Add solution #1986

Author: JoshCrozier

README.md

@@ -1,4 +1,4 @@
-# 1,670 LeetCode solutions in JavaScript
+# 1,671 LeetCode solutions in JavaScript
 
 [https://leetcodejavascript.com](https://leetcodejavascript.com)
 
@@ -1524,6 +1524,7 @@
 1982|[Find Array Given Subset Sums](./solutions/1982-find-array-given-subset-sums.js)|Hard|
 1984|[Minimum Difference Between Highest and Lowest of K Scores](./solutions/1984-minimum-difference-between-highest-and-lowest-of-k-scores.js)|Easy|
 1985|[Find the Kth Largest Integer in the Array](./solutions/1985-find-the-kth-largest-integer-in-the-array.js)|Medium|
+1986|[Minimum Number of Work Sessions to Finish the Tasks](./solutions/1986-minimum-number-of-work-sessions-to-finish-the-tasks.js)|Medium|
 1996|[The Number of Weak Characters in the Game](./solutions/1996-the-number-of-weak-characters-in-the-game.js)|Medium|
 2000|[Reverse Prefix of Word](./solutions/2000-reverse-prefix-of-word.js)|Easy|
 2011|[Final Value of Variable After Performing Operations](./solutions/2011-final-value-of-variable-after-performing-operations.js)|Easy|

solutions/1986-minimum-number-of-work-sessions-to-finish-the-tasks.js

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+/**
+ * 1986. Minimum Number of Work Sessions to Finish the Tasks
+ * https://leetcode.com/problems/minimum-number-of-work-sessions-to-finish-the-tasks/
+ * Difficulty: Medium
+ *
+ * There are n tasks assigned to you. The task times are represented as an integer array tasks of
+ * length n, where the ith task takes tasks[i] hours to finish. A work session is when you work
+ * for at most sessionTime consecutive hours and then take a break.
+ *
+ * You should finish the given tasks in a way that satisfies the following conditions:
+ * - If you start a task in a work session, you must complete it in the same work session.
+ * - You can start a new task immediately after finishing the previous one.
+ * - You may complete the tasks in any order.
+ *
+ * Given tasks and sessionTime, return the minimum number of work sessions needed to finish all the
+ * tasks following the conditions above.
+ *
+ * The tests are generated such that sessionTime is greater than or equal to the maximum element
+ * in tasks[i].
+ */
+
+/**
+ * @param {number[]} tasks
+ * @param {number} sessionTime
+ * @return {number}
+ */
+var minSessions = function(tasks, sessionTime) {
+  const n = tasks.length;
+  const dp = new Array(1 << n).fill(n + 1);
+  dp[0] = 0;
+
+  for (let mask = 1; mask < 1 << n; mask++) {
+    let time = 0;
+    for (let i = 0; i < n; i++) {
+      if (mask & (1 << i)) {
+        time += tasks[i];
+      }
+    }
+
+    for (let subset = mask; subset; subset = (subset - 1) & mask) {
+      if (subset === mask) continue;
+      let subsetTime = 0;
+      for (let i = 0; i < n; i++) {
+        if (subset & (1 << i)) {
+          subsetTime += tasks[i];
+        }
+      }
+      if (subsetTime <= sessionTime) {
+        dp[mask] = Math.min(dp[mask], dp[mask ^ subset] + 1);
+      }
+    }
+
+    if (time <= sessionTime) {
+      dp[mask] = 1;
+    }
+  }
+
+  return dp[(1 << n) - 1];
+};