Problem Description

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

Example 2:

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

My Idea

The idea here is to implement the basic version of binary search, where we compare with the middle element and depending on weather it’s value is > or < than our target we restrict our problem space to the left or right half of the array respectively. Thus halving the searched area with every step and yielding a time complexity of O(log n).

My solution

# Time complexity: O(log n)
def search(nums:list[int], target:int) -> int:
    l=0
    r=len(nums)-1

    while l<=r:
        mid = (l+r)//2
        if nums[mid]>target:
            r=mid-1
        elif nums[mid]<target:
            l=mid+1
        else:
            return mid

    return -1