Add solution #1425

Author: JoshCrozier

README.md

@@ -1,4 +1,4 @@
-# 1,298 LeetCode solutions in JavaScript
+# 1,299 LeetCode solutions in JavaScript
 
 [https://leetcodejavascript.com](https://leetcodejavascript.com)
 
@@ -1090,6 +1090,7 @@
 1422|[Maximum Score After Splitting a String](./solutions/1422-maximum-score-after-splitting-a-string.js)|Easy|
 1423|[Maximum Points You Can Obtain from Cards](./solutions/1423-maximum-points-you-can-obtain-from-cards.js)|Medium|
 1424|[Diagonal Traverse II](./solutions/1424-diagonal-traverse-ii.js)|Medium|
+1425|[Constrained Subsequence Sum](./solutions/1425-constrained-subsequence-sum.js)|Hard|
 1431|[Kids With the Greatest Number of Candies](./solutions/1431-kids-with-the-greatest-number-of-candies.js)|Easy|
 1436|[Destination City](./solutions/1436-destination-city.js)|Easy|
 1437|[Check If All 1's Are at Least Length K Places Away](./solutions/1437-check-if-all-1s-are-at-least-length-k-places-away.js)|Easy|

solutions/1425-constrained-subsequence-sum.js

@@ -0,0 +1,42 @@
+/**
+ * 1425. Constrained Subsequence Sum
+ * https://leetcode.com/problems/constrained-subsequence-sum/
+ * Difficulty: Hard
+ *
+ * Given an integer array nums and an integer k, return the maximum sum of a non-empty subsequence
+ * of that array such that for every two consecutive integers in the subsequence, nums[i] and
+ * nums[j], where i < j, the condition j - i <= k is satisfied.
+ *
+ * A subsequence of an array is obtained by deleting some number of elements (can be zero) from the
+ * array, leaving the remaining elements in their original order.
+ */
+
+/**
+ * @param {number[]} nums
+ * @param {number} k
+ * @return {number}
+ */
+var constrainedSubsetSum = function(nums, k) {
+  const maxSums = [...nums];
+  const deque = [];
+  let result = nums[0];
+
+  for (let i = 0; i < nums.length; i++) {
+    while (deque.length && deque[0] < i - k) {
+      deque.shift();
+    }
+
+    if (deque.length) {
+      maxSums[i] = Math.max(maxSums[i], maxSums[deque[0]] + nums[i]);
+    }
+
+    while (deque.length && maxSums[deque[deque.length - 1]] <= maxSums[i]) {
+      deque.pop();
+    }
+
+    deque.push(i);
+    result = Math.max(result, maxSums[i]);
+  }
+
+  return result;
+};