Maximice los paseos con los tokens dados Java
from heapq import *
def process(arr,n):
count = 0
heap = []
for i in range(len(arr)):
heappush(heap,(arr[i],-(len(arr)-i))) # Constructing min-heap with second index as negative of maximum number of rides
while(n>0 and heap):
cost,no_of_rides = heappop(heap)
no_of_rides = -1 * no_of_rides # Changing maximum no_of_rides from negative to positive
div = n//cost
# If the amount of money is not sufficient to calculate the last number of rides user could take
if(div<no_of_rides):
count += div
break
# Else decrement the number of tokens by minimum cost * maximum no_of_rides
else:
count += no_of_rides
n -= no_of_rides*cost
return count;
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