Job Sequencing Problem with Deadlines

In this problem, there is a list of jobs given. In the list, the deadline and the profits are also given for each job. Every job will take a single unit of time, so the minimum deadline for a job is 1. If only one job can be scheduled at a time, then maximize the profit.

To solve this problem, all subset of the set of jobs are generated to check whether the individual subset is feasible or not. Also, keep track on maximum profit for all feasible subset that has generated.

The time complexity of this algorithm is O(n^2)

Input and Output

A list of jobs with job id, deadline and profit. And the number of jobs n.
{('a', 2, 100), ('b', 1, 19), ('c', 2, 27), ('d', 1, 25), ('e', 3, 15)}
n = 5
Following is maximum profit sequence of job sequence: c a e


jobSequence(jobList, n)

Input − The list of jobs and the number of jobs present in the list.

Output − The sequence, how jobs are taken.

   sort the jobs in jobList according to their profit create a list of job sequence and slot to track free time slots
   initially make all slots are free
   for all given jobs i do
      for all jobs in the list from ending of the list do
         if slot[j] is free then
            jobSequence[j] := i
            make slot[j] := fill
            break the loop

   for all slots when it is not free do
      print id of job using jobList[jobSequence[i]]


using namespace std;

struct Job {
   char id;
   int deadLine;
   int profit;

bool comp(Job j1, Job j2) {
   return (j1.profit > j2.profit);    //compare jobs based on profit

int min(int a, int b) {
   return (a<b)?a:b;

void jobSequence(Job jobList[], int n) {
   sort(jobList, jobList+n, comp);    //sort jobList on profit

   int jobSeq[n];     // To store result (Sequence of jobs)
   bool slot[n];      // To keep track of free time slots

   for (int i=0; i<n; i++)
      slot[i] = false; //initially all slots are free

   for (int i=0; i<n; i++) {    //for all given jobs
      for (int j=min(n, jobList[i].deadLine)-1; j>=0; j--) {   //search from last free slot
         if (slot[j]==false) {
            jobSeq[j] = i;   // Add this job to job sequence
            slot[j] = true;  // mark this slot as occupied

   for (int i=0; i<n; i++)
      if (slot[i])
         cout << jobList[jobSeq[i]].id << " ";    //display the sequence

int main() {
   Job jobList[] = {{'a',2,100}, {'b',1,19}, {'c',2,27},{'d',1,25},{'e',3,15}};
   int n = 5;
   cout << "Following is maximum profit sequence of job sequence: ";
   jobSequence(jobList, n);


Following is maximum profit sequence of job sequence: c a e
karthikeya Boyini
karthikeya Boyini

I love programming (: That's all I know