IBM OA 在 10 分鐘內有 2 個簡單的代碼問題。老師只用了 10 分鐘就完成了這兩個問題。讓我們一起來看看吧。IBM 社招的 OA，10 分鐘 2 道題目，兩道題都是 Easy 題，老師只花了 10 分鐘就寫完兩道，讓我們一起來看下吧.
Question 1
There is a list of items in the shopping cart, each having a cost associated with it.
There are n items, the cost of the ith item is i dollars and m items have already been bought represented in the array arr. Currently, there are k dollars, find the maximum number of distinct items one can have in total after purchasing any number of items from that money.
为了解决这个问题，我们必须在给定约束的情况下可以购买的不同物品的最大数量。考虑到已经购买的物品数量和有限的支出，我们的目标应该是最大限度地增加我们的不同购买总数。
To solve this problem, so we need identify the maximum number of distinct items we can purchase given our constraints. Due to given a set number of alreadypurchased items and limited money to spend, the goal should be to maximize our total number of distinct purchases.
Steps to solve the problem:
- Identify the Remaining Items: Out of all available products, we need to determine which are still available for purchase (i.e. not yet purchased by someone else).
- Sort the Remaining Items by Cost: In order to maximize our purchase capacity, it’s often beneficial to start buying cheaper items first.
- Purchase Items Until the Budget Runs Out: Start purchasing cheap items until either we run out of funds or cannot purchase any more without exceeding budget constraints.
- Calculating the Total Distinct Items: Add up the quantities of already purchased and newly acquired items, because of arriving at an estimate for total distinct items.
Python Code
def findMaxDistinctItems(n, arr, k):
# Convert arr to a set for O(1) average time complexity lookups
already_bought = set(arr)
# Calculate remaining items
remaining_items = [i for i in range(1, n+1) if i not in already_bought]
# Sort remaining items based on their cost (which is equal to their value)
remaining_items.sort()
total_distinct_items = len(already_bought)
current_budget = k
# Try to buy items starting from the cheapest one
for item_cost in remaining_items:
if current_budget >= item_cost:
current_budget -= item_cost
total_distinct_items += 1
else:
break
return total_distinct_items
# Example usage
n = 10
m = 3
k = 10
arr = [1, 3, 8]
print(findMaxDistinctItems(n, arr, k)) # Output: 5Question 2
A Domain Name System (DNS) translates domain names to IP addresses which are then used by browsers to load internet resources. For quicker DNS lookups,browsers often store a number of recent DNS queries in a DNS cache. Retrieving data from the cache is often faster than retrieving it from a DNS server. Thistask aims to simulate DNS resolution and determine the time taken to process different URLs.
為了解決這個問題，我們可以使用最近最少使用（LRU）緩存策略。 這種方法確保了當緩存達到其最大大小，並請求新的 URL 時，會驅逐最近使用最少的 URL，為新 URL 騰出空間。 我們可以使用 Python 的集合。 OrderedDict 有效地實現了這一點。
To address this issue, the Least Recently Used (LRU) cache strategy can provide an effective solution. When our cache reaches its maximum capacity and anew URL request comes through, when evicting older URLs to make room for newer ones is decided upon in accordance with this plan – we can usePython collections.OrderedDict to implement this efficiently.
Steps to solve the problem:
- Initialize the Cache: Use an
OrderedDictto maintain the cache where the keys are URLs and values can be arbitrary since we only need to track the presence and order of URLs. - Process Each URL: For each URL in the list:
- If the URL is already in the cache, it takes
cache_timeunits to resolve, and we need to move this URL to the end to mark it as recently used. - Or if the URL is not in the cache, it takes
server_timeunits to resolve. If the cache is full, remove the least recently used URL. Add the new URL to the cache.
- If the URL is already in the cache, it takes
- Calculate Total Time: Sum the time taken to resolve each URL.
Python Code
from collections import OrderedDict
def dns_resolution_time(cache_size, cache_time, server_time, urls):
cache = OrderedDict()
total_time = 0
for url in urls:
if url in cache:
# URL is in cache
total_time += cache_time
# Move the accessed URL to the end to mark it as recently used
cache.move_to_end(url)
else:
# URL is not in cache
total_time += server_time
if len(cache) >= cache_size:
# Cache is full, remove the least recently used item
cache.popitem(last=False)
# Add the new URL to the cache
cache[url] = None # The value doesn't matter, we're just using keys for URLs
return total_time
# Example usage
cache_size = 3
cache_time = 2
server_time = 5
urls = [
"http://www.hackerrank.com",
"http://www.google.com",
"http://www.yahoo.com",
"http://www.gmail.com",
"http://www.yahoo.com",
"http://www.hackerrank.com",
"http://www.gmail.com"
]
print(dns_resolution_time(cache_size, cache_time, server_time, urls)) # Output: 24Read More
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經過我們的面試輔助，OA 代寫，候選人通過這些面試題的解析和溝通，面試官不僅瞭解了候選人的程式設計能力，也看到了我在解決問題過程中清晰的思路和有效的溝通技巧。 這些不僅有助於應對 IBM 的面試，同時也能提升我們解決實際程式設計問題的能力。 祝大家面試順利！
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