feat: solve No.1834

Author: BaffinLee

1801-1900/1834. Single-Threaded CPU.md

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+# 1834. Single-Threaded CPU
+
+- Difficulty: Medium.
+- Related Topics: Array, Sorting, Heap (Priority Queue).
+- Similar Questions: Parallel Courses III, Minimum Time to Complete All Tasks.
+
+## Problem
+
+You are given `n`​​​​​​ tasks labeled from `0` to `n - 1` represented by a 2D integer array `tasks`, where `tasks[i] = [enqueueTimei, processingTimei]` means that the `i​​​​​​th`​​​​ task will be available to process at `enqueueTimei` and will take `processingTimei` to finish processing.
+
+You have a single-threaded CPU that can process **at most one** task at a time and will act in the following way:
+
+
+	
+- If the CPU is idle and there are no available tasks to process, the CPU remains idle.
+	
+- If the CPU is idle and there are available tasks, the CPU will choose the one with the **shortest processing time**. If multiple tasks have the same shortest processing time, it will choose the task with the smallest index.
+	
+- Once a task is started, the CPU will **process the entire task** without stopping.
+	
+- The CPU can finish a task then start a new one instantly.
+
+
+Return **the order in which the CPU will process the tasks.**
+
+ 
+Example 1:
+
+```
+Input: tasks = [[1,2],[2,4],[3,2],[4,1]]
+Output: [0,2,3,1]
+Explanation: The events go as follows: 
+- At time = 1, task 0 is available to process. Available tasks = {0}.
+- Also at time = 1, the idle CPU starts processing task 0. Available tasks = {}.
+- At time = 2, task 1 is available to process. Available tasks = {1}.
+- At time = 3, task 2 is available to process. Available tasks = {1, 2}.
+- Also at time = 3, the CPU finishes task 0 and starts processing task 2 as it is the shortest. Available tasks = {1}.
+- At time = 4, task 3 is available to process. Available tasks = {1, 3}.
+- At time = 5, the CPU finishes task 2 and starts processing task 3 as it is the shortest. Available tasks = {1}.
+- At time = 6, the CPU finishes task 3 and starts processing task 1. Available tasks = {}.
+- At time = 10, the CPU finishes task 1 and becomes idle.
+```
+
+Example 2:
+
+```
+Input: tasks = [[7,10],[7,12],[7,5],[7,4],[7,2]]
+Output: [4,3,2,0,1]
+Explanation: The events go as follows:
+- At time = 7, all the tasks become available. Available tasks = {0,1,2,3,4}.
+- Also at time = 7, the idle CPU starts processing task 4. Available tasks = {0,1,2,3}.
+- At time = 9, the CPU finishes task 4 and starts processing task 3. Available tasks = {0,1,2}.
+- At time = 13, the CPU finishes task 3 and starts processing task 2. Available tasks = {0,1}.
+- At time = 18, the CPU finishes task 2 and starts processing task 0. Available tasks = {1}.
+- At time = 28, the CPU finishes task 0 and starts processing task 1. Available tasks = {}.
+- At time = 40, the CPU finishes task 1 and becomes idle.
+```
+
+ 
+**Constraints:**
+
+
+	
+- `tasks.length == n`
+	
+- `1 <= n <= 105`
+	
+- `1 <= enqueueTimei, processingTimei <= 109`
+
+
+
+## Solution
+
+```javascript
+/**
+ * @param {number[][]} tasks
+ * @return {number[]}
+ */
+var getOrder = function(tasks) {
+    var arr = tasks.map((task, i) => [task[0], task[1], i]).sort((a, b) => a[0] - b[0]);
+
+    var queue = new MinPriorityQueue();
+    var i = 0;
+    var res = [];
+    var lastTime = 0;
+    while (i < arr.length || queue.size()) {
+        if (!queue.size() && arr[i][0] > lastTime) {
+            lastTime = arr[i][0];
+        }
+
+        while (arr[i] && arr[i][0] <= lastTime) {
+            var priority = arr[i][1] + (arr[i][2] / Math.pow(10, `${arr.length}`.length));
+            queue.enqueue(arr[i], priority);
+            i++;
+        }
+
+        var task = queue.dequeue().element;
+        res.push(task[2]);
+        lastTime += task[1];
+    }
+
+    return res;
+};
+```
+
+**Explain:**
+
+nope.
+
+**Complexity:**
+
+* Time complexity : O(n).
+* Space complexity : O(n).