# Minimum possible travel cost among N cities

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

**i**city would cost of

^{th}**C[i]**dollars to travel

**1**unit of distance. In other words, cost to travel from the

**i**city to the

^{th}**j**city is

^{th}**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

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**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 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` |

## 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>` |

**Output:**

18

**Time Complexity:** O(N)**Auxiliary Space: **O(1)