Binary Search

Intuition

Binary search is an efficient array search algorithm. It works by narrowing down the search range by half each time. If you have looked up a word in a physical dictionary, you've already used binary search in real life. Let's look at a simple example:

Given a sorted array of integers and an integer called target, find the element that equals the target and return its index. If the element is not found, return -1.

The key observation here is that the array is sorted. We pick a random element in the array and compare it to the target.

  • If we happen to pick the element that equals the target (how lucky!), then bingo. We don't need to do any more work; return its index.
  • If the element is smaller than the target, then we know the target cannot be found in the section to the left of the current element since everything to the left is even smaller. So we discard the current element and everything on the left from the search range.
  • If the element is larger than the target, then we know the target cannot be found in the section to the right of the current element since everything to the right is even larger. So we discard the current element and everything on the right from the search range.

We repeat this process until we find the target. Instead of picking a random element, we always pick the middle element in the current search range. This way, we can discard half of the options and shrink the search range by half each time. This gives us O(log(N)) runtime.

Try it yourself

Implementation

The search range is represented by the left and right indices that start from both ends of the array and move towards each other as we search. When moving the index, we discard elements and shrink the search range.

Time Complexity: O(log(n))

1from typing import List
2
3def binary_search(arr: List[int], target: int) -> int:
4    left, right = 0, len(arr) - 1
5    while left <= right:  # <= because left and right could point to the same element, < would miss it
6        mid = (left + right) // 2 # double slash for integer division in python 3, we don't have to worry about integer `left + right` overflow since python integers can be arbitrarily large
7        # found target, return its index
8        if arr[mid] == target:
9            return mid
10        # middle less than target, discard left half by making left search boundary `mid + 1`
11        if arr[mid] < target:
12            left = mid + 1
13        # middle greater than target, discard right half by making right search boundary `mid - 1`
14        else:
15            right = mid - 1
16    return -1 # if we get here we didn't hit above return so we didn't find target
17
18if __name__ == '__main__':
19    arr = [int(x) for x in input().split()]
20    target = int(input())
21    res = binary_search(arr, target)
22    print(res)
23
1import java.util.Arrays;
2import java.util.List;
3import java.util.Scanner;
4import java.util.stream.Collectors;
5
6class Solution {
7    public static int binarySearch(List<Integer> arr, int target) {
8        int left = 0;
9        int right = arr.size() - 1;
10
11        while (left <= right) { // <= here because left and right could point to the same element, < would miss it
12            int mid = left + (right - left) / 2; // use `(right - left) /2` to prevent `left + right` potential overflow
13            // found target, return its index
14            if (arr.get(mid) == target) return mid;
15            if (arr.get(mid) < target) {
16                // middle less than target, discard left half by making left search boundary `mid + 1`
17                left = mid + 1;
18            } else {
19                // middle greater than target, discard right half by making right search boundary `mid - 1`
20                right = mid - 1;
21            }
22        }
23        return -1; // if we get here we didn't hit above return so we didn't find target
24    }
25
26    public static List<String> splitWords(String s) {
27        return s.isEmpty() ? List.of() : Arrays.asList(s.split(" "));
28    }
29
30    public static void main(String[] args) {
31        Scanner scanner = new Scanner(System.in);
32        List<Integer> arr = splitWords(scanner.nextLine()).stream().map(Integer::parseInt).collect(Collectors.toList());
33        int target = Integer.parseInt(scanner.nextLine());
34        scanner.close();
35        int res = binarySearch(arr, target);
36        System.out.println(res);
37    }
38}
39
1using System;
2using System.Collections.Generic;
3using System.Linq;
4
5class Solution
6{
7    public static int BinarySearch(List<int> arr, int target)
8    {
9        int left = 0;
10        int right = arr.Count - 1;
11        while (left <= right) // <= here because left and right could point to the same element, < would miss it
12        {
13            int mid = left + (right - left) / 2; // use `(right - left) /2` to prevent `left + right` potential overflow
14            // found target, return its index
15            if (arr[mid] == target) return mid;
16            if (arr[mid] < target)
17            {
18                // middle less than target, discard left half by making left search boundary `mid + 1`
19                left = mid + 1;
20            }
21            else
22            {
23                // middle greater than target, discard right half by making right search boundary `mid - 1`
24                right = mid - 1;
25            }
26        }
27        return -1; // if we get here we didn't hit above return so we didn't find target
28    }
29
30    public static List<string> SplitWords(string s)
31    {
32      return string.IsNullOrEmpty(s) ? new List<string>() : s.Trim().Split(' ').ToList();
33    }
34
35    public static void Main()
36    {
37        List<int> arr = SplitWords(Console.ReadLine()).Select(int.Parse).ToList();
38        int target = int.Parse(Console.ReadLine());
39        int res = BinarySearch(arr, target);
40        Console.WriteLine(res);
41    }
42}
43
1function binarySearch(arr, target) {
2    let left = 0;
3    let right = arr.length - 1;
4
5    while (left <= right) {  // <= because left and right could point to the same element, < would miss it
6        let mid = left + Math.floor((right - left) / 2);  // use `(right - left) /2` to prevent `left + right` potential overflow
7        if (arr[mid] === target) return mid;  // found target, return its index
8        if (arr[mid] < target) {
