Add solution #689

Author: JoshCrozier

README.md

@@ -520,6 +520,7 @@
 686|[Repeated String Match](./0686-repeated-string-match.js)|Easy|
 687|[Longest Univalue Path](./0687-longest-univalue-path.js)|Medium|
 688|[Knight Probability in Chessboard](./0688-knight-probability-in-chessboard.js)|Medium|
+689|[Maximum Sum of 3 Non-Overlapping Subarrays](./0689-maximum-sum-of-3-non-overlapping-subarrays.js)|Hard|
 693|[Binary Number with Alternating Bits](./0693-binary-number-with-alternating-bits.js)|Easy|
 695|[Max Area of Island](./0695-max-area-of-island.js)|Medium|
 696|[Count Binary Substrings](./0696-count-binary-substrings.js)|Easy|

solutions/0689-maximum-sum-of-3-non-overlapping-subarrays.js

@@ -0,0 +1,55 @@
+/**
+ * 689. Maximum Sum of 3 Non-Overlapping Subarrays
+ * https://leetcode.com/problems/maximum-sum-of-3-non-overlapping-subarrays/
+ * Difficulty: Hard
+ *
+ * Given an integer array nums and an integer k, find three non-overlapping subarrays of
+ * length k with maximum sum and return them.
+ *
+ * Return the result as a list of indices representing the starting position of each
+ * interval (0-indexed). If there are multiple answers, return the lexicographically
+ * smallest one.
+ */
+
+/**
+ * @param {number[]} nums
+ * @param {number} k
+ * @return {number[]}
+ */
+var maxSumOfThreeSubarrays = function(nums, k) {
+  const n = nums.length;
+  let [sumFirst, sumSecond, sumThird] = [0, 0, 0];
+  let [maxFirst, maxFirstSecond, maxTotal] = [0, 0, 0];
+  let [startFirst, startFirstBest, startSecondBest] = [0, 0, k];
+  let result = [0, k, 2 * k];
+
+  for (let i = 0; i < k; i++) {
+    sumFirst += nums[i];
+    sumSecond += nums[i + k];
+    sumThird += nums[i + 2 * k];
+  }
+  [maxFirst, maxFirstSecond, maxTotal] = [
+    sumFirst, sumFirst + sumSecond,
+    sumFirst + sumSecond + sumThird
+  ];
+
+  for (let i = 1; i <= n - 3 * k; i++) {
+    sumFirst += nums[i + k - 1] - nums[i - 1];
+    sumSecond += nums[i + 2 * k - 1] - nums[i + k - 1];
+    sumThird += nums[i + 3 * k - 1] - nums[i + 2 * k - 1];
+
+    if (sumFirst > maxFirst) [maxFirst, startFirst] = [sumFirst, i];
+    if (maxFirst + sumSecond > maxFirstSecond) {
+      [maxFirstSecond, startFirstBest, startSecondBest] = [maxFirst + sumSecond, startFirst, i + k];
+    }
+    if (maxFirstSecond + sumThird > maxTotal) {
+      [maxTotal, ...result] = [
+        maxFirstSecond + sumThird,
+        startFirstBest,
+        startSecondBest, i + 2 * k
+      ];
+    }
+  }
+
+  return result;
+};