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:
- Start with the first city with a cost of C[1].
- 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.
- Calculate cost as abs(j – i) * C[i] and add it to the total cost so far.
- 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 lastint 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 codeint main(){ vector<int> cost{ 4, 7, 8, 3, 4 }; int n = cost.size(); cout << minCost(cost, n); return 0;} |
Java
// Java implementation of the approachclass GFG{// Function to return the minimum cost to// travel from the first city to the laststatic 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 codepublic 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 lastdef 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 codecost = [ 4, 7, 8, 3, 4]n = len(cost)print(minCost(cost, n))# This code is contributed by Mohit Kumar |
C#
// C# implementation of the approachusing System;class GFG{// Function to return the minimum cost to// travel from the first city to the laststatic 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 codepublic 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 |
Javascript
<script>// javascript implementation of the approach // Function to return the minimum cost to // travel from the first city to the last function minCost(cost , n) { // To store the total cost var totalCost = 0; // Start from the first city var boardingBus = 0; for (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 var cost = [ 4, 7, 8, 3, 4 ]; var n = cost.length; document.write(minCost(cost, n));// This code contributed by umadevi9616</script> |
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