9            // middle less than target, discard left half by making left search boundary `mid + 1`
10            left = mid + 1;
11        } else {
12            // middle greater than target, discard right half by making right search boundary `mid - 1`
13            right = mid - 1;
14        }
15    }
16    // if we get here we didn't hit above return so we didn't find target
17    return -1;
18}
19
20function splitWords(s) {
21    return s == "" ? [] : s.split(' ');
22}
23
24function* main() {
25    const arr = splitWords(yield).map((v) => parseInt(v));
26    const target = parseInt(yield);
27    const res = binarySearch(arr, target);
28    console.log(res);
29}
30
31class EOFError extends Error {}
32{
33    const gen = main();
34    const next = (line) => gen.next(line).done && process.exit();
35    let buf = '';
36    next();
37    process.stdin.setEncoding('utf8');
38    process.stdin.on('data', (data) => {
39        const lines = (buf + data).split('\n');
40        buf = lines.pop();
41        lines.forEach(next);
42    });
43    process.stdin.on('end', () => {
44        buf && next(buf);
45        gen.throw(new EOFError());
46    });
47}
48
1#include <algorithm> // copy
2#include <iostream> // cin, cout, streamsize
3#include <iterator> // back_inserter, istream_iterator
4#include <limits> // numeric_limits
5#include <sstream> // istringstream
6#include <string> // getline, string
7#include <vector> // vector
8
9int binary_search(std::vector<int> arr, int target) {
10    int left = 0;
11    int right = arr.size() - 1;
12
13    while (left <= right) { // <= here because left and right could point to the same element, < would miss it
14        int mid = left + (right - left) / 2; // use `(right - left) /2` to prevent `left + right` potential overflow
15        // found target, return its index
16        if (arr.at(mid) == target) return mid;
17        if (arr.at(mid) < target) {
18            // middle less than target, discard left half by making left search boundary `mid + 1`
19            left = mid + 1;
20        } else {
21            // middle greater than target, discard right half by making right search boundary `mid - 1`
22            right = mid - 1;
23        }
24    }
25    return -1; // if we get here we didn't hit above return so we didn't find target
26}
27
28template<typename T>
29std::vector<T> get_words() {
30    std::string line;
31    std::getline(std::cin, line);
32    std::istringstream ss{line};
33    std::vector<T> v;
34    std::copy(std::istream_iterator<T>{ss}, std::istream_iterator<T>{}, std::back_inserter(v));
35    return v;
36}
37
38void ignore_line() {
39    std::cin.ignore(std::numeric_limits<std::streamsize>::max(), '\n');
40}
41
42int main() {
43    std::vector<int> arr = get_words<int>();
44    int target;
45    std::cin >> target;
46    ignore_line();
47    int res = binary_search(arr, target);
48    std::cout << res << '\n';
49}
50
1package main
2
3import (
4    "bufio"
5    "fmt"
6    "os"
7    "strconv"
8    "strings"
9)
10
11func binarySearch(arr []int, target int) int {
12    left := 0
13    right := len(arr) - 1
14
15    for left <= right { // <= here because left and right could point to the same element, < would miss it
16        mid := left + (right - left) / 2 // use `(right - left) /2` to prevent `left + right` potential overflow
17        // found target, return its index
18        if arr[mid] == target {
19            return mid;
20        }
21        if arr[mid] < target {
22            // middle less than target, discard left half by making left search boundary `mid + 1`
23            left = mid + 1
24        } else {
25            // middle greater than target, discard right half by making right search boundary `mid - 1`
26            right = mid - 1
27        }
28    }
29    return -1 // if we get here we didn't hit above return so we didn't find target
30}
31
32func arrayAtoi(arr []string) []int {
33    res := []int{}
34    for _, x := range arr {
35        v, _ := strconv.Atoi(x)
36        res = append(res, v)
37    }
38    return res
39}
40
41func splitWords(s string) []string {
42    if s == "" {
43        return []string{}
44    }
45    return strings.Split(s, " ")
46}
47
48func main() {
49    scanner := bufio.NewScanner(os.Stdin)
50    scanner.Scan()
51    arr := arrayAtoi(splitWords(scanner.Text()))
52    scanner.Scan()
53    target, _ := strconv.Atoi(scanner.Text())
54    res := binarySearch(arr, target)
55    fmt.Println(res)
56}
57
1#lang racket
2
3(define (binary-search arr target)
4  (let search ([l 0]
5               [r (sub1 (vector-length arr))])
6    ;; <= because left and right could point to the same element, < would miss it
7    (if (<= l r)
8        (let* ([m (quotient (+ l r) 2)]
9               [v (vector-ref arr m)])
10          (cond
11            ;; found target, return its index
12            [(= v target) m]
13            ;; middle less than target, discard left half by making left search boundary `m + 1`
14            [(< v target) (search (add1 m) r)]
15            ;; middle greater than target, discard right half by making right search boundary `m - 1`
16            [else (search l (sub1 r))]))
17        ;; if we get here we didn't hit above guard so we didn't find target
18        -1)))
19
20(define arr (list->vector (map string->number (string-split (read-line)))))
21(define target (string->number (read-line)))
22(define res (binary-search arr target))
23(displayln res)
24
1import Data.Array (Array)
2import qualified Data.Array.IArray as A
3import Data.Array.IArray (bounds, listArray)
4
5binarySearch :: Array Int Int -> Int -> Int
6binarySearch arr target = uncurry search $ bounds arr where
7  search l r
8    -- <= because left and right could point to the same element, < would miss it
9    | l <= r = case compare (arr A.! m) target of
10        -- found target, return its index
11        EQ -> m
12        -- middle less than target, discard left half by making left search boundary `m + 1`
13        LT -> search (succ m) r
14        -- middle greater than target, discard right half by making right search boundary `m - 1`
15        GT -> search l (pred m)
16    -- if we get here we didn't hit above guard so we didn't find target
17    | otherwise = -1
18    -- use `(r - l) / 2` to prevent `l + r` potential overflow
19    where m = l + (r - l) `div` 2
20
21main :: IO ()
22main = do
23  arr' <- map read . words <$> getLine
24  let arr = listArray (0, length arr' - 1) arr'
25  target <- readLn
26  let res = binarySearch arr target
27  print res
28

