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Concise algorithm solution in Speedy category for Bigger Price by Igor_Sekretarev
from heapq import heappop, heappush, heappushpop
def bigger_price(limit: int, data: list) -> list:
items = []
for item in data:
if len(items) < limit:
heappush(items, (item['price'], item))
else:
heappushpop(items, (item['price'], item))
most_expensive = []
while items:
_, item = heappop(items)
most_expensive.append(item)
most_expensive.reverse()
return most_expensive
if __name__ == '__main__':
from pprint import pprint
print('Example:')
pprint(bigger_price(2, [
{"name": "bread", "price": 100},
{"name": "wine", "price": 138},
{"name": "meat", "price": 15},
{"name": "water", "price": 1}
]))
# These "asserts" using for self-checking and not for auto-testing
assert bigger_price(2, [
{"name": "bread", "price": 100},
{"name": "wine", "price": 138},
{"name": "meat", "price": 15},
{"name": "water", "price": 1}
]) == [
{"name": "wine", "price": 138},
{"name": "bread", "price": 100}
], "First"
assert bigger_price(1, [
{"name": "pen", "price": 5},
{"name": "whiteboard", "price": 170}
]) == [{"name": "whiteboard", "price": 170}], "Second"
print('Done! Looks like it is fine. Go and check it')
April 24, 2021
