Add solution #1335

Author: JoshCrozier

README.md

@@ -1,4 +1,4 @@
-# 1,236 LeetCode solutions in JavaScript
+# 1,237 LeetCode solutions in JavaScript
 
 [https://leetcodejavascript.com](https://leetcodejavascript.com)
 
@@ -1013,6 +1013,7 @@
 1332|[Remove Palindromic Subsequences](./solutions/1332-remove-palindromic-subsequences.js)|Easy|
 1333|[Filter Restaurants by Vegan-Friendly, Price and Distance](./solutions/1333-filter-restaurants-by-vegan-friendly-price-and-distance.js)|Medium|
 1334|[Find the City With the Smallest Number of Neighbors at a Threshold Distance](./solutions/1334-find-the-city-with-the-smallest-number-of-neighbors-at-a-threshold-distance.js)|Medium|
+1335|[Minimum Difficulty of a Job Schedule](./solutions/1335-minimum-difficulty-of-a-job-schedule.js)|Hard|
 1342|[Number of Steps to Reduce a Number to Zero](./solutions/1342-number-of-steps-to-reduce-a-number-to-zero.js)|Easy|
 1351|[Count Negative Numbers in a Sorted Matrix](./solutions/1351-count-negative-numbers-in-a-sorted-matrix.js)|Easy|
 1352|[Product of the Last K Numbers](./solutions/1352-product-of-the-last-k-numbers.js)|Medium|

solutions/1335-minimum-difficulty-of-a-job-schedule.js

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+/**
+ * 1335. Minimum Difficulty of a Job Schedule
+ * https://leetcode.com/problems/minimum-difficulty-of-a-job-schedule/
+ * Difficulty: Hard
+ *
+ * You want to schedule a list of jobs in d days. Jobs are dependent (i.e To work on the ith job,
+ * you have to finish all the jobs j where 0 <= j < i).
+ *
+ * You have to finish at least one task every day. The difficulty of a job schedule is the sum of
+ * difficulties of each day of the d days. The difficulty of a day is the maximum difficulty of a
+ * job done on that day.
+ *
+ * You are given an integer array jobDifficulty and an integer d. The difficulty of the ith job
+ * is jobDifficulty[i].
+ *
+ * Return the minimum difficulty of a job schedule. If you cannot find a schedule for the jobs
+ * return -1.
+ */
+
+/**
+ * @param {number[]} jobDifficulty
+ * @param {number} d
+ * @return {number}
+ */
+var minDifficulty = function(jobDifficulty, d) {
+  const n = jobDifficulty.length;
+  if (n < d) return -1;
+
+  const dp = new Array(d + 1).fill().map(() => new Array(n + 1).fill(Infinity));
+  dp[0][0] = 0;
+
+  for (let days = 1; days <= d; days++) {
+    for (let i = days; i <= n; i++) {
+      let maxDifficulty = 0;
+      for (let j = i - 1; j >= days - 1; j--) {
+        maxDifficulty = Math.max(maxDifficulty, jobDifficulty[j]);
+        dp[days][i] = Math.min(dp[days][i], dp[days - 1][j] + maxDifficulty);
+      }
+    }
+  }
+
+  return dp[d][n] === Infinity ? -1 : dp[d][n];
+};