Suppose in a shoe shop there are n different shoe with different sizes present in an array called size and another list of pairs for m customers called demand is given, where demand[i] contains (shoe_size, money) so a customer with demand i has a demand of shoe whose size is shoe_size and he/she can pay given amount of money. We have to find how much money the shopkeeper can earn by selling these shoes.
So, if the input is like shoes = [2,3,4,5,6,8,7,6,5,18] demand = [(6,55), (6,45), (6,55), (4,40), (18,60), (10,50)], then the output will be 200 because
first one will buy shoe with size 6 at cost 55
second one will buy shoe with size 6 at cost 45
There is no shoe with size 6 in the stock
fourth one will buy shoe with size 4 at cost 40
fifth one will buy shoe with size 18 at cost 60
sixth one will not get shoe because there is no shoe of size 10
Total earn 55 + 45 + 40 + 60 = 200.
To solve this, we will follow these steps −
- n := size of demand
- sizes := a map containing frequencies of shoes based on size
- earn := 0
- for i in range 0 to n - 1, do
- (sz, price) := demand[i]
- if shoe with size sz is present in sizes, then
- sizes[sz] := sizes[sz] - 1
- earn := earn + price
- return earn
Example
Let us see the following implementation to get better understanding
from collections import Counter
def solve(shoes, demand):
n = len(demand)
sizes = Counter(shoes)
earn = 0
for i in range(n):
sz, price = demand[i]
if sizes[sz]:
sizes[sz] -= 1
earn += price
return earn
shoes = [2,3,4,5,6,8,7,6,5,18]
demand = [(6,55), (6,45), (6,55), (4,40), (18,60), (10,50)]
print(solve(shoes, demand))Input
[2,3,4,5,6,8,7,6,5,18], [(6,55), (6,45), (6,55), (4,40), (18,60), (10,50)]
Output
200