binary_search.py

"""
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 <= 104
-104 < nums[i], target < 104
All the integers in nums are unique.
nums is sorted in ascending order.

"""

from typing import List


class Solution:
    def search(self, nums: List[int], target: int) -> int:

        left = 0
        right = len(nums) - 1
        while left <= right:
            mid = (left + right) // 2
            if nums[mid] == target:
                return mid
            elif nums[mid] > target:
                right = mid - 1
            else:
                left = mid + 1

        return -1


if __name__ == "__main__":
    solution = Solution()
    # print(solution.search(nums=[-1, 0, 3, 5, 9, 12], target=9))
    print(solution.search(nums=[-1, 0, 3, 5, 9, 12], target=2))