CPU Scheduling - Google Top Interview Questions

Problem Statement :

You are given a two-dimensional list of integers tasks. 
Each element contains [id, queue_time, duration] which represents a CPU task with id id, queued at time queue_time, that will run for duration amount of time. All tasks have unique ids.

Given that the CPU scheduler will run a job that is currently in the queue with the lowest duration first, return the order of job ids that will be processed. 
If there's more than one job with the lowest duration, then it will run the job with lowest id first.


n ≤ 100,000 where n is the length of tasks

Example 1


tasks = [

    [0, 1, 5],

    [1, 1, 5],

    [2, 2, 2]



[0, 2, 1]


At time 1 we run task 0 since it's the only one that's queued. It runs for 5 units of time. Then we have 
tasks 1 and 2 that are queued. We run task 2 first because it has lower duration.

Example 2


tasks = [

    [0, 1, 5],

    [1, 1, 5]



[0, 1]


Both tasks are queued up at the same time and both of them have the same duration. Since task 0 has 
lower id, we run it first.

Solution :


                        Solution in C++ :

bool timesort(vector<int>& a, vector<int>& b) {
    return a[1] < b[1];

vector<int> solve(vector<vector<int>>& tasks) {
    vector<int> ret;
    priority_queue<pair<int, int>, vector<pair<int, int>>, greater<pair<int, int>>> queued;
    int t = 0;
    int n = tasks.size();
    sort(tasks.rbegin(), tasks.rend(), timesort);
    while (ret.size() < n) {
        if (queued.size() == 0) t = max(t, tasks.back()[1]);
        while (tasks.size() && tasks.back()[1] <= t) {
            queued.emplace(tasks.back()[2], tasks.back()[0]);
        t += queued.top().first;
    return ret;

                        Solution in Java :

import java.util.*;

class Solution {
        Lowest duration first.

        Return the order of jobs ids that will be processed

        Sort based on duration first

        Basically, having 2 priority queue. Set the first priority queue based on the  queueTime,
       the second priority queue based on the duration. We always put the first priority queue as
       the single source of truth. The second priority queue will execute the task from the first
       priority queue.

        We will always pop the first priority queue and insert that value to the second priority
       queue. The second priority queue execute the task and calculate the next execution time. This
       is where the current time unit get a bit tricky. We get the end of the range starting from
       the first queue time.

        We will poll all the elements in the first priority queue until it the peek value that is <=

        Each of the poll element from the 2nd PQ we will insert to the result array, and return that
       result array at the end. We will keep polling until both PQ is empty.

        Then, we will increment the queue range based on the maximum of the following two task:
            1. current end range + task[2] which is the duration of the current task that is finish
            2. the current queue task + the current task duration

            For instance:
            [ [0, 1, 5], [1, 1, 5], [2, 2, 2] ]
                                                --> secondPQ has [[0,1,5],[1,1,5]]
                                                --> initial endRange = 1 (queueTime for [0, 1, 5])
                                                --> initial execution endRange = 1 + 5 (which ends
       at 6)
                                                --> secondPQ has [[2,2,2],[1,1,5]]
                                                --> second execution endRange = max (6 +2, 2+2)
                                                --> secondPQ has [1,1,5]
                                                --> execute [1,1,5]

        - If there are multiple 0 duration with different queue time [[0,1,0], [1,3,0],[2,5,0]] -->
        - If duration is shorter in the later queue time, and longer in the beginning queue time
       [[0,1,6],[1,3,5],[2,5,4],[3,6,2]] --> [0,3,2,1]
            - queue time doesn't interleave [[0,1,4],[1,3,3],[2,4,3],[3,6,2]] --> [0,1,3,2]
        - If the duration of the initial queue time is smaller than the duration in the later queue
       time [[0,1,1],[1,2,2],[2,3,3]] --> [0,1,2]
        - If the duration of the initial queue time is interleave with the duration of the later
       queue time. [[0,1,6],[1,3,5],[2,5,7],[3,6,3]] --> [0, 3, 1, 2]
        - If the both duration and queue time is the same but different id, will be queue up from id
            [[0, 1, 5],[1, 1, 5]] -- [0,1]

        - Cases where the from duration needs to be the initial queue time: tasks = [[0, 1, 2],[1,
       3, 1],[2, 2, 3],[3, 1, 0]]
        - Cases if the duration is the same and the queueTime is the same, we set it to the id in

    public int[] solve(int[][] tasks) {
            Having a pq to store all the values of queueTime.
            Pop the queue time and get all the value of that queue time into the q task.
            Then the queue task will calculate the next max value, and will pop all the
            value from the pq from that max value
        PriorityQueue<int[]> queueTimeQ = new PriorityQueue<>(new Comparator<int[]>() {
            public int compare(int[] a, int[] b) {
                if (a[1] == b[1]) {
                    return (a[2] != b[2]) ? a[2] - b[2] : a[0] - b[0];


                    return a[1] - b[1];

        PriorityQueue<int[]> qExecution = new PriorityQueue<>(new Comparator<int[]>() {
            public int compare(int[] a, int[] b) {
                if (a[2] == b[2]) {
                    return a[0] - b[0];

                    return a[2] - b[2];

        int[] res = new int[tasks.length];

        // all the queuedTime
        // TreeSet<Integer> treeSet = new TreeSet<>();
        for (int i = 0; i < tasks.length; i++) {

        int i = 0;
        int to =
            (!queueTimeQ.isEmpty()) ? queueTimeQ.peek()[1] : 0; // set it to the initialQueueTime

        while (!queueTimeQ.isEmpty()) {

            while (!qExecution.isEmpty()) {
                int[] task = qExecution.poll();
                res[i++] = task[0];

                // get the to value, which is from + the res[2], we also need to check if the
                // initial task is more than the end task we need to set whoever the end range is
                // the largest
                to = Math.max(to + task[2], task[1] + task[2]);

                // check for queue time that is less than or equal to `to` in the queueTimeQ
                while (!queueTimeQ.isEmpty() && queueTimeQ.peek()[1] <= to)

        return res;

                        Solution in Python : 
class Solution:
    def solve(self, tasks):
        tasks.sort(key=lambda x: (x[1], x[2], x[0]))
        idx = 0
        time = tasks[0][1]
        heap = []
        res = []
        while idx < len(tasks):
            while idx < len(tasks) and tasks[idx][1] <= time:
                heappush(heap, (tasks[idx][2], tasks[idx][0]))
                idx += 1
            if not heap:
                time = tasks[idx][1]
            dur, ID = heappop(heap)
            time += dur
        while heap:
        return res

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