LeetCode Python/Java/C++/JS  >  704. Binary Search  >  Solved in Java, Python, C++, JavaScript, C#, Go, Ruby  >  GitHub

    LeetCode link: 704. Binary Search, difficulty: Easy.

    Given an array of integers nums which is sorted in ascending order, and an integer target, write a function to search target in nums. If target exists, then return its index. Otherwise, return -1.

    You must write an algorithm with O(log n) runtime complexity.

    Example 1:

    Input: nums = [-1,0,3,5,9,12], target = 9

    Output: 4

    Explanation: 9 exists in `nums` and its index is 4

    Example 2:

    Input: nums = [-1,0,3,5,9,12], target = 2

    Output: -1

    Explanation: 2 does not exist in `nums` so return -1

    Constraints:

    • 1 <= nums.length <= 10000
    • 104 < nums[i], target < 10000
    • All the integers in nums are unique.
    • nums is sorted in ascending order.
    Topics

    Array

    Intuition
    PreviousNext

    Because it is an already sorted array, by using the middle value for comparison, half of the numbers can be eliminated each time.

    👋 Hi, I'm the developer behind coder.id — Build your programmer brand. Create your portfolio, resume, and blog in minutes, and land your dream offer!

    Step-by-Step Solution

    The fastest and easiest way is to use the three indices left, right, and middle.
    If nums[middle] > target, then right = middle - 1, otherwise, left = middle + 1.

    Complexity

    Time complexity

    O(log N)

    Space complexity

    O(1)

    Solution in languages: Java Python C++ JavaScript C# Go Ruby

    Java # 

    class Solution {
    public int search(int[] nums, int target) {
        var left = 0;
        var right = nums.length - 1;

        while (left <= right) {
            var middle = (left + right) / 2;

            if (nums[middle] == target) {
                return middle;
            }

            if (nums[middle] > target) {
                right = middle - 1;
            } else {
                left = middle + 1;
            }
        }

        return -1;
    }
}

    Python # 

    class Solution:
    def search(self, nums: List[int], target: int) -> int:
        left = 0
        right = len(nums) - 1

        while left <= right:
            middle = (left + right) // 2

            if nums[middle] == target:
                return middle

            if nums[middle] > target:
                right = middle - 1
            else:
                left = middle + 1

        return -1

    C++ # 

    class Solution {
public:
    int search(vector<int>& nums, int target) {
        auto left = 0;
        int right = nums.size() - 1; // Should not use 'auto' here because 'auto' will make this variable become `unsigned long` which has no `-1`.

        while (left <= right) {
            auto middle = (left + right) / 2;

            if (nums[middle] == target) {
                return middle;
            }

            if (nums[middle] > target) {
                right = middle - 1;
            } else {
                left = middle + 1;
            }
        }

        return -1;
    }
};

    JavaScript # 

    var search = function (nums, target) {
  let left = 0
  let right = nums.length - 1

  while (left <= right) {
    const middle = Math.floor((left + right) / 2)

    if (nums[middle] == target) {
      return middle
    }

    if (nums[middle] > target) {
      right = middle - 1
    } else {
      left = middle + 1
    }
  }

  return -1
};

    C# # 

    public class Solution
{
    public int Search(int[] nums, int target)
    {
        int left = 0;
        int right = nums.Length - 1;

        while (left <= right)
        {
            int middle = (left + right) / 2;

            if (nums[middle] == target)
            {
                return middle;
            }

            if (nums[middle] > target)
            {
                right = middle - 1;
            } 
            else
            {
                left = middle + 1;
            }
        }

        return -1;
    }
}

    Go # 

    func search(nums []int, target int) int {
    left := 0
    right := len(nums) - 1

    for left <= right {
        middle := (left + right) / 2

        if nums[middle] == target {
            return middle
        }

        if nums[middle] > target {
            right = middle - 1
        } else {
            left = middle + 1
        }
    }

    return -1
}

    Ruby # 

    def search(nums, target)
  left = 0
  right = nums.size - 1

  while left <= right
    middle = (left + right) / 2

    return middle if nums[middle] == target

    if nums[middle] > target
      right = middle - 1
    else
      left = middle + 1
    end
  end

  -1
end
    Solution in languages: Java Python C++ JavaScript C# Go Ruby

    Other languages

    Welcome to contribute code to GitHub repo -> 704. Binary Search. Thanks!
    PreviousNext

    Level Up Your Developer Identity

    • 📄 Resume, Portfolio & Blog
    • 🌐 Custom Domain Binding Supported
    • ✨ Premium Subdomain: https://yourname.coder.id

    Build Your Developer Brand at coder.id →