LeetCode 458. Poor Pigs

本文探讨了如何使用最少数量的猪来找出含有毒药的水桶。通过算法设计,解决了在限定时间内确定哪只猪会因中毒而死亡的问题。文章还提供了一种通用算法解决类似问题,并附带代码实现。

摘要生成于 C知道 ,由 DeepSeek-R1 满血版支持, 前往体验 >

Problem Statement

(Source) There are 1000 buckets, one and only one of them contains poison, the rest are filled with water. They all look the same. If a pig drinks that poison it will die within 15 minutes. What is the minimum amount of pigs you need to figure out which bucket contains the poison within one hour.

Answer this question, and write an algorithm for the follow-up general case.

Follow-up:

If there are n buckets and a pig drinking poison will die within m minutes, how many pigs (x) you need to figure out the “poison” bucket within p minutes? There is exact one bucket with poison.

Tags: Brain Teaser, Bit Manipulation.

Solution

class Solution(object):
    def poorPigs(self, buckets, minutesToDie, minutesToTest):
        """
        :type buckets: int
        :type minutesToDie: int
        :type minutesToTest: int
        :rtype: int
        """
        if buckets == 1: return 0
        test_rounds = minutesToTest / minutesToDie + 1
        return int(math.ceil(math.log(buckets, test_rounds)))

Reference:
(1) https://discuss.leetcode.com/topic/67066/c-solution-with-explanation

评论
添加红包

请填写红包祝福语或标题

红包个数最小为10个

红包金额最低5元

当前余额3.43前往充值 >
需支付:10.00
成就一亿技术人!
领取后你会自动成为博主和红包主的粉丝 规则
hope_wisdom
发出的红包
实付
使用余额支付
点击重新获取
扫码支付
钱包余额 0

抵扣说明:

1.余额是钱包充值的虚拟货币,按照1:1的比例进行支付金额的抵扣。
2.余额无法直接购买下载,可以购买VIP、付费专栏及课程。

余额充值