Binary search is one of the fundamental algorithms in computer science. In order to explore it, we’ll first build up a theoretical backbone, then use that to implement the algorithm properly and avoid those nasty off-by-one errors everyone’s been talking about.

Finding a value in a sorted sequence In its simplest form, binary search is
used to quickly find a value in a sorted sequence (consider a sequence an
ordinary array for now). We’ll call the sought value the target value for clarity.

Binary search maintains a contiguous subsequence of the starting sequence
where the target value is surely located. This is called the search space. The
search space is initially the entire sequence. At each step, the algorithm
compares the median value in the search space to the target value. Based on the comparison and because the sequence is sorted, it can then eliminate half of the search space. By doing this repeatedly, it will eventually be left with a search space consisting of a single element, the target value.


Consider the following sequence of integers sorted in ascending order and say we are looking for the number 55:
0 5 13 19 22 41 55 68 72

  • We are interested in the location of the target value in the sequence so we will represent the search space as indices into the sequence. Initially, the search space contains indices 1 through 11. Since the search space is really an interval, it suffices to store just two numbers, the low and high indices.
  • As described above, we now choose the median value, which is the value at index 6 (the midpoint between 1 and 11): this value is 41 and it is smaller than the target value. From this we conclude not only that the element at index 6 is not the target value, but also that no element at indices between 1 and 5 can be the target value, because all elements at these indices are smaller than 41, which is smaller than the target value.
  • This brings the search space down to indices 7 through 11:
    55 68 72 81 98
  • similar to that we chop off the second half of the search space and this are left with:
    55 68
  • Depending on how we choose the median of an even number of elements we will either find 55 in the next step or chop off 68 to get a search space of only one element. Either way, we conclude that the index where the target value is located is 7.
  • If the target value was not present in the sequence, binary search would empty the search space entirely. This condition is easy to check and handle.

Here is some code to go with the description:

binary_search(A, target):
lo = 1, hi = size(A)
while lo <= hi:
mid = lo + (hi-lo)/2
if A[mid] == target:
return mid
else if A[mid] < target:
lo = mid+1
hi = mid-1
// target was not found


Since each comparison binary search uses halves the search space, we can
assert and easily prove that binary search will never use more than (in bigoh notation) O(log N) comparisons to find the target value.

The logarithm is an awfully slowly growing function. In case you’re not
aware of just how efficient binary search is, consider looking up a name in a phone book containing a million names. Binary search lets you systematically find any given name using at most 21 comparisons. If you could manage a list containing all the people in the world sorted by name, you could find any person in less than 35 steps.

Binary search program here we will write the code for binary search in two different ways first is recursive binary search and other is iterative binary search.

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