Binary search algorithm
Generally, to find a value in unsorted array, we should look through elements of an array one by one, until searched value is found. In case of searched value is absent from array, we go through all elements. In average, complexity of such an algorithm is proportional to the length of the array.
Situation changes significantly, when array is sorted. If we know it, random access capability can be utilized very efficiently to find searched value quick. Cost of searching algorithm reduces to binary logarithm of the array length. For reference, log_{2}(1 000 000) ≈ 20. It means, that in worst case, algorithm makes 20 steps to find a value in sorted array of a million elements or to say, that it doesn't present it the array.
Algorithm
Algorithm is quite simple. It can be done either recursively or iteratively:
 get the middle element;
 if the middle element equals to the searched value, the algorithm stops;
 otherwise, two cases are possible:
 searched value is less, than the middle element. In this case, go to the step 1 for the part of the array, before middle element.
 searched value is greater, than the middle element. In this case, go to the step 1 for the part of the array, after middle element.
Examples
Example 1. Find 6 in {1, 5, 6, 18, 19, 25, 46, 78, 102, 114}.
Step 1 (middle element is 19 > 6): 1 5 6 18 19 25 46 78 102 114
Step 2 (middle element is 5 < 6): 1 5 6 18 19 25 46 78 102 114
Step 3 (middle element is 6 == 6): 1 5 6 18 19 25 46 78 102 114
Example 2. Find 103 in {1, 5, 6, 18, 19, 25, 46, 78, 102, 114}.
Step 1 (middle element is 19 < 103): 1 5 6 18 19 25 46 78 102 114
Step 2 (middle element is 78 < 103): 1 5 6 18 19 25 46 78 102 114
Step 3 (middle element is 102 < 103): 1 5 6 18 19 25 46 78 102 114
Step 4 (middle element is 114 > 103): 1 5 6 18 19 25 46 78 102 114
Step 5 (searched value is absent): 1 5 6 18 19 25 46 78 102 114
Complexity analysis
Huge advantage of this algorithm is that it's complexity depends on the array size logarithmically in worst case. In practice it means, that algorithm will do at most log_{2}(n) iterations, which is a very small number even for big arrays. It can be proved very easily. Indeed, on every step the size of the searched part is reduced by half. Algorithm stops, when there are no elements to search in. Therefore, solving following inequality in whole numbers:
n / 2^{iterations} > 0
resulting in
iterations <= log_{2}(n).
It means, that binary search algorithm time complexity is O(log_{2}(n)).
Code snippets.
You can see recursive solution for Java and iterative for C++ below.Java
/**
* searches for a value in sorted array
*
* @param array
* array to search in
* @param value
* searched value
* @param left
* index of left boundary
* @param right
* index of right boundary
* @return position of searched value, if it presents in the array or 1, if
* it is absent
*/
int binarySearch(int[] array, int value, int left, int right) {
if (left > right)
return 1;
int middle = (left + right) / 2;
if (array[middle] == value)
return middle;
else if (array[middle] > value)
return binarySearch(array, value, left, middle  1);
else
return binarySearch(array, value, middle + 1, right);
}
C++
/*
* searches for a value in sorted array
* arr is an array to search in
* value is searched value
* left is an index of left boundary
* right is an index of right boundary
* returns position of searched value, if it presents in the array
* or 1, if it is absent
*/
int binarySearch(int arr[], int value, int left, int right) {
while (left <= right) {
int middle = (left + right) / 2;
if (arr[middle] == value)
return middle;
else if (arr[middle] > value)
right = middle  1;
else
left = middle + 1;
}
return 1;
}
Visualizers
Two responses to "Binary Search Tutorial"

on Oct 20, 2009 said:
The last value is never readed. For example...
int[] a = { 0, 3, 1, 7, 2, 8, 12, 53, 64, 95, 6, 71, 84, 9 };
and i want to search number 9
it returns that 9 is not in the array
Algorithm requires that source array is sorted in order to work correct.

on Nov 5, 2008 said:
Hi, greetings from Argentina. I don't know whether this site is too old or very new. Anyway, I believe there is a mistake with the binary search. I noticed that when "if (arr[middle] > value)" is true, it means that you discard the first half of your array, considering 0,1,2,...,n. Then, in that case, the next sentence should be left=middle + 1 instead of right=middle1, which makes you consider only the first half of the array. Best wishes, Andres from Buenos Aires, Argentina.
There is no mistake. If condition "value < arr[middle]" is true, it means, that value may present only in the first part of an array. In this case second part of an array is discarded and search continued in the first part. Thanks for your reply.
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