This is kind of job scheduling problem and thus firstly we need to sort the intervals based on starting time.

After sorting, we basically has two person to assign the jobs without overlapping. But since after sorting the position of the jobs will be changed so, we need to track their original indexes for final output.

WE have two person to assign jobs , Jaime as J and Cameron as C

We would declare two stacks namely Stack J and Stack C for them respectively.

Also, result will be our output which is a character array of length equal to number of jobs.

We will assign job based on their sorted order.

We will first try to assign job to J, if it's not possible then will try to assign job to C, if that's also not possible then it's impossible to assign jobs.

If job is possible to assign, we will push the job to the respective stack and we will mark the original index of the job with respective name ('C' or 'J' depending on which stack we assigned)

To keep track of original job index we create a structure like below:

class job{
    public:
    int index; //original index in input list
    int start; //start time  
    int end; //end time
}

Then sort the input list based on start time. For details refer to the comparator function in my code.

Let the sorted list be jobs with length n

1. Initialize two stacks as empty first
2. Initialize the character array of length n( n be number of jobs)
3. Push the first job in stack C and mark result[jobs[0].index] = 'C'

Remember jobs[0].index may not be 0 as we have sorted the original list and this will return the original index from the input. (this the purpose of the structure we created).

1. for(int i=1;i<n;i++)
       if  jobs[i].start>stack C top.end
           Push jobs[i] to stack C
           Mark result[jobs[i].index] as 'C'
       Else
           if  jobs[i].start>stack J top.end
               Push jobs[i] to stack j
               Mark result[jobs[i].index] as 'J'
           Else
               It's impossible to assign jobs ,so return "Impossible"
       end if-else
   End for
   

2. If it was possible to assign jobs, the result array has the final answer.

Dry run with example:

For the first example,

Number of jobs=3

Jobs are:
360 480
420 540
600 660

After sorting based on start time
360 480 ,index =0
420 540, index =1
600 660, index =2

It was pre-sorted basically

Now,
Assign job_0 to stack C and mark result[0]='C'  
For the next job we can't assign to stack c AS there's overlapping
Assign it to stack J and mark result[1]='J'
The last job can be assigned to stack C and thus final result will be "CJC"

C++ Implementation:

#include <bits/stdc++.h>
using namespace std;

class job {
public:
    int index;
    int start;
    int end;
    job()
    {
        index = 0;
        start = 0;
        end = 0;
    }
};

bool mycomp(job a, job b)
{
    if (a.start < b.start)
        return true;
    else if (a.start > b.start)
        return false;
    else
        return (a.index < b.index);
}

int main()
{
    int t;
 
    cin >> t;
 
    for (int test_case = 1; test_case <= t; test_case++) {
        int n;
 
        cin >> n;
 
        vector<job> arr(n);
 
        int lstart, lend;
        for (int i = 0; i < n; i++) {
            cin >> lstart >> lend;
            arr[i].index = i;
            arr[i].start = lstart;
            arr[i].end = lend;
        }
        sort(arr.begin(), arr.end(), mycomp);

        stack<job> st1;
        stack<job> st2;
 
        char result[n];
        st1.push(arr[0]);
 
        result[arr[0].index] = 'C';
 
        bool flag = false;
        for (int i = 1; i < n; i++) {
            job pq = st1.top();
            if (arr[i].start < pq.end) {
                if (st2.empty()) {
                    st2.push(arr[i]);
                    result[arr[i].index] = 'J';
                }
                else {
                    job pv = st2.top();
                    if (arr[i].start < pv.end) {
                        flag = true;
                        break;
                    }
                    else {
                        st2.push(arr[i]);
                        result[arr[i].index] = 'J';
                    }
                }
            }
            else {
                st1.push(arr[i]);
                result[arr[i].index] = 'C';
            }
        }

        if (flag == true) {
            cout << "Case #" << test_case << ": "
                 << "IMPOSSIBLE" << endl;
        }
        else {
            string s = "";
            for (int i = 0; i < n; i++)
                s += string(1, result[i]);

            cout << "Case #" << test_case << ": " << s << endl;
        }
    }
    
    return 0;
}

Output:

4
3
360 480
420 540
600 660
Case #1: CJC
3
0 1440
1 3
2 4
Case #2: IMPOSSIBLE
5
99 150
1 100
100 301
2 5
150 250
Case #3: JCCJJ
2
0 720
720 1440
Case #4: CC