Given n appointments, find all conflicting appointments.
Examples:
Input: appointments[] = { {1, 5} {3, 7}, {2, 6}, {10, 15}, {5, 6}, {4, 100}}
Output: Following are conflicting intervals
[3,7] Conflicts with [1,5]
[2,6] Conflicts with [1,5]
[5,6] Conflicts with [3,7]
[4,100] Conflicts with [1,5]
An appointment is conflicting, if it conflicts with any of the previous appointments in array.
Simple Solution is to one by one process all appointments from second appointment to last. For every appointment i, check if it conflicts with i-1, i-2, … 0. The time complexity of this method is O(n2).