Add solution #3147

Author: JoshCrozier

README.md

@@ -2660,6 +2660,7 @@
 3135|[Equalize Strings by Adding or Removing Characters at Ends](./solutions/3135-equalize-strings-by-adding-or-removing-characters-at-ends.js)|Medium|
 3136|[Valid Word](./solutions/3136-valid-word.js)|Easy|
 3141|[Maximum Hamming Distances](./solutions/3141-maximum-hamming-distances.js)|Hard|
+3147|[Taking Maximum Energy From the Mystic Dungeon](./solutions/3147-taking-maximum-energy-from-the-mystic-dungeon.js)|Medium|
 3151|[Special Array I](./solutions/3151-special-array-i.js)|Easy|
 3155|[Maximum Number of Upgradable Servers](./solutions/3155-maximum-number-of-upgradable-servers.js)|Medium|
 3157|[Find the Level of Tree with Minimum Sum](./solutions/3157-find-the-level-of-tree-with-minimum-sum.js)|Medium|

solutions/3147-taking-maximum-energy-from-the-mystic-dungeon.js

@@ -0,0 +1,39 @@
+/**
+ * 3147. Taking Maximum Energy From the Mystic Dungeon
+ * https://leetcode.com/problems/taking-maximum-energy-from-the-mystic-dungeon/
+ * Difficulty: Medium
+ *
+ * In a mystic dungeon, n magicians are standing in a line. Each magician has an attribute
+ * that gives you energy. Some magicians can give you negative energy, which means taking
+ * energy from you.
+ *
+ * You have been cursed in such a way that after absorbing energy from magician i, you will
+ * be instantly transported to magician (i + k). This process will be repeated until you
+ * reach the magician where (i + k) does not exist.
+ *
+ * In other words, you will choose a starting point and then teleport with k jumps until you
+ * reach the end of the magicians' sequence, absorbing all the energy during the journey.
+ *
+ * You are given an array energy and an integer k. Return the maximum possible energy you can gain.
+ *
+ * Note that when you are reach a magician, you must take energy from them, whether it is negative
+ * or positive energy.
+ */
+
+/**
+ * @param {number[]} energy
+ * @param {number} k
+ * @return {number}
+ */
+var maximumEnergy = function(energy, k) {
+  const n = energy.length;
+  const dp = new Array(n).fill(0);
+  let result = -Infinity;
+
+  for (let i = n - 1; i >= 0; i--) {
+    dp[i] = energy[i] + (i + k < n ? dp[i + k] : 0);
+    result = Math.max(result, dp[i]);
+  }
+
+  return result;
+};