Maximize jobs that can be completed under given constraint

Given an integer N denoting number of jobs and a matrix ranges[] consisting of a range [start day, end day] for each job within which it needs to be completed, the task is to find the maximum possible jobs that can be completed.


Input: N = 5, Ranges = {{1, 5}, {1, 5}, {1, 5}, {2, 3}, {2, 3}}
Output: 5
Explanation: Job 1 on day 1, Job 4 on day 2, Job 5 on day 3, Job 2 on day 4, Job 3 on day 5

Input: N=6, Ranges = {{1, 3}, {1, 3}, {2, 3}, {2, 3}, {1, 4}, {2, 5}}
Output: 5

Approach: The above problem can be solved using a Priority Queue. Follow the steps below to solve the problems:

  • Find the minimum and maximum day in the range of jobs.
  • Sort all jobs in increasing order of start day.
  • Iterate from the minimum to maximum day, and for every ith day, select the job having least end day which can be completed on that day.
  • In order to perform the above step, maintain a Min Heap, and on every ith day, insert the jobs that can be completed on that day, into the Min Heap sorted by end day. If any job can completed on the ith day, consider the one with the lowest end day and increase the count of jobs completed.
  • Repeat this process for all the days and finally print the count of jobs completed.

Below is an implementation of the above approach:






// C++ Program to implement the
// above approach
#include <bits/stdc++.h>
using namespace std;
// Function to find maxiumum
// number of jobs
int find_maximum_jobs(
    int N,
    vector<pair<int, int> > ranges)
    // Min Heap
    priority_queue<int, vector<int>,
                   greater<int> >
    // Sort ranges by start day
    sort(ranges.begin(), ranges.end());
    // Stores the minimum and maximum
    // day in the ranges
    int min_day = ranges[0].first;
    int max_day = 0;
    for (int i = 0; i < N; i++)
            = max(max_day,
    int index = 0, count_jobs = 0;
    // Iterating from min_day to max_day
    for (int i = min_day; i <= max_day; i++) {
        // Insert the end day of the jobs
        // which can be completed on
        // i-th day in a priority queue
        while (index < ranges.size()
               && ranges[index].first <= i) {
        // Pop all jobs whose end day
        // is less than current day
        while (!queue.empty()
               && < i)
        // If queue is empty, no job
        // can be completed on
        // the i-th day
        if (queue.empty())
        // Increment the count of
        // jobs completed
        // Pop the job with
        // least end day
    // Return the jobs
    // on the last day
    return count_jobs;
// Driver Code
int main()
    int N = 5;
    vector<pair<int, int> > ranges;
    ranges.push_back({ 1, 5 });
    ranges.push_back({ 1, 5 });
    ranges.push_back({ 1, 5 });
    ranges.push_back({ 2, 3 });
    ranges.push_back({ 2, 3 });
    cout << find_maximum_jobs(N, ranges);




Time Complexity: O(Xlog(N)), where X is the difference between maximum and minimum day and N is the number of jobs.
Auxiliary Space: O(N2)


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