amazon OA1
7道debug题20分钟
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Print Pattern 变量量ch++ -> print++
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Even odd pattern 加括号
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Manchester == -> !=
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Reverse Array += -> -=
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Sort Array > -> <
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Get Digit sum parity result %2 == 0 -> !=
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Distinct Element Count f==1 -> f==0
另一个人的 new grad OA1
20分钟7道debug,题⽬目为:
1 sort array
2 奇数偶数
3 check palindromic num->temp
4 distinct element count
5 digitalSumPairy if (result % 2 !=0)->if (result % 2 ==0)
6 countproduct for loop ⾥里里⾯面限制条件错了了改下就好
7 找出出现k次的元素
OA2:
两道coding题,70min。
第一题是利口240,在2d matrix里找⼀个数。
第二题是找连通图(2d array)⾥的所有bridge(critical connections), 使得这个graph变成两个disconnected parts.
new grad OA 3
- Work Simulation: ⽬前没有在⾯经里看到原题,佛系吧,还是那几个原则,
manager和customer⾄上,requirement⽐deadline重要,⻓远利益⽐目前利益重要
- Logic:很多原题,正好做完
Logic前⾯面基本都是⼩⼟刀的原题,后面的应用题有一两个新题
应⽤题我记得的原题是factory那个,新题具体细节太长了有点忘,但是很简单
很straightforward那种,只不不过时间很紧,前⾯面的简单题⽐比如说找规律的争取大概⼗五秒一个吧。。。这样后来的长题才有时间…
贴几个别人的:
亚麻OA1 new grad
刚刚做完OA1,和前⾯发的⾯经⼀样,20min 7道debug题。顺序记不清了了,以下是我遇到的题目:
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ArrayOperation
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CheckGrade
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printPattern
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occurence
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countDigit
-
EvenOddPattern
-
replaceValues
2020 new grad OA1+OA2
8/15 网上海投
OA1 debug 七题
OA2 70min 兩题 两个hashmap 统计⻛格 + 蠡⼝一三巴
OA2 2020 new grad
三部分OA中的第⼆部分 包含两题算法跟一堆亚麻文化测试
楼主遇到的两题是利口 贰肆凌 武漆贰
社招OA
第⼀题:merge files -->⽤ PriorityQueue 就可以;第二题:Remove
Obstacle(背景:机器人去除障碍物) —>BFS搞定。
8⽉25号完成的OA, 过后recuiter 让我提供几个时间说要电话面试,昨天我把时间发给了recuiter, 过了⼀天 到现在recuiter还没有回信。。这种情况应该怎么办?
我是不是在等两天,然后催一下呢?希望有经验的人给点建议。
ps: 投的是加拿⼤大温哥华的职位。
new grad OA1
以前投过亚麻挂了,之前没有投2020 grad full time,是系统把我从student portal⾥捞起来,我就apply了,昨天刚做的oa1,发上来。题⽬顺序不记得了了,都是原题,oa1只有debug题,没有逻辑题。祝⼤大家好运。
countOccurrence
返回Value在Array⾥面出现的次数
while loop⾥面缺少i++
printPattern(出现了两次)
for loop缺少 {}
selectionSort
<>改⼀一下
reverseArray
把len++改成len–
CheckGrade
两个else if⾥里里的‘||’改为‘&&’
countDigits
加⼀句temp=num; 然后把return⾥所有的num改为temp, 这样num就不会被改变了
lllll
Min Abs Diff Between 2 Elements
You are given an array arr
of n
elements of positive integers. Your task is to determine the minimum absolute difference between any 2 elements of the array. You can perform the following two types of operations on the array elements any number of times:
- If element
e
is even, then you can replace it bye / 2
- If element
e
is odd, then you can replace it by2 * e
Determine the minimum absolute difference between any two elements after performing this operation any number of times (possibly zero) on any element of the array.
Example 1:
Input: [1, 2]
Output: 0
Constraints:
- 1<=
T
<=50 - 2<=
n
<=50000 - 1<=
arr[i]
<=10^8 for each validi
Time limit: 1 sec
It is guaranteed that summation of n
over all test cases does not exceed 250000.