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# Maximum CPU Load from the given list of jobs

• Difficulty Level : Expert
• Last Updated : 26 Aug, 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}},

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}

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.*;` `import` `java.util.Arrays;`   `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``] < li.get(li.size() - ``1``)[``1``]) {` `                ``li.get(li.size() - ``1``)[``1``] = Math.max(` `                    ``p[``1``], li.get(li.size() - ``1``)[``1``]);` `                ``li.get(li.size() - ``1``)[``2``]` `                    ``= p[``2``] + li.get(li.size() - ``1``)[``2``];` `            ``}` `            ``else` `{` `                ``li.add(p);` `            ``}`   `            ``ans = Math.max(ans, li.get(li.size() - ``1``)[``2``]);` `        ``}`   `        ``return` `ans;` `    ``}`   `    ``// Driver Code` `    ``public` `static` `void` `main(String[] args)` `    ``{` `        ``// Given Process` `        ``int``[][] process = ``new` `int``[][] { { ``1``, ``4``, ``3` `},` `                                        ``{ ``7``, ``9``, ``6` `},` `                                        ``{ ``2``, ``5``, ``4` `} };`   `        ``// Function call` `        ``int` `ans = maxCpuLoad(process);` `        ``System.out.print(ans);` `    ``}` `}`

Approach 3:

The idea is simple. We have suppose n intervals, so we have 2n endpoints ( here endpoint is the end of an interval and its value is the time associated with it). We can take an endpoint and combine it with its load value associated with it and with a flag which states whether it is a starting point or ending point of an interval. Then we can just sort the endpoints in increasing order(if there is tie in value of endpoints then we will break the tie by putting the endpoint which is starting at first place as compared to the endpoint which is ending; if both the endpoints are starting or ending then we will break the tie arbitrarily).

After sorting, we will proceed through the endpoints using for loop. And if we have an endpoint which is starting point of an interval then we will add the load value associated with it in a variable say, count. We will also take maximum of the count values and store it in a variable called result.

But when we get an endpoint which is ending then we will decrease the load value associated with it from count.

At the end we will return result.

Lets take an example: suppose the jobs are {1, 4, 3}, {2, 5, 4}, {7, 9, 6}.

our sorted endpoints will be 1(start), 2(start), 4(end), 5(end), 7(start), 9(end) .

and the corresponding loads will be 3, 4, 3, 4, 6, 6.

start traversing the endpoints:

so after traversing first endpoint which is 1(start) we have count+=3 (here 3 is the load associated with it) so count =3. Since the 1 is starting point so we will update the result. So result=max(result,count) so, result=3.

After traversing 2(start) we have count+=4, so count=7, result=max(result,count)=7.

After traversing 4(end) we have count-=3(we have subtracted because it is ending point) so count=4. result will not be updated since we are decreasing the count.

After traversing 5(end) we have count-=4 so count=0.

After traversing 7(start) we have count+=6 so count=6, result=max(result,count)=7.

After traversing 9(end) we have count-=6 so count=0.

Our result will be 7.

## C++14

 `// C++ implementation to find the` `// maximum CPU Load from the given array of jobs`   `#include ` `using` `namespace` `std;`   `// the job struct will have s(starting time) , e(ending time) , load(cpu load)` `struct` `Job {` `    ``int` `s, e, load;` `};`   `// endpoint struct will have val(the time associated with the endpoint),` `// load(cpu load associated with the endpoint),` `// a flag isStart which will tell if the endpoint is starting or ending point of` `// an interval` `struct` `Endpoint {` `    ``int` `val, load;` `    ``bool` `isStart;` `};`   `// custom comparator function which is used by the c++ sort stl` `bool` `comp(``const` `Endpoint& a, ``const` `Endpoint& b) {` `    ``if` `(a.val != b.val)` `        ``return` `a.val < b.val;` `    ``return` `a.isStart == ``true` `&& b.isStart == ``false``;` `}` `//function to find maximum cpu load` `int` `maxCpuLoad(vector v)` `{` `    ``int` `count = 0; ``// count will contain the count of current loads` `      ``int` `result = 0; ``// result will contain maximum of all counts`   `      ``// this data array will contain all the endpoints combined with their load values` `      ``// and flags telling whether they are starting or ending point` `    ``vector data;`   `    ``for` `(``int` `i = 0; i < v.size(); i++) {` `        ``data.emplace_back(Endpoint{ v[i].s, v[i].load, ``true``});` `        ``data.emplace_back(Endpoint{ v[i].e, v[i].load, ``false``});` `    ``}`   `    ``sort(data.begin(), data.end(), comp);`   `    ``for` `(``int` `i = 0; i < data.size(); i++) {` `        ``if` `(data[i].isStart == ``true``) {` `            ``count += data[i].load;` `            ``result = max(result, count);` `        ``}` `        ``else` `            ``count -= data[i].load;`   `    ``}`   `    ``return` `result;` `}` `//Driver code to test maxCpuLoad function` `int` `main() {` `    ``vector v = {` `        ``{6, 7, 10},` `          ``{2, 4, 11}, ` `          ``{8, 12, 15}` `    ``};` `    ``cout << maxCpuLoad(v);` `    ``return` `0;` `}` `// this code is contributed by Mohit Puri`

Output

`15`

Time Complexity : O(nlogn) for sorting the data array.

Space Complexity: O(n) which is the size of data array

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