Gas Station

 There are n gas stations along a circular route, where the amount of gas at the ith station is gas[i].


You have a car with an unlimited gas tank and it costs cost[i] of gas to travel from the ith station to its next (i + 1)th station. You begin the journey with an empty tank at one of the gas stations.


Given two integer arrays gas and cost, return the starting gas station's index if you can travel around the circuit once in the clockwise direction, otherwise return -1. If there exists a solution, it is guaranteed to be unique.


 


Example 1:


Input: gas = [1,2,3,4,5], cost = [3,4,5,1,2]

Output: 3

Explanation:

Start at station 3 (index 3) and fill up with 4 unit of gas. Your tank = 0 + 4 = 4

Travel to station 4. Your tank = 4 - 1 + 5 = 8

Travel to station 0. Your tank = 8 - 2 + 1 = 7

Travel to station 1. Your tank = 7 - 3 + 2 = 6

Travel to station 2. Your tank = 6 - 4 + 3 = 5

Travel to station 3. The cost is 5. Your gas is just enough to travel back to station 3.

Therefore, return 3 as the starting index.

Example 2:


Input: gas = [2,3,4], cost = [3,4,3]

Output: -1

Explanation:

You can't start at station 0 or 1, as there is not enough gas to travel to the next station.

Let's start at station 2 and fill up with 4 unit of gas. Your tank = 0 + 4 = 4

Travel to station 0. Your tank = 4 - 3 + 2 = 3

Travel to station 1. Your tank = 3 - 3 + 3 = 3

You cannot travel back to station 2, as it requires 4 unit of gas but you only have 3.

Therefore, you can't travel around the circuit once no matter where you start.

 

class Solution {
    public int canCompleteCircuit(int[] gas, int[] cost) {

        int n = gas.length;

        int totalgas = 0;   // Total amount of gas available in all stations
        int totalcost = 0;  // Total amount of gas required to travel through all stations

        // Calculate total gas available
        for (int val : gas) {
            totalgas += val;
        }

        // Calculate total cost (gas needed) for the full trip
        for (int val : cost) {
            totalcost += val;
        }

        // If total gas is less than total cost, we can never complete the circuit
        if (totalgas < totalcost) {
            return -1;
        }

        int start = 0;         // Starting station index
        int currentgas = 0;    // Current gas in tank during travel

        // Traverse each station
        for (int i = 0; i < n; i++) {

            // Add gas from current station and subtract cost to go to next station
            currentgas = currentgas + gas[i] - cost[i];

            // If currentgas becomes negative, we cannot reach next station from 'start'
            if (currentgas < 0) {
                // So we reset the starting station to the next station
                start = i + 1;
                // Reset the current gas in tank as we are choosing new start
                currentgas = 0;
            }
        }

        // The point where we can complete the full circle
        return start;
    }
}

Comments

Post a Comment

Popular posts from this blog

Two Sum II - Input Array Is Sorted

 Given a 1-indexed array of integers numbers that is already sorted in non-decreasing order , find two numbers such that they add up to a specific target number. Let these two numbers be numbers[index1] and numbers[index2] where 1 <= index1 < index2 <= numbers.length. Return the indices of the two numbers, index1 and index2, added by one as an integer array [index1, index2] of length 2. The tests are generated such that there is exactly one solution. You may not use the same element twice. Your solution must use only constant extra space .   Example 1: Input: numbers = [2,7,11,15], target = 9 Output: [1,2] Explanation: The sum of 2 and 7 is 9. Therefore, index1 = 1, index2 = 2. We return [1, 2]. Example 2: Input: numbers = [2,3,4], target = 6 Output: [1,3] Explanation: The sum of 2 and 4 is 6. Therefore index1 = 1, index2 = 3. We return [1, 3]. Example 3: Input: numbers = [-1,0], target = -1 Output: [1,2] Explanation: The sum of -1 and 0 is -1. Therefore index1 = 1,...
Read more

Comparable Vs. Comparator in Java

  In Java , Comparable and Comparator both are interfaces used to sort objects. Comparable It defines  natural ordering  within a class. It is implemented  inside  the class being sorted. It uses the compareTo() method. It allows sorting  by one one . Comparable provides a  single sorting sequence . In other words, we can sort the collection on the basis of a single element such as id, name, and price. Comparable  affects the original class , i.e., the actual class is modified. Comparable is present in  java.lang  package. We can sort the list elements of Comparable type by  Collections.sort(List)  method. Only one compareTo() method can be implemented, restricting flexibility. Comparator It defines  custom sorting logic It is implemented  in a separate class  or using lambda expressions . It uses the compare() method. It allows sorting  by multiple criteria . The Comparator provides  multiple sorting...
Read more

Increasing Triplet Subsequence

 Given an integer array nums, return true if there exists a triple of indices (i, j, k) such that i < j < k and nums[i] < nums[j] < nums[k]. If no such indices exists, return false.   Example 1: Input: nums = [1,2,3,4,5] Output: true Explanation: Any triplet where i < j < k is valid. Example 2: Input: nums = [5,4,3,2,1] Output: false Explanation: No triplet exists. Example 3: Input: nums = [2,1,5,0,4,6] Output: true Explanation: The triplet (3, 4, 5) is valid because nums[3] == 0 < nums[4] == 4 < nums[5] == 6. Brute Force : Failed for some testcase class Solution {     public boolean increasingTriplet ( int [] nums ) {         int n = nums . length ;         int i = 0 ;         int j =i+ 1 ;         int k =j+ 1 ;         while (i <n && j<n && k<n)         {         if (nums...
Read more