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Minimum possible travel cost among N cities
  • Difficulty Level : Hard
  • Last Updated : 23 Dec, 2019

There are N cities situated on a straight road and each is separated by a distance of 1 unit. You have to reach the (N + 1)th city by boarding a bus. The ith city would cost of C[i] dollars to travel 1 unit of distance. In other words, cost to travel from the ith city to the jth city is abs(i – j ) * C[i] dollars. The task is to find the minimum cost to travel from city 1 to city (N + 1) i.e. beyond the last city.

Examples:

Input: C[] = {3, 5, 4}
Output: 9
The bus boarded from the first city has the minimum
cost of all so it will be used to travel (N + 1) unit.

Input: C[] = {4, 7, 8, 3, 4}
Output: 18
Board the bus at the first city then change
the bus at the fourth city.
(3 * 4) + (2 * 3) = 12 + 6 = 18

Approach: The approach is very simple, just travel by the bus which has the lowest cost so far. Whenever a bus with an even lower cost is found, change the bus from that city. Following are the steps to solve:



  1. Start with the first city with a cost of C[1].
  2. Travel to the next city until a city j having cost less than the previous city (by which we are travelling, let’s say city i) is found.
  3. Calculate cost as abs(j – i) * C[i] and add it to the total cost so far.
  4. Repeat the previous steps until all the cities have been traversed.

Below is the implementation of the above approach:

C++




// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
  
// Function to return the minimum cost to
// travel from the first city to the last
int minCost(vector<int>& cost, int n)
{
  
    // To store the total cost
    int totalCost = 0;
  
    // Start from the first city
    int boardingBus = 0;
  
    for (int i = 1; i < n; i++) {
  
        // 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
int main()
{
    vector<int> cost{ 4, 7, 8, 3, 4 };
    int n = cost.size();
  
    cout << minCost(cost, n);
  
    return 0;
}


Java




// Java implementation of the approach
class GFG
{
  
// Function to return the minimum cost to
// travel from the first city to the last
static int minCost(int []cost, int n)
{
  
    // To store the total cost
    int totalCost = 0;
  
    // Start from the first city
    int boardingBus = 0;
  
    for (int i = 1; i < n; i++)
    {
  
        // 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
public static void main(String[] args)
{
    int []cost = { 4, 7, 8, 3, 4 };
    int n = cost.length;
  
    System.out.print(minCost(cost, n));
}
}
  
// This code is contributed by PrinciRaj1992


Python3




# 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))
  
# This code is contributed by Mohit Kumar 


C#




// C# implementation of the approach
using System;
  
class GFG
{
  
// Function to return the minimum cost to
// travel from the first city to the last
static int minCost(int []cost, int n)
{
  
    // To store the total cost
    int totalCost = 0;
  
    // Start from the first city
    int boardingBus = 0;
  
    for (int i = 1; i < n; i++)
    {
  
        // 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
public static void Main(String[] args)
{
    int []cost = { 4, 7, 8, 3, 4 };
    int n = cost.Length;
  
    Console.Write(minCost(cost, n));
}
}
  
// This code is contributed by 29AjayKumar


Output:

18

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