INFOSYS EXAM SOLUTIONS
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Repost from INFOSYS EXAM SOLUTIONS
# Python3 implementation of the approach
# Function to return the minimum cost to
# travel from the first city to the last
def minCost(cost, n):
# To store the total cost
totalCost = 0
# Start from the first city
boardingBus = 0
for i in range(1, n):
# If found any city with cost less than
# that of the previous boarded
# bus then change the bus
if (cost[boardingBus] > cost[i]):
# Calculate the cost to travel from
# the currently boarded bus
# till the current city
totalCost += ((i - boardingBus) *
cost[boardingBus])
# Update the currently boarded bus
boardingBus = i
# Finally calculate the cost for the
# last boarding bus till the (N + 1)th city
totalCost += ((n - boardingBus) *
cost[boardingBus])
return totalCost
# Driver code
cost = [ 4, 7, 8, 3, 4]
n = len(cost)
print(minCost(cost, n))
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