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++ 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 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);
}
} |
# Python3 program for the above approach # Function to calculate the minimum cost # required to reach the end of Line def 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 & M N,K,M = 10 , 3 , 4
# Given arrays a = [ 0 , 1 , 4 , 6 ]
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 shinjanpatra |
// C# program for the above approach using 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. |
<script> // Javascript program for the above approach // Function to calculate the minimum cost // required to reach the end of Line function 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 & M var N = 10, K = 3, M = 4;
// Given arrays var a = [0, 1, 4, 6 ];
var 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 rutvik_56. </script> |
-1
Time Complexity: O(N * logN)
Auxiliary Space: O(N)