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Minimize cost to reach end of an N-length straight path
  • Difficulty Level : Medium
  • Last Updated : 19 Mar, 2021
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Given an integer K denoting the fuel tank capacity of a car running at the cost of 1 liter / mtr on a straight path of length N meters and two arrays a[] and b[], each of size M, where a[i] denotes the location of the ith station and b[i] denotes the cost of 1-liter of fuel at that station. The task is to find the minimum cost required to reach the end of the line, starting from 0. If it is not possible to reach the end, then print -1.

Examples:

Input: N = 10, K = 10, M = 2, a[] = {0, 1}, b[] = {5, 2}
Output: 23
Explanation:
At the 0th location, fill the car tank with 1 liter of fuel at cost 5. Then, at 1st location, fill 9 liters of fuel at cost 18. 
Therefore, the minimum cost required is 23.

Input: N = 10, K = 5, M = 3, a[] = {0, 3, 5}, b[] = {5, 9, 3}
Output: 40

Approach: The idea is to use Priority Queue and HashMap to store the fuel stations in order to get the fuel station with minimum cost. Follow the steps below to solve the problem:



  • Initialize a HashMap, say map, to store the index of the station and its respective rate of fuel.
  • Initialize a Priority Queue, say pq, to store the station’s index and cost of fuel and liters of fuel that is being filled.
  • Initialize two variables, say cost, to store the minimum cost required, and set flag = false to check if there are any filling stations or not.
  • Iterate over the range [1, N]:
    • Check if there is a station or not. If found to be true, then insert it into pq.
    • Remove all the stations where fuel cannot be filled.
    • If pq is empty, then it is not possible to reach the end of line. Therefore, set flag = true.
    • Store the least cost station and update the cost and the number of liters pumped from that particular station.
    • Insert it again into pq.
  • If flag is true, then print “-1”. It means there are no filling stations. Therefore, it is not possible to reach end of line.
  • Otherwise, print the minimum cost to reach the end of the line.

Below is the implementation of the above approach:

C++




// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
 
// For priority_queue
struct Compare {
  bool operator()(array<int, 3> a, array<int, 3> b)
  {
    return a[1] > b[1];
  }
};
 
// Function to calculate the minimum cost
// required to reach the end of Line
void minCost(int N, int K, int M, int a[], int b[])
{
  // Checks if possible to
  // reach end or not
  bool flag = true;
 
  // Stores the stations and
  // respective rate of fuel
  unordered_map<int, int> map;
  for (int i = 0; i < M; i++) {
    map[a[i]] = b[i];
    if (i == M - 1 && K < N - a[i]) {
      flag = false;
      break;
    }
    else if (i < M - 1 && K < a[i + 1] - a[i]) {
      flag = false;
      break;
    }
  }
 
  if (!flag) {
    cout << -1;
    return;
  }
 
  // Stores the station index and cost of fuel and
  // litres of petrol which is being fueled
  priority_queue<array<int, 3>, vector<array<int, 3> >,
  Compare>
    pq;
  int cost = 0;
  flag = false;
 
  // Iterate through the entire line
  for (int i = 0; i < N; i++) {
 
    // Check if there is a station at current index
    if (map.find(i) != map.end()) {
      array<int, 3> arr = { i, map[i], 0 };
      pq.push(arr);
    }
 
    // Remove all the stations where
    // fuel cannot be pumped
    while (pq.size() > 0 && pq.top()[2] == K)
      pq.pop();
 
    // If there is no station left to fill fuel
    // in tank, it is not possible to reach end
    if (pq.size() == 0) {
      flag = true;
      break;
    }
 
    // Stores the best station
    // visited so far
    array<int, 3> best_bunk = pq.top();
    pq.pop();
 
    // Pump fuel from the best station
    cost += best_bunk[1];
 
    // Update the count of litres
    // taken from that station
    best_bunk[2]++;
 
    // Update the bunk in queue
    pq.push(best_bunk);
  }
  if (flag) {
    cout << -1 << "\n";
    return;
  }
 
  // Print the cost
  cout << cost << "\n";
}
 
// Driven Program
int main()
{
 
  // Given value of N, K & M
  int N = 10, K = 3, M = 4;
 
  // Given arrays
  int a[] = { 0, 1, 4, 6 };
  int b[] = { 5, 2, 2, 4 };
 
  // Function call to calculate minimum
  // cost to reach end of the line
  minCost(N, K, M, a, b);
 
  return 0;
}
 
// This code is contributed by Kingash.

Java




// Java program for the above approach
 
import java.util.*;
public class Main {
 
    // Function to calculate the minimum cost
    // required to reach the end of Line
    static void minCost(int N, int K, int M,
                        int[] a, int[] b)
    {
        // Checks if possible to
        // reach end or not
        boolean flag = true;
 
        // Stores the stations and
        // respective rate of fuel
        HashMap<Integer, Integer> map = new HashMap<>();
        for (int i = 0; i < M; i++) {
            map.put(a[i], b[i]);
            if (i == M - 1 && K < N - a[i]) {
                flag = false;
                break;
            }
            else if (i < M - 1 && K < a[i + 1] - a[i]) {
                flag = false;
                break;
            }
        }
 
        if (!flag) {
            System.out.println("-1");
            return;
        }
 
        // Stores the station index and cost of fuel and
        // litres of petrol which is being fueled
        PriorityQueue<int[]> pq
            = new PriorityQueue<>((c, d) -> c[1] - d[1]);
        int cost = 0;
        flag = false;
 
        // Iterate through the entire line
        for (int i = 0; i < N; i++) {
 
            // Check if there is a station at current index
            if (map.containsKey(i))
                pq.add(new int[] { i, map.get(i), 0 });
 
            // Remove all the stations where
            // fuel cannot be pumped
            while (pq.size() > 0 && pq.peek()[2] == K)
                pq.poll();
 
            // If there is no station left to fill fuel
            // in tank, it is not possible to reach end
            if (pq.size() == 0) {
                flag = true;
                break;
            }
 
            // Stores the best station
            // visited so far
            int best_bunk[] = pq.poll();
 
            // Pump fuel from the best station
            cost += best_bunk[1];
 
            // Update the count of litres
            // taken from that station
            best_bunk[2]++;
 
            // Update the bunk in queue
            pq.add(best_bunk);
        }
        if (flag) {
            System.out.println("-1");
            return;
        }
 
        // Print the cost
        System.out.println(cost);
    }
 
    // Driver Code
    public static void main(String args[])
    {
        // Given value of N, K & M
        int N = 10, K = 3, M = 4;
 
        // Given arrays
        int a[] = { 0, 1, 4, 6 };
        int b[] = { 5, 2, 2, 4 };
 
        // Function call to calculate minimum
        // cost to reach end of the line
        minCost(N, K, M, a, b);
    }
}

 
 

Output: 
-1

 

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

 

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