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.

**We strongly recommend to minimize the browser and try this yourself first.**

A **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(n^{2}).

We can use **Interval Tree** to solve this problem in O(nLogn) time. Following is detailed algorithm.

1) Create an Interval Tree, initially with the first appointment. 2) Do following for all other appointments starting from the second one. a) Check if the current appointment conflicts with any of the existing appointments in Interval Tree. If conflicts, then print the current appointment. This step can be done O(Logn) time. b) Insert the current appointment in Interval Tree. This step also can be done O(Logn) time.

Following is C++ implementation of above idea.

`// C++ program to print all conflicting appointments in a ` `// given set of appointments ` `#include <iostream> ` `using` `namespace` `std; ` ` ` `// Structure to represent an interval ` `struct` `Interval ` `{ ` ` ` `int` `low, high; ` `}; ` ` ` `// Structure to represent a node in Interval Search Tree ` `struct` `ITNode ` `{ ` ` ` `Interval *i; ` `// 'i' could also be a normal variable ` ` ` `int` `max; ` ` ` `ITNode *left, *right; ` `}; ` ` ` `// A utility function to create a new Interval Search Tree Node ` `ITNode * newNode(Interval i) ` `{ ` ` ` `ITNode *temp = ` `new` `ITNode; ` ` ` `temp->i = ` `new` `Interval(i); ` ` ` `temp->max = i.high; ` ` ` `temp->left = temp->right = NULL; ` `}; ` ` ` `// A utility function to insert a new Interval Search Tree ` `// Node. This is similar to BST Insert. Here the low value ` `// of interval is used tomaintain BST property ` `ITNode *insert(ITNode *root, Interval i) ` `{ ` ` ` `// Base case: Tree is empty, new node becomes root ` ` ` `if` `(root == NULL) ` ` ` `return` `newNode(i); ` ` ` ` ` `// Get low value of interval at root ` ` ` `int` `l = root->i->low; ` ` ` ` ` `// If root's low value is smaller, then new interval ` ` ` `// goes to left subtree ` ` ` `if` `(i.low < l) ` ` ` `root->left = insert(root->left, i); ` ` ` ` ` `// Else, new node goes to right subtree. ` ` ` `else` ` ` `root->right = insert(root->right, i); ` ` ` ` ` `// Update the max value of this ancestor if needed ` ` ` `if` `(root->max < i.high) ` ` ` `root->max = i.high; ` ` ` ` ` `return` `root; ` `} ` ` ` `// A utility function to check if given two intervals overlap ` `bool` `doOVerlap(Interval i1, Interval i2) ` `{ ` ` ` `if` `(i1.low < i2.high && i2.low < i1.high) ` ` ` `return` `true` `; ` ` ` `return` `false` `; ` `} ` ` ` `// The main function that searches a given interval i ` `// in a given Interval Tree. ` `Interval *overlapSearch(ITNode *root, Interval i) ` `{ ` ` ` `// Base Case, tree is empty ` ` ` `if` `(root == NULL) ` `return` `NULL; ` ` ` ` ` `// If given interval overlaps with root ` ` ` `if` `(doOVerlap(*(root->i), i)) ` ` ` `return` `root->i; ` ` ` ` ` `// If left child of root is present and max of left child ` ` ` `// is greater than or equal to given interval, then i may ` ` ` `// overlap with an interval is left subtree ` ` ` `if` `(root->left != NULL && root->left->max >= i.low) ` ` ` `return` `overlapSearch(root->left, i); ` ` ` ` ` `// Else interval can only overlap with right subtree ` ` ` `return` `overlapSearch(root->right, i); ` `} ` ` ` `// This function prints all conflicting appointments in a given ` `// array of apointments. ` `void` `printConflicting(Interval appt[], ` `int` `n) ` `{ ` ` ` `// Create an empty Interval Search Tree, add first ` ` ` `// appointment ` ` ` `ITNode *root = NULL; ` ` ` `root = insert(root, appt[0]); ` ` ` ` ` `// Process rest of the intervals ` ` ` `for` `(` `int` `i=1; i<n; i++) ` ` ` `{ ` ` ` `// If current appointment conflicts with any of the ` ` ` `// existing intervals, print it ` ` ` `Interval *res = overlapSearch(root, appt[i]); ` ` ` `if` `(res != NULL) ` ` ` `cout << ` `"["` `<< appt[i].low << ` `","` `<< appt[i].high ` ` ` `<< ` `"] Conflicts with ["` `<< res->low << ` `","` ` ` `<< res->high << ` `"]\n"` `; ` ` ` ` ` `// Insert this appointment ` ` ` `root = insert(root, appt[i]); ` ` ` `} ` `} ` ` ` ` ` `// Driver program to test above functions ` `int` `main() ` `{ ` ` ` `// Let us create interval tree shown in above figure ` ` ` `Interval appt[] = { {1, 5}, {3, 7}, {2, 6}, {10, 15}, ` ` ` `{5, 6}, {4, 100}}; ` ` ` `int` `n = ` `sizeof` `(appt)/` `sizeof` `(appt[0]); ` ` ` `cout << ` `"Following are conflicting intervals\n"` `; ` ` ` `printConflicting(appt, n); ` ` ` `return` `0; ` `} ` |

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]

Note that the above implementation uses simple Binary Search Tree insert operations. Therefore, time complexity of the above implementation is more than O(nLogn). We can use Red-Black Tree or AVL Tree balancing techniques to make the above implementation O(nLogn).

This article is contributed by **Anmol**. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above