "turtle") or "not found")

# outputs:

index of tiger :  5
index of dog   :  1
index of turtle:  not found

These are the steps involved in finding tiger:

• Our initial search range is between the very first and last items, hence low is 0 and high is 6. We calculate the middle to be 3, so we set the index accordingly.

Initial Step

0  cat        <-- low
1  dog
2  elephant
3  lion       <-- index
4  sparrow
5  tiger
6  wombat     <-- high

• lion is less than tiger, so we now set low to index + 1, thus excluding lion and the items before it from our next search. We calculate the new middle to be 5, and set the index.

Next Step

0  cat
1  dog
2  elephant
3  lion
4  sparrow      <-- low
5  tiger        <-- index
6  wombat       <-- high

• tiger is indeed what we are looking for, so we have finished our search.

That’s the binary search in a nutshell.