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20250404.py
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104 lines (79 loc) · 2.39 KB
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'''
https://leetcode.com/problems/binary-search/description/
Binary Search
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
import math
## First pass:
# class Solution:
# def search(self, nums: List[int], target: int) -> int:
# if len(nums) == 1:
# return 0 if nums[0] == target else -1
#
# start = 0
# end = len(nums)-1
#
# while start <= end:
# if nums[start] == target:
# return start
# if nums[end] == target:
# return end
# start += 1
# end -= 1
# return -1
#
# Note: Time complexity O(n) | Space complexity: O(1)
class Solution:
def search(self, nums: List[int], target: int) -> int:
start = 0
end = len(nums) - 1
while start <= end:
middle = math.floor((start + end) / 2)
if nums[middle] == target:
return middle
if nums[middle] > target:
end = middle - 1
else:
start = middle + 1
return -1
solution = Solution()
print(solution.search([-1,0,3,5,9,12], 9)) # -> 4
print(solution.search([-1,0,3,5,9,12], 2)) # -> -1
print(solution.search([5], 5)) # -> 0
print(solution.search([5], -5)) # -> -1
print(solution.search([-1,0,5], 0)) # -> 1
'''
Pseudocode:
- Initialize start as zero
- Initialize end as length of nums list minus one
- While start is less than or equal to end:
- Declare middle as start + end divided by two and rounded down
- If nums at index middle is equal to target
- Return middle
- If nums at index middle is greater than target:
- Reassign end as middle - 1
- Else:
- Reassign start as middle + 1
- Return -1 as default
- Time complexity: O(log n)
- Space complexity: O(1)
- Note:
- I needed help on the O(log n) solution
- Keep practicing and studying:
- https://builtin.com/software-engineering-perspectives/nlogn
'''