Amazon Coding challenge on Time Slots

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Here is the Amazon coding challenge with solution. Please do comment below if you find any issues with the solution. Ready to learn and evolve. Knowledge is Power. 

Given M busy-time slots of the team members in the Kindle team, can you print all the available time slots when all of them can schedule a meeting for a duration of K minutes.
The event time will be of the form HH MM (where 0 <= HH <= 23 and 0 <= MM <= 59). K will be in the form minutes.



Input Format:

M K [M number of busy time slots , K is the duration in minutes]
This is followed by M lines with 4 numbers on each line.
Each line will be of the form StartHH StartMM EndHH EndMM  [Eg: 9am-11am time slot will be given as 9 00 11 00]
An event time slot is of the form [Start Time, End Time) which means the start time is inclusive but not the end time;
So, an event of the form 10 00  11 00 => implies that the meeting starts at 10:00 and ends at 11:00. Hence, another meeting can start at 11:00.

Sample Input:
5 120
16 00 17 00
10 30 14 30
20 45 22 15
10 00 13 15
09 00 11 00

Sample Output:
00 00 09 00
17 00 20 45


/* Enter your code here. Read input from STDIN. Print output to STDOUT */

#include
#include
using namespace std;

int main()
{
    int bTimeSlots, duraInMin;   
    cin >> bTimeSlots >> duraInMin;
   
    int startHH, stopHH, startMM, stopMM;
    const int minPerday = 1440;
    int storeBooking[minPerday] = {0};
   
    for (int i=0;i        cin >> startHH >> startMM >> stopHH >> stopMM;
       
        if (startHH>24 or stopHH > 24 or startMM > 59 or stopMM>59)
            exit(EXIT_FAILURE);
       
        int startMeet = startHH * 60 + startMM;
        int stopMeet = stopHH * 60 + stopMM;
       
        for (int j=startMeet;j<=stopMeet;j++){
            storeBooking[j]=1;
        }       
    }
   
    int timeInterval = 0;
    int freeTimeStart=0, freeTimeStop;
   
    for (int k=0;k       
        if (storeBooking[k]==1){
            if ((k/duraInMin) > 0 && storeBooking[k-1]==0){
               
                if ((k - freeTimeStart) >= duraInMin){
               
                cout << setfill('0') << setw(2) << freeTimeStart/60
                    << " "
                    << setfill('0') << setw(2) << freeTimeStart%60
                    << " "
                    << setfill('0') << setw(2) << k/60       
                    << " "
                    << setfill('0') << setw(2) << k%60   
                    << "\n";
                }
               
            }
          
            freeTimeStart = k;
        }else if(k==minPerday -1){
            if ((k - freeTimeStart) >= duraInMin){
                int finalHH;
                if (((k+1)/60)==24){ finalHH=0; }else{finalHH = (k+1)/60; }
               
                cout << setfill('0') << setw(2) << freeTimeStart/60
                    << " "
                    << setfill('0') << setw(2) << freeTimeStart%60
                    << " "
                    << setfill('0') << setw(2) << finalHH   
                    << " "
                    << setfill('0') << setw(2) << (k+1)%60   
                    << "\n";
            }
        }
    }
      
   
}