Add solution #813

Author: JoshCrozier

Diff for: README.md

@@ -621,6 +621,7 @@
 810|[Chalkboard XOR Game](./0810-chalkboard-xor-game.js)|Hard|
 811|[Subdomain Visit Count](./0811-subdomain-visit-count.js)|Medium|
 812|[Largest Triangle Area](./0812-largest-triangle-area.js)|Easy|
+813|[Largest Sum of Averages](./0813-largest-sum-of-averages.js)|Medium|
 819|[Most Common Word](./0819-most-common-word.js)|Easy|
 821|[Shortest Distance to a Character](./0821-shortest-distance-to-a-character.js)|Easy|
 824|[Goat Latin](./0824-goat-latin.js)|Easy|

Diff for: solutions/0813-largest-sum-of-averages.js

@@ -0,0 +1,50 @@
+/**
+ * 813. Largest Sum of Averages
+ * https://leetcode.com/problems/largest-sum-of-averages/
+ * Difficulty: Medium
+ *
+ * You are given an integer array nums and an integer k. You can partition the array into at most
+ * k non-empty adjacent subarrays. The score of a partition is the sum of the averages of each
+ * subarray.
+ *
+ * Note that the partition must use every integer in nums, and that the score is not necessarily
+ * an integer.
+ *
+ * Return the maximum score you can achieve of all the possible partitions. Answers within 10-6
+ * of the actual answer will be accepted.
+ */
+
+/**
+ * @param {number[]} nums
+ * @param {number} k
+ * @return {number}
+ */
+var largestSumOfAverages = function(nums, k) {
+  const prefixSum = new Array(nums.length + 1).fill(0);
+  for (let i = 0; i < nums.length; i++) {
+    prefixSum[i + 1] = prefixSum[i] + nums[i];
+  }
+
+  const dp = new Array(nums.length).fill(0).map(() => new Array(k + 1).fill(0));
+
+  for (let i = 0; i < nums.length; i++) {
+    dp[i][1] = calculateAverage(i, nums.length - 1);
+  }
+
+  for (let partitions = 2; partitions <= k; partitions++) {
+    for (let start = 0; start < nums.length - partitions + 1; start++) {
+      for (let end = start; end < nums.length - partitions + 1; end++) {
+        dp[start][partitions] = Math.max(
+          dp[start][partitions],
+          calculateAverage(start, end) + dp[end + 1][partitions - 1]
+        );
+      }
+    }
+  }
+
+  return dp[0][k];
+
+  function calculateAverage(i, j) {
+    return (prefixSum[j + 1] - prefixSum[i]) / (j - i + 1);
+  }
+};