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...ode_java/src/main/java/LeetCodeJava/DynamicProgramming/MinimumNumberOfRefuelingStops.java
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| package LeetCodeJava.DynamicProgramming; | ||
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| // https://leetcode.com/problems/minimum-number-of-refueling-stops/description/ | ||
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| import java.util.Collections; | ||
| import java.util.PriorityQueue; | ||
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| /** | ||
| * 871. Minimum Number of Refueling Stops Hard Topics Companies A car travels from a starting | ||
| * position to a destination which is target miles east of the starting position. | ||
| * | ||
| * <p>There are gas stations along the way. The gas stations are represented as an array stations | ||
| * where stations[i] = [positioni, fueli] indicates that the ith gas station is positioni miles east | ||
| * of the starting position and has fueli liters of gas. | ||
| * | ||
| * <p>The car starts with an infinite tank of gas, which initially has startFuel liters of fuel in | ||
| * it. It uses one liter of gas per one mile that it drives. When the car reaches a gas station, it | ||
| * may stop and refuel, transferring all the gas from the station into the car. | ||
| * | ||
| * <p>Return the minimum number of refueling stops the car must make in order to reach its | ||
| * destination. If it cannot reach the destination, return -1. | ||
| * | ||
| * <p>Note that if the car reaches a gas station with 0 fuel left, the car can still refuel there. | ||
| * If the car reaches the destination with 0 fuel left, it is still considered to have arrived. | ||
| * | ||
| * <p>Example 1: | ||
| * | ||
| * <p>Input: target = 1, startFuel = 1, stations = [] Output: 0 Explanation: We can reach the target | ||
| * without refueling. Example 2: | ||
| * | ||
| * <p>Input: target = 100, startFuel = 1, stations = [[10,100]] Output: -1 Explanation: We can not | ||
| * reach the target (or even the first gas station). Example 3: | ||
| * | ||
| * <p>Input: target = 100, startFuel = 10, stations = [[10,60],[20,30],[30,30],[60,40]] Output: 2 | ||
| * Explanation: We start with 10 liters of fuel. We drive to position 10, expending 10 liters of | ||
| * fuel. We refuel from 0 liters to 60 liters of gas. Then, we drive from position 10 to position 60 | ||
| * (expending 50 liters of fuel), and refuel from 10 liters to 50 liters of gas. We then drive to | ||
| * and reach the target. We made 2 refueling stops along the way, so we return 2. | ||
| * | ||
| * <p>Constraints: | ||
| * | ||
| * <p>1 <= target, startFuel <= 109 0 <= stations.length <= 500 1 <= positioni < positioni+1 < | ||
| * target 1 <= fueli < 109 | ||
| */ | ||
| public class MinimumNumberOfRefuelingStops { | ||
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| // V0 | ||
| // TODO : implement | ||
| // public int minRefuelStops(int target, int startFuel, int[][] stations) { | ||
| // | ||
| // } | ||
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| // V1 | ||
| // IDEA : DP | ||
| // https://leetcode.com/problems/minimum-number-of-refueling-stops/editorial/ | ||
| public int minRefuelStops_1(int target, int startFuel, int[][] stations) { | ||
| int N = stations.length; | ||
| long[] dp = new long[N + 1]; | ||
| dp[0] = startFuel; | ||
| for (int i = 0; i < N; ++i) | ||
| for (int t = i; t >= 0; --t) | ||
| if (dp[t] >= stations[i][0]) dp[t + 1] = Math.max(dp[t + 1], dp[t] + (long) stations[i][1]); | ||
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| for (int i = 0; i <= N; ++i) if (dp[i] >= target) return i; | ||
| return -1; | ||
| } | ||
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| // V2 | ||
| // IDEA : HEAP | ||
| // https://leetcode.com/problems/minimum-number-of-refueling-stops/editorial/ | ||
| public int minRefuelStops(int target, int tank, int[][] stations) { | ||
| // pq is a maxheap of gas station capacities | ||
| PriorityQueue<Integer> pq = new PriorityQueue(Collections.reverseOrder()); | ||
| int ans = 0, prev = 0; | ||
| for (int[] station : stations) { | ||
| int location = station[0]; | ||
| int capacity = station[1]; | ||
| tank -= location - prev; | ||
| while (!pq.isEmpty() && tank < 0) { // must refuel in past | ||
| tank += pq.poll(); | ||
| ans++; | ||
| } | ||
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| if (tank < 0) return -1; | ||
| pq.offer(capacity); | ||
| prev = location; | ||
| } | ||
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| // Repeat body for station = (target, inf) | ||
| { | ||
| tank -= target - prev; | ||
| while (!pq.isEmpty() && tank < 0) { | ||
| tank += pq.poll(); | ||
| ans++; | ||
| } | ||
| if (tank < 0) return -1; | ||
| } | ||
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| return ans; | ||
| } | ||
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| } |