Calculating mid

Note that when calculating mid, if the number of elements is even, there are two elements in the middle. We usually follow the convention of picking the first one, equivalent to doing integer division (left + right) / 2.

In most programming languages, we calculate mid with left + floor((right-left) / 2) to avoid potential integer overflow. However, in Python, we do not need to worry about left + right integer overflow because Python3 integers can be arbitrarily large.

Deducing binary search

It's essential to understand and deduce the algorithm yourself instead of memorizing it. In an actual interview, interviewers may ask you additional questions to test your understanding, so simply memorizing the algorithm may not be enough to convince the interviewer.

Key elements in writing a correct binary search:

1. When to terminate the loop

Make sure the while loop has an equality comparison. Otherwise, we'd skip the loop and miss the potential match for the edge case of a one-element array.

2. Whether/how to update left and right boundary in the if conditions

Consider which side to discard. If arr[mid] is already smaller than the target, we should discard everything on the left by making left = mid + 1.

3. Should I discard the current element?

For vanilla binary search, we can discard it since it can't be the final answer if it is not equal to the target. There might be situations where you would want to think twice before discarding the current element. We'll discuss this in the next module.

When to use binary search

Interestingly, binary search works beyond sorted arrays. You can use binary search whenever you make a binary decision to shrink the search range. We will see this in the following modules.

P.S. Did you notice AlgoMonster's logo is an illustration of binary search? :P