Add solution #871

Author: JoshCrozier

README.md

@@ -678,6 +678,7 @@
 868|[Binary Gap](./0868-binary-gap.js)|Easy|
 869|[Reordered Power of 2](./0869-reordered-power-of-2.js)|Medium|
 870|[Advantage Shuffle](./0870-advantage-shuffle.js)|Medium|
+871|[Minimum Number of Refueling Stops](./0871-minimum-number-of-refueling-stops.js)|Hard|
 872|[Leaf-Similar Trees](./0872-leaf-similar-trees.js)|Easy|
 873|[Length of Longest Fibonacci Subsequence](./0873-length-of-longest-fibonacci-subsequence.js)|Medium|
 875|[Koko Eating Bananas](./0875-koko-eating-bananas.js)|Medium|

solutions/0871-minimum-number-of-refueling-stops.js

@@ -0,0 +1,48 @@
+/**
+ * 871. Minimum Number of Refueling Stops
+ * https://leetcode.com/problems/minimum-number-of-refueling-stops/
+ * Difficulty: Hard
+ *
+ * A car travels from a starting position to a destination which is target miles east of the
+ * starting position.
+ *
+ * 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.
+ *
+ * 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.
+ *
+ * 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.
+ *
+ * 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.
+ */
+
+/**
+ * @param {number} target
+ * @param {number} startFuel
+ * @param {number[][]} stations
+ * @return {number}
+ */
+var minRefuelStops = function(target, startFuel, stations) {
+  const dp = [startFuel];
+  const stationCount = stations.length;
+
+  for (let i = 0; i < stationCount; i++) {
+    dp.push(0);
+    for (let stops = i; stops >= 0; stops--) {
+      if (dp[stops] >= stations[i][0]) {
+        dp[stops + 1] = Math.max(dp[stops + 1], dp[stops] + stations[i][1]);
+      }
+    }
+  }
+
+  for (let stops = 0; stops <= stationCount; stops++) {
+    if (dp[stops] >= target) return stops;
+  }
+
+  return -1;
+};