Amazon AS Intern OA（含题目和解题参考）

刚刚做了亚麻 AS Intern的OA
. .и

• 在Hackerrank上
• 70min 2题
• 把地里所有相关的帖子整理了一遍发现就五道题，然后把答案都写了一遍。
• 最后发现我抽到的题目不在那五道题里面哈哈哈哈
  

Anyway
我抽到的是这两题，感觉过于简单了，不知道是不是因为亚麻打算摆烂不招人了

• Reverse Array Queries 这个直接搜stackoverflow 里面有一样的 不过本来也很简单
• Validating Strings with RegEx 这个直接搜 利口 里面有讨论这题的  答案应该是这个 r'^(.).*\1$|[ab]$'
  

下面是我整理的5道历史题目和我自己结合资料做的答案

求加米！！（

• Code Question 1
  

A list of unique product IDs needs to be extracted from a database of products available on Amazon's website. In the database, multiple versions of the products IDs exist such that the order of the characters varies (e.g, code and ecod are the same product). To create the unique list, all anagrams must be removed. Two strings are anagrams if they are permutations of each other. In other words, both strings have the same size and the same characters. For example, ""aaagmnrs"" is an anagram of "anagrams"". Given an list of strings, remove each string that is an anagram of an earlie

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ositive integer values that evenly divide into a number) and then return the p element of the list, sorted ascending. If there is no p" element return 0.

Example

n 20

p =3

The factors of 20 in ascending order are (1, 2, 4 5, 10, 20). Using 1- based indexing, if p =3, then 4 is returned. If p > 6, 0 would be

n = 20
p = 3

1. <div>def ith_factor(n,p):</div><div>  factor = [[i, n//i] for i in range(1, int(n**0.5) + 1) if n % i == 0]</div><div>  factor = sorted(sum(factor, []))</div><div>  if p > len(factor):</div><div>    return 0</div><div>  else:</div><div>    return(factor[p-1])</div><div>
   
2. </div><div>ith_factor(20,3)</div>

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4

[url]https://assets.leetcode.com/users/images/b2d78610-ec5a-4130-b798-41375791c8ee_1647994025.476528.png[/url]

这题原来的答案不太对 我按照讨论里的重新修改了下～

1. <div>def droppedRequests(requestTime):</div><div>  requestDrop = 0</div><div>  request01 = []</div><div>  request10 = []</div><div>  request60 = []</div><div>
   
2. </div><div>  for i in range(len(requestTime)):</div><div>    currRequest = requestTime[i]</div><div>
   
3. </div><div>    currTime = currRequest</div><div>    pre01 = max(currTime - 1, 1)</div><div>    pre10 = max(currTime - 9, 1)</div><div>    pre60 = max(currTime - 59, 1)</div><div>
   
4. </div><div>    drop01, drop10, drop60 = False, False, False</div><div>
   
5. </div><div>    while request01 and request01[0] < currTime:</div><div>      request01.pop(0)</div><div>    while request10 and request10[0] < pre10:</div><div>      request10.pop(0)</div><div>    while request60 and request60[0] < pre60:</div><div>      request60.pop(0)</div><div>    </div><div>    request01.append(currRequest)</div><div>    request10.append(currRequest)</div><div>    request60.append(currRequest)</div><div>
   
6. </div><div>    if len(request01) > 3:</div><div>      drop01 = True</div><div>    if len(request10) > 20:</div><div>      drop10 = True</div><div>    if len(request60) > 60:</div><div>      drop60 = True</div><div>    </div><div>    if drop01 or drop10 or drop60:</div><div>      requestDrop += 1</div><div>  </div><div>  return requestDrop</div><div>a = [1,2,3,4]</div><div>a[1:3] = a[2:0:-1]</div><div>a</div><div>[1, 3, 2, 4]</div><div>droppedRequests([1,1,1,1,2,2,2,3,3,3,4,4,4,5,5,5,6,6,6,7,7,7,7,11,11,11,11])</div>

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匿名用户 发表于 2022-11-8 18:02
大兄弟啊大兄弟！ 我强烈怀疑你做的是test啊， 我刚试了test 就是你说的那两题啊，那只是practice demo啊.. ...

好家伙！！！谢谢大兄弟！！我今天ddl刚刚赶去看了还真是我晕，我这也太眼花了随便点了个link就做了。不过刚刚去做了下，也很简单大概十分钟就能做完，就是上面的merge interval和一题新的easy题。感谢大兄弟的提醒哈哈哈哈

大兄弟啊大兄弟！ 我强烈怀疑你做的是test啊， 我刚试了test 就是你说的那两题啊，那只是practice demo啊........

新手发帖还不太熟练，把code里的<div>去掉应该就成

请问什么时候投的呀

kc777 发表于 2022-11-7 23:23
请问什么时候投的呀

大概是10月中？

十月中投的AS intern现在还没有消息...

OhAbia 发表于 2022-11-8 18:56
好家伙！！！谢谢大兄弟！！我今天ddl刚刚赶去看了还真是我晕，我这也太眼花了随便点了个link就做了。不 ...

兄弟 这么有缘 一起加个微信准备vo  可好？sky9527223 我的话

请问Throttling Gateway这道题原来的答案不太对是哪些test cases没有pass?谢谢