** 100 min
70 min 2 coding problems
10-20min workstyle assessments**
题目集合!!!
【超全超详细】Amazon new grad OA2 新鲜面经&资料大集合
Amazon OA2
亚马逊】【OA】整理60道
新鲜亚麻 NG OA2
题库
Amazon Online Assessment Questions 2021
Amazon OA question complilation
Input:
5,
[2, 4, 6, 8, 10]
Output:
2
Amazon Online Assessment (OA) - Fill The Truck
from typing import List
def fill_the_truck(num: int, boxes: List[int], unit_size: int, units_per_box: List[int], truck_size: int) -> int:
# WRITE YOUR BRILLIANT CODE HERE
maxUnits = 0
space = truck_size
for i, units in enumerate(units_per_box):
if space < boxes[i]:
return maxUnits + space * units
maxUnits += boxes[i] * units
space -= boxes[i]
return maxUnits
if __name__ == '__main__':
num = int(input())
boxes = [int(x) for x in input().split()]
unit_size = int(input())
units_per_box = [int(x) for x in input().split()]
truck_size = int(input())
res = fill_the_truck(num, boxes, unit_size, units_per_box, truck_size)
print(res)
Similar LeetCode 1710. Maximum Units on a Truck
class Solution:
def maximumUnits(self, boxTypes: List[List[int]], truckSize: int) -> int:
boxTypes.sort(key = lambda x:x[1], reverse = True) # sort list based on second
# item (numberOfUnitsPerBox) in decreasing
maxUnits = 0
for box in boxTypes:
if truckSize < box[0]: # truck不够装了,剩下多少全装满,然后关门
return maxUnits + truckSize * box[1]
maxUnits += box[0] *box[1]
truckSize -= box[0] # 更新truck剩余空间
return maxUnits
Amazon | OA 2020 | Highest Profit
Amazon Online Assessment (OA) - Find The Highest Profit
Input
numSuppliers
: an integer representing the number of suppliers
inventory
: a list of long integers representing the value of the item at a given supplier
order
: a long integer representing the number of items to be ordered.
Output
Return a long integer representing the highest profit that can be generated for the given product.
from typing import List
from collections import Counter
def find_profit(num_suppliers: int, inventory: List[int], order: int) -> int:
# WRITE YOUR BRILLIANT CODE HERE
d = Counter()
for i in range(num_suppliers):
d[inventory[i]] += 1 # {不同的利润的商品(来自各个suplliers):有多少个}
cur = max(d) # 哪个利润最高
profit = 0
while order > 0 and cur > 0:
freq = d[cur] # 这个利润有几个商品
if order >= freq:
order -= freq
profit += cur * freq
elif order < freq:
profit += order * cur
break
cur -= 1 # 这个价位卖光啦
d[cur] += freq # 之前买的利润一起下降1
return profit
if __name__ == '__main__':
num_suppliers = int(input())
inventory = [int(x) for x in input().split()]
order = int(input())
res = find_profit(num_suppliers, inventory, order)
print(res)
class Solution:
def slowestKey(self, releaseTimes: List[int], keysPressed: str) -> str:
l = len(releaseTimes)
res = keysPressed[0]
maxTime = releaseTimes[0]
if l == 1:
return res
if l == 0:
return None
for i in range(1, l):
time = releaseTimes[i] - releaseTimes[i - 1]
if time == maxTime:
res = max(res, keysPressed[i])
if time > maxTime:
res = keysPressed[i]
maxTime = time
return res
class Solution:
def twoSum(self, nums: List[int], target: int) -> List[int]:
memo = []
for i, num in enumerate(nums):
if target - num in memo:
return [i,nums.index(target - num)]
memo.append(num)
class Solution:
def robotSim(self, commands: List[int], obstacles: List[List[int]]) -> int:
obs = set(map(tuple,obstacles)) # 提升in的查找速度(键值对应)
dx = [0, 1, 0, -1] # dx和dy组合起来代表
dy = [1, 0, -1, 0] # [north, east, south, west]
# 向左移动:左转(逆时针);向右移动:右转
di = 0
x = y = 0
ans = 0
for c in commands:
if c == -2:
di = (di - 1) % 4
elif c == -1:
di = (di + 1) % 4
else:
for i in range(c):
if (x + dx[di], y + dy[di]) not in obs:
x += dx[di]
y += dy[di]
ans = max(ans, x*x + y*y)
return ans
Count of ways to split a given number into prime segments
Input: str = “3175”
Output: 3
Explanation:
There are 3 ways to split this string into prime numbers which are (31, 7, 5), (3, 17, 5), (317, 5).
