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.

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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