Add solution #1031

Author: JoshCrozier

README.md

@@ -1,4 +1,4 @@
-# 1,102 LeetCode solutions in JavaScript
+# 1,103 LeetCode solutions in JavaScript
 
 [https://leetcodejavascript.com](https://leetcodejavascript.com)
 
@@ -838,6 +838,7 @@
 1028|[Recover a Tree From Preorder Traversal](./solutions/1028-recover-a-tree-from-preorder-traversal.js)|Hard|
 1029|[Two City Scheduling](./solutions/1029-two-city-scheduling.js)|Medium|
 1030|[Matrix Cells in Distance Order](./solutions/1030-matrix-cells-in-distance-order.js)|Easy|
+1031|[Maximum Sum of Two Non-Overlapping Subarrays](./solutions/1031-maximum-sum-of-two-non-overlapping-subarrays.js)|Medium|
 1037|[Valid Boomerang](./solutions/1037-valid-boomerang.js)|Easy|
 1038|[Binary Search Tree to Greater Sum Tree](./solutions/1038-binary-search-tree-to-greater-sum-tree.js)|Medium|
 1041|[Robot Bounded In Circle](./solutions/1041-robot-bounded-in-circle.js)|Medium|

solutions/1031-maximum-sum-of-two-non-overlapping-subarrays.js

@@ -0,0 +1,48 @@
+/**
+ * 1031. Maximum Sum of Two Non-Overlapping Subarrays
+ * https://leetcode.com/problems/maximum-sum-of-two-non-overlapping-subarrays/
+ * Difficulty: Medium
+ *
+ * Given an integer array nums and two integers firstLen and secondLen, return the maximum sum
+ * of elements in two non-overlapping subarrays with lengths firstLen and secondLen.
+ *
+ * The array with length firstLen could occur before or after the array with length secondLen,
+ * but they have to be non-overlapping.
+ *
+ * A subarray is a contiguous part of an array.
+ */
+
+/**
+ * @param {number[]} nums
+ * @param {number} firstLen
+ * @param {number} secondLen
+ * @return {number}
+ */
+var maxSumTwoNoOverlap = function(nums, firstLen, secondLen) {
+  const prefixSums = getPrefixSums(nums);
+  let maxFirst = 0;
+  let maxSecond = 0;
+  let result = 0;
+
+  for (let i = firstLen; i <= nums.length - secondLen; i++) {
+    maxFirst = Math.max(maxFirst, prefixSums[i] - prefixSums[i - firstLen]);
+    const secondSum = prefixSums[i + secondLen] - prefixSums[i];
+    result = Math.max(result, maxFirst + secondSum);
+  }
+
+  for (let i = secondLen; i <= nums.length - firstLen; i++) {
+    maxSecond = Math.max(maxSecond, prefixSums[i] - prefixSums[i - secondLen]);
+    const firstSum = prefixSums[i + firstLen] - prefixSums[i];
+    result = Math.max(result, maxSecond + firstSum);
+  }
+
+  return result;
+
+  function getPrefixSums(arr) {
+    const sums = [0];
+    for (let i = 0; i < arr.length; i++) {
+      sums.push(sums[i] + arr[i]);
+    }
+    return sums;
+  }
+};