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: 7
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) = 7Below is the implementation of the above approach:
C++
// C++ implementation to find the// maximum CPU Load from the given// lists of the jobs#include <algorithm>#include <iostream>#include <queue>#include <vector>using namespace std;// Blueprint of the jobclass 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<Job>& 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<Job, vector<Job>, 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 Codeint main(int argc, char* argv[]){ vector<Job> 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 jobsfrom heapq import *# Blueprint of the jobclass 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 jobsdef 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 Codeif __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))) |
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// approachimport 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]<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); }} |
7
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