Add solution #826

Author: JoshCrozier

README.md

@@ -634,6 +634,7 @@
 823|[Binary Trees With Factors](./0823-binary-trees-with-factors.js)|Medium|
 824|[Goat Latin](./0824-goat-latin.js)|Easy|
 825|[Friends Of Appropriate Ages](./0825-friends-of-appropriate-ages.js)|Medium|
+826|[Most Profit Assigning Work](./0826-most-profit-assigning-work.js)|Medium|
 827|[Making A Large Island](./0827-making-a-large-island.js)|Hard|
 830|[Positions of Large Groups](./0830-positions-of-large-groups.js)|Easy|
 831|[Masking Personal Information](./0831-masking-personal-information.js)|Medium|

solutions/0826-most-profit-assigning-work.js

@@ -0,0 +1,48 @@
+/**
+ * 826. Most Profit Assigning Work
+ * https://leetcode.com/problems/most-profit-assigning-work/
+ * Difficulty: Medium
+ *
+ * You have n jobs and m workers. You are given three arrays: difficulty, profit, and worker where:
+ * - difficulty[i] and profit[i] are the difficulty and the profit of the ith job, and
+ * - worker[j] is the ability of jth worker (i.e., the jth worker can only complete a job with
+ *   difficulty at most worker[j]).
+ *
+ * Every worker can be assigned at most one job, but one job can be completed multiple times.
+ * - For example, if three workers attempt the same job that pays $1, then the total profit will
+ *   be $3. If a worker cannot complete any job, their profit is $0.
+ *
+ * Return the maximum profit we can achieve after assigning the workers to the jobs.
+ */
+
+/**
+ * @param {number[]} difficulty
+ * @param {number[]} profit
+ * @param {number[]} worker
+ * @return {number}
+ */
+var maxProfitAssignment = function(difficulty, profit, worker) {
+  const jobs = difficulty.map((d, i) => [d, profit[i]]);
+  jobs.sort((a, b) => a[0] - b[0]);
+
+  const n = jobs.length;
+  const bestProfit = new Array(n);
+  let maxProfit = 0;
+
+  for (let i = 0; i < n; i++) {
+    maxProfit = Math.max(maxProfit, jobs[i][1]);
+    bestProfit[i] = [jobs[i][0], maxProfit];
+  }
+
+  return worker.reduce((total, ability) => {
+    let left = 0;
+    let right = n - 1;
+
+    while (left <= right) {
+      const mid = (left + right) >> 1;
+      bestProfit[mid][0] <= ability ? left = mid + 1 : right = mid - 1;
+    }
+
+    return total + (right >= 0 ? bestProfit[right][1] : 0);
+  }, 0);
+};