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_queuestruct 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 Linevoid 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 Programint 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 approachimport 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); }} |
Python3
# Python3 program for the above approach# Function to calculate the minimum cost# required to reach the end of Linedef minCost(N, K, M, a, b): # Checks if possible to # reach end or not flag = True # Stores the stations and # respective rate of fuel map = {} for i in range(M): map[a[i]] = b[i] if (i == M - 1 and K < N - a[i]): flag = False break elif (i < M - 1 and K < a[i + 1] - a[i]): flag = False break if (flag == 0): print(-1) return # Stores the station index and cost of fuel and # litres of petrol which is being fueled pq = [] cost = 0 flag = False # Iterate through the entire line for i in range(N): # Check if there is a station at current index if (i in map): arr = [ i, map[i], 0 ] pq.append(arr) pq.sort() # Remove all the stations where # fuel cannot be pumped while (len(pq) > 0 and pq[-1][2] == K): pq.pop() # If there is no station left to fill fuel # in tank, it is not possible to reach end if (len(pq) == 0): flag = True break # Stores the best station # visited so far best_bunk = pq[len(pq)-1] pq.pop() # Pump fuel from the best station cost += best_bunk[1] # Update the count of litres # taken from that station best_bunk[2] += 1 # Update the bunk in queue pq.append(best_bunk) pq.sort() if (flag): print( -1) return # Print the cost print(cost)# Driven Program# Given value of N, K & MN,K,M = 10,3,4# Given arraysa = [0, 1, 4, 6]b = [5, 2, 2, 4]# Function call to calculate minimum# cost to reach end of the lineminCost(N, K, M, a, b)# This code is contributed by shinjanpatra |
C#
// C# program for the above approachusing System;using System.Linq;using System.Collections.Generic;class GFG { // 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 bool flag = true; // Stores the stations and // respective rate of fuel Dictionary<int, int> map = new Dictionary<int, int>(); 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) { Console.WriteLine("-1"); return; } // Stores the station index and cost of fuel and // litres of petrol which is being fueled List<int[]> pq = new List<int[]>(); 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[i], 0 }); pq = pq.OrderBy(c => c[1]).ToList(); // Remove all the stations where // fuel cannot be pumped while (pq.Count > 0 && pq[0][2] == K) pq.RemoveAt(0); // If there is no station left to fill fuel // in tank, it is not possible to reach end if (pq.Count == 0) { flag = true; break; } // Stores the best station // visited so far int[] best_bunk = pq[0]; // 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) { Console.WriteLine("-1"); return; } // Print the cost Console.WriteLine(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); }}// This code is contributed by phasing17. |
Javascript
<script>// Javascript program for the above approach// Function to calculate the minimum cost// required to reach the end of Linefunction minCost(N, K, M, a, b){ // Checks if possible to // reach end or not var flag = true; // Stores the stations and // respective rate of fuel var map = new Map(); for (var i = 0; i < M; i++) { map.set(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) { document.write(-1); return; } // Stores the station index and cost of fuel and // litres of petrol which is being fueled var pq = []; var cost = 0; flag = false; // Iterate through the entire line for (var i = 0; i < N; i++) { // Check if there is a station at current index if (map.has(i)) { var arr = [ i, map.get(i), 0 ]; pq.push(arr); } pq.sort(); // Remove all the stations where // fuel cannot be pumped while (pq.length > 0 && pq[pq.length-1][2] == K) pq.pop(); // If there is no station left to fill fuel // in tank, it is not possible to reach end if (pq.length == 0) { flag = true; break; } // Stores the best station // visited so far var best_bunk = pq[pq.length-1]; 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); pq.sort(); } if (flag) { document.write( -1 + "<br>"); return; } // Print the cost document.write(cost + "<br>");}// Driven Program// Given value of N, K & Mvar N = 10, K = 3, M = 4;// Given arraysvar a = [0, 1, 4, 6 ];var b = [5, 2, 2, 4];// Function call to calculate minimum// cost to reach end of the lineminCost(N, K, M, a, b);// This code is contributed by rutvik_56.</script> |
Output:
-1
Time Complexity: O(N * logN)
Auxiliary Space: O(N)
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