MOD = 1000000007
# Function to check whether a number is a prime number or not
def isPrime(number):
num = int(number)
i = 2
while i * i <= num:
if ((num % i) == 0):
return False
i += 1
if num > 1:
return True
else: # i is not a prime number
return False
# Function to find the count of ways to split string into prime numbers
def countPrimeStrings(number, i):
# 1 based indexing
if (i == 0):
return 1
cnt = 0
# Consider every suffix up to 6 digits
for j in range(1, 7):
# Number should not have a leading zero and it should be a prime number
if (i - j >= 0 and number[i - j] != '0' and isPrime(number[i - j : i])):
cnt += countPrimeStrings(number, i - j)
cnt %= MOD
# Return the final result
return cnt
# Driver code
if __name__ == "__main__":
s1 = "3175"
l = len(s1)
print (countPrimeStrings(s1, l))
# This code is contributed by Chitranayal
Count of substrings of length K with exactly K distinct characters
Count number of substrings with exactly k distinct characters
def countkDist(str1, k):
n = len(str1)
res = 0
# To store count of characters from 'a' to 'z'
cnt = [0] * 27
# Consider all substrings beginning with str[i]
for i in range(0, n):
dist_count = 0
cnt = [0] * 27 # Initializing array with 0
for j in range(i, n): # Consider all substrings between str[i..j]
# If this is a new character for this substring, increment dist_count.
if(cnt[ord(str1[j]) - 97] == 0):
dist_count += 1
# Increment count of current character
cnt[ord(str1[j]) - 97] += 1
# If distinct character count becomes k,
# then increment result.
if(dist_count == k):
res += 1
if(dist_count > k):
break
return res
# Driver Code
if __name__ == "__main__":
str1 = "abcbaa"
k = 3
print("Total substrings with exactly", k,
"distinct characters : ", end = "")
print(countkDist(str1, k))
# This code is contributed by
# Sairahul Jella
Amazon | OA 2019 | Substrings with exactly K distinct chars
class Solution:
# max-heap
def kClosest(self, points: List[List[int]], K: int) -> List[List[int]]:
h = []
heapq.heapify(h)
# 将所有点的距离放入最小堆
for i, point in enumerate(points):
distance = point[0] ** 2 + point[1] ** 2
heapq.heappush(h, (distance, i))
ans = []
# 弹出最小堆的K个元素
for i in range(K):
p = heapq.heappop(h)
ans.append(points[p[1]])
return ans
Amazon Online Assessment 2 - SDE 1( New Graduate ) 2021 ( Coding - 2 Questions ) - With Solutions
Amazon Online Assessment (OA) - Fetch Items To Display
Amazon | OA 2019 | Subtree with Maximum Average
Throttling Gateway Hackerrank/Citadel
Amazon Online Assessment (OA) - Throttling Gateway 要买会员
Input: logs = [“a1 9 2 3 1”,“g1 act car”,“zo4 4 7”,“ab1 off key dog”,“a8 act zoo”]
Output: [“g1 act car”,“a8 act zoo”,“ab1 off key dog”,“a1 9 2 3 1”,“zo4 4 7”]
Amazon | OA 2019 | Optimal Utilization
Given 2 lists a and b. Each element is a pair of integers where the first integer represents the unique id and the second integer represents a value. Your task is to find an element from a and an element form b such that the sum of their values is less or equal to target and as close to target as possible. Return a list of ids of selected elements. If no pair is possible, return an empty list.
Given two non-empty binary trees s and t, check whether tree t has exactly the same structure and node values with a subtree of s. A subtree of s is a tree consists of a node in s and all of this node’s descendants. The tree s could also be considered as a subtree of itself.
Amazon OA | storage optimization
Amazon Leadership Principles
Question:
I believe planning ahead would lead to a better outcome once in a while people find errors in my work
<-----o------------------------------------o---------------------------------------o----------------------------------o-----------------------------------o---->
*most like me *more like me *default *more like me *most like me
60道workstyle assessment