704.Binary_Search_(Easy).md

704. Binary Search (Easy)

Date and Time: Jun 25, 2024, 15:45 (EST)

Link: https://leetcode.com/problems/binary-search/

Question:

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

Edge Case:

Input: nums = [5], target = 5

Output: 0

KeyPoints:

We should use Binary Search, which will reach $O(log\ n)$ runtime. Look at the edge case, so we set the base case to be while l <= r. Then we compare the middle element with the target to see if middle element is less than target, we set l = m + 1.

My Solution:

class Solution:
    def search(self, nums: List[int], target: int) -> int:
        l, r = 0, len(nums) - 1
        while l <= r:
            m = (l + r) // 2
            if nums[m] > target:
                r = m - 1
            elif nums[m] < target:
                l = m + 1
            else:
                return m
        return -1

Time Complexity: $O(log\ n)$