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Maximum CPU Load from the given list of jobs
• Difficulty Level : Expert
• Last Updated : 25 Jan, 2021

Given an array of jobs with different time requirements, where each job consists of start time, end time and CPU load. The task is to find the maximum CPU load at any time if all jobs are running on the same machine.

Examples:

Input: jobs[] = {{1, 4, 3}, {2, 5, 4}, {7, 9, 6}}
Output:
Explanation:
In the above-given jobs, there are two jobs which overlaps.
That is, Job [1, 4, 3] and [2, 5, 4] overlaps for the time period in [2, 4]
Hence, the maximum CPU Load at this instant will be maximum (3 + 4 = 7).

Input: jobs[] = {{6, 7, 10}, {2, 4, 11}, {8, 12, 15}}
Output: 15
Explanation:
Since, There are no jobs that overlaps.
Maximum CPU Load will be – max(10, 11, 15) = 15

This problem is generally the application of the Merge Intervals
Approach: The idea is to maintain min-heap for the jobs on the basis of their end times. Then, for each instance find the jobs which are complete and remove them from the Min-heap. That is, Check that the end-time of the jobs in the min-heap had ended before the start time of the current job. Also at each instance, find the maximum CPU Load on the machine by taking the sum of all the jobs that are present in the min-heap.

