Add solution #1696

JoshCrozier · JoshCrozier · commit df95f977fe5f · 2025-04-26T15:17:09.000-05:00
diff --git a/README.md b/README.md
@@ -1,4 +1,4 @@
-# 1,482 LeetCode solutions in JavaScript
+# 1,483 LeetCode solutions in JavaScript
 
 [https://leetcodejavascript.com](https://leetcodejavascript.com)
 
@@ -1307,6 +1307,7 @@
 1691|[Maximum Height by Stacking Cuboids](./solutions/1691-maximum-height-by-stacking-cuboids.js)|Hard|
 1694|[Reformat Phone Number](./solutions/1694-reformat-phone-number.js)|Easy|
 1695|[Maximum Erasure Value](./solutions/1695-maximum-erasure-value.js)|Medium|
+1696|[Jump Game VI](./solutions/1696-jump-game-vi.js)|Medium|
 1716|[Calculate Money in Leetcode Bank](./solutions/1716-calculate-money-in-leetcode-bank.js)|Easy|
 1718|[Construct the Lexicographically Largest Valid Sequence](./solutions/1718-construct-the-lexicographically-largest-valid-sequence.js)|Medium|
 1726|[Tuple with Same Product](./solutions/1726-tuple-with-same-product.js)|Medium|
diff --git a/solutions/1696-jump-game-vi.js b/solutions/1696-jump-game-vi.js
@@ -0,0 +1,46 @@
+/**
+ * 1696. Jump Game VI
+ * https://leetcode.com/problems/jump-game-vi/
+ * Difficulty: Medium
+ *
+ * You are given a 0-indexed integer array nums and an integer k.
+ *
+ * You are initially standing at index 0. In one move, you can jump at most k steps forward without
+ * going outside the boundaries of the array. That is, you can jump from index i to any index in
+ * the range [i + 1, min(n - 1, i + k)] inclusive.
+ *
+ * You want to reach the last index of the array (index n - 1). Your score is the sum of all
+ * nums[j] for each index j you visited in the array.
+ *
+ * Return the maximum score you can get.
+ */
+
+/**
+ * @param {number[]} nums
+ * @param {number} k
+ * @return {number}
+ */
+var maxResult = function(nums, k) {
+  const n = nums.length;
+  const maxScores = new Array(n).fill(-Infinity);
+  const deque = [];
+
+  maxScores[0] = nums[0];
+  deque.push(0);
+
+  for (let i = 1; i < n; i++) {
+    while (deque.length && deque[0] < i - k) {
+      deque.shift();
+    }
+
+    maxScores[i] = nums[i] + maxScores[deque[0]];
+
+    while (deque.length && maxScores[i] >= maxScores[deque[deque.length - 1]]) {
+      deque.pop();
+    }
+
+    deque.push(i);
+  }
+
+  return maxScores[n - 1];
+};
