# Amazon Coding challenge on Time Slots

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";
}
}
}

}