For Example:

```Given Jobs be {{1, 4, 3}, {2, 5, 4}, {7, 9, 6}}
Min-Heap - {}

Instance 1:
The job {1, 4, 3} is inserted into the min-heap
Min-Heap - {{1, 4, 3}},
Total CPU Load  = 3

Instance 2:
The job {2, 5, 4} is inserted into the min-heap.
While the job {1, 4, 3} is still in the CPU,
because end-time of Job 1 is greater than
the start time of the new job {2, 5, 4}.
Min-Heap - {{1, 4, 3}, {2, 5, 4}}
Total CPU Load = 4 + 3 = 7

Instance 3:
The job {7, 9, 6} is inserted into the min-heap.
After popping up all the other jobs because their
end time is less than the start time of the new job
Min Heap - {7, 9, 6}
Total CPU Load =  6

Maximum CPU Load = max(3, 7, 6) = 7```

Below is the implementation of the above approach:

## C++

 `// C++ implementation to find the``// maximum CPU Load from the given``// lists of the jobs` `#include ``#include ``#include ``#include ` `using` `namespace` `std;` `// Blueprint of the job``class` `Job {``    ``public``:``        ``int` `start = 0;``        ``int` `end = 0;``        ``int` `cpuLoad = 0;``        ` `        ``// Constructor function for``        ``// the CPU Job``        ``Job(``int` `start, ``int` `end,``                   ``int` `cpuLoad)``        ``{``            ``this``->start = start;``            ``this``->end = end;``            ``this``->cpuLoad = cpuLoad;``        ``}``};` `class` `MaximumCPULoad {``    ` `    ``// Structure to compare two``    ``// CPU Jobs by their end time``    ``public``:``        ``struct` `endCompare {``            ``bool` `operator()(``const` `Job& x,``            ``const` `Job& y) {``                ``return` `x.end > y.end;``            ``}``        ``};``    ` `    ``// Function to find the maximum``    ``// CPU Load at any instance of``    ``// the time for given jobs``    ``static` `int` `findMaxCPULoad(``                 ``vector& jobs)``    ``{``        ``// Condition when there are``        ``// no jobs then CPU Load is 0``        ``if` `(jobs.empty()) {``            ``return` `0;``        ``}``        ` `        ``// Sorting all the jobs``        ``// by their start time``        ``sort(jobs.begin(), jobs.end(),``          ``[](``const` `Job& a, ``const` `Job& b) {``              ``return` `a.start < b.start;``          ``});` `        ``int` `maxCPULoad = 0;``        ``int` `currentCPULoad = 0;``        ` `        ``// Min-heap implemented using the``        ``// help of the priority queue``        ``priority_queue,``                    ``endCompare> minHeap;``                    ` `        ``// Loop to iterate over all the``        ``// jobs from the given list``        ``for` `(``auto` `job : jobs) {``            ` `            ``// Loop to remove all jobs from``            ``// the heap which is ended``            ``while` `(!minHeap.empty() &&``             ``job.start > minHeap.top().end) {``                ``currentCPULoad -=``                   ``minHeap.top().cpuLoad;``                ``minHeap.pop();``            ``}``            ` `            ``// Add the current Job to CPU``            ``minHeap.push(job);``            ``currentCPULoad += job.cpuLoad;``            ``maxCPULoad = max(maxCPULoad,``                            ``currentCPULoad);``        ``}``        ``return` `maxCPULoad;``    ``}``};` `// Driver Code``int` `main(``int` `argc, ``char``* argv[])``{``    ``vector input = { { 1, 4, 3 },``            ``{ 7, 9, 6 }, { 2, 5, 4 } };``    ``cout << ``"Maximum CPU load at any time: "``         ``<< MaximumCPULoad::findMaxCPULoad(input)``         ``<< endl;` `}`

## Python3

 `# Python implementation to find the``# maximum CPU Load from the given``# lists of the jobs` `from` `heapq ``import` `*` `# Blueprint of the job``class` `job:``    ` `    ``# Constructor of the Job``    ``def` `__init__(``self``, start,\``              ``end, cpu_load):``        ``self``.start ``=` `start``        ``self``.end ``=` `end``        ``self``.cpu_load ``=` `cpu_load``    ` `    ``# Operator overloading for the``    ``# Object that is Job``    ``def` `__lt__(``self``, other):` `        ``# min heap based on job.end``        ``return` `self``.end < other.end` `# Function to find the maximum``# CPU Load for the given list of jobs``def` `find_max_cpu_load(jobs):``    ` `    ``# Sort the jobs by start time``    ``jobs.sort(key ``=` `lambda` `x: x.start)``    ``max_cpu_load, current_cpu_load ``=` `0``, ``0``    ` `    ``# Min-Heap``    ``min_heap ``=` `[]``    ` `    ``# Loop to iterate over the list``    ``# of the jobs given for the CPU``    ``for` `j ``in` `jobs:``        ` `        ``# Remove all the jobs from``        ``# the min-heap which ended``        ``while``(``len``(min_heap) > ``0` `and``\``          ``j.start >``=` `min_heap[``0``].end):``            ``current_cpu_load ``-``=` `min_heap[``0``].cpu_load``            ``heappop(min_heap)``    ` `        ``# Add the current job``        ``# into min_heap``        ``heappush(min_heap, j)``        ``current_cpu_load ``+``=` `j.cpu_load``        ``max_cpu_load ``=` `max``(max_cpu_load,``                       ``current_cpu_load)``    ``return` `max_cpu_load` `# Driver Code``if` `__name__ ``=``=` `"__main__"``:``    ``jobs ``=` `[job(``1``, ``4``, ``3``), job(``2``, ``5``, ``4``),\``                         ``job(``7``, ``9``, ``6``)]``                         ` `    ``print``(``"Maximum CPU load at any time: "` `+``\``               ``str``(find_max_cpu_load(jobs)))`
Output
`Maximum CPU load at any time: 7`

Performance Analysis:

• Time complexity: O(N*logN)
• Auxiliary Space: O(N)

Approach 2  – Without using heap, Merge the overlapping intervals.

This can also be solved by using idea of Merge Intervals
The idea is fairly straight forward – > Merge the overlapping intervals and add their load.
Below is the Java code for the same.

## Java

 `// JAVA Implementation of the above``// approach` `import` `java.util.Arrays;``import` `java.util.*;` `public` `class` `MaximumCpuLoad {``    ` `private` `static` `int` `maxCpuLoad(``int``[][] process) {``    ``Arrays.sort(process,(a,b)->'``    ``{``        ``return` `a[``0``]-b[``0``];``    ``});``    ` `     ` `    ``// list of intervals``    ``List<``int``[]> li = ``new` `LinkedList<``int``[]>();``    ` `     ` `    ``// variable to store the result``    ``int` `ans=``0``;``     ` `    ``// Merge intervals``    ``for``(``int``[] p : process)``    ``{``        ``if``(!li.isEmpty() && p[``0``]
Output
`7`

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