Python binary search

Halve a sorted search interval after each comparison.

Python

def binary_search(values, target):
    low, high = 0, len(values) - 1
    while low <= high:
        mid = low + (high - low) // 2
        if values[mid] == target:
            return mid
        if values[mid] < target:
            low = mid + 1
        else:
            high = mid - 1
    return -1

print(binary_search([2, 4, 6, 8, 10], 8))
print(binary_search([2, 4, 6, 8, 10], 7))
assert binary_search([], 1) == -1
assert binary_search([2], 2) == 0

Output

3
-1

Explanation

Binary search requires input sorted in ascending order under the same comparison used by the search. The inclusive interval [low, high] contains every remaining candidate.

If the middle value is smaller than the target, eliminate the left half including mid. Otherwise eliminate the right half. When low exceeds high the interval is empty.

Explore the dry run

Guided walkthroughReady

1def binary_search(values, target):

2 low, high = 0, len(values) - 1

3 while low <= high:

4 mid = low + (high - low) // 2

5 if values[mid] == target:

6 return mid

7 if values[mid] < target:

8 low = mid + 1

9 else:

10 high = mid - 1

11 return -1

13print(binary_search([2, 4, 6, 8, 10], 8))

14print(binary_search([2, 4, 6, 8, 10], 7))

15assert binary_search([], 1) == -1

16assert binary_search([2], 2) == 0

Follow the example

Start to inspect the source, values, and output. These prepared steps explain the published sample; grouped loop steps describe multiple iterations.

View the complete dry-run table

Prepared trace of the published program. Some rows group repeated iterations.
What happens	State after step	Output so far
Initialize the first search interval.	`low = 0, high = 4, target = 8`	—
Inspect index 2.	`mid = 2, values[mid] = 6`	—
6 < 8, so discard indices 0 through 2.	`low = 3, high = 4`	—
Inspect index 3.	`mid = 3, values[mid] = 8`	—
Return and print the matching index.	`mid = 3, values[mid] = 8`	3
For target 7, inspect 6 then 8; the interval becomes low=3, high=2.	`empty interval`	3
Print the missing sentinel.	`empty interval`	3 -1

Cost and common mistakes

O(log n) worst-case time and O(1) auxiliary space; sorting beforehand has a separate cost.

Unsorted data invalidates the algorithm.
With duplicates this version returns a matching index, not necessarily the first.

Try it yourself

Trace a search for 2 in the same list.

Reveal answer and reasoning

Inspect indices 2 then 0 and return 0.

Python binary search

Explanation

Follow the example

Cost and common mistakes

Try it yourself

Further reading