Given n appointments, find all conflicting appointments
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 the 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 the 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).
We can use Interval Tree to solve this problem in O(nLogn) time. Following is a detailed algorithm.
- Create an Interval Tree, initially with the first appointment.
- Do following for all other appointments starting from the second one.
- 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.
- nsert the current appointment in Interval Tree. This step also can be done O(Logn) time.
Following is the implementation of the above idea.
C++
// C++ program to print all conflicting appointments in a // given set of appointments #include <bits/stdc++.h> 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; return temp; }; // 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 appointments. 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; } |
Java
// Java program to print all conflicting // appointments in a given set of appointments class GfG{ // Structure to represent an interval static class Interval { int low, high; } static class ITNode { // 'i' could also be a normal variable Interval i; int max; ITNode left, right; } // A utility function to create a new node static Interval newNode( int l, int h) { Interval temp = new Interval(); temp.low = l; temp.high = h; return temp; } // A utility function to create a new node static ITNode newNode(Interval i) { ITNode temp = new ITNode(); temp.i = i; temp.max = i.high; temp.left = temp.right = null ; return temp; } // 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 to // maintain BST property static 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 static boolean 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. static 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 appointments. static 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 ) System.out.print( "[" + appt[i].low + "," + appt[i].high + "] Conflicts with [" + res.low + "," + res.high + "]\n" ); // Insert this appointment root = insert(root, appt[i]); } } // Driver code public static void main(String[] args) { Interval appt[] = new Interval[ 6 ]; appt[ 0 ] = newNode( 1 , 5 ); appt[ 1 ] = newNode( 3 , 7 ); appt[ 2 ] = newNode( 2 , 6 ); appt[ 3 ] = newNode( 10 , 15 ); appt[ 4 ] = newNode( 5 , 6 ); appt[ 5 ] = newNode( 4 , 100 ); int n = appt.length; System.out.print( "Following are conflicting intervals\n" ); printConflicting(appt, n); } } // This code is contributed by tushar_bansal |
Python3
# Python3 program to print all conflicting # appointments in a given set of appointments # Structure to represent an interval class Interval: def __init__( self ): self .low = None self .high = None # Structure to represent a node # in Interval Search Tree class ITNode: def __init__( self ): self . max = None self .i = None self .left = None self .right = None def newNode(j): #print(j) temp = ITNode() temp.i = j temp. max = j[ 1 ] return temp # A utility function to check if # given two intervals overlap def doOVerlap(i1, i2): if (i1[ 0 ] < i2[ 1 ] and i2[ 0 ] < i1[ 1 ]): return True return False # Function to create a new node def insert(node, data): global succ # If the tree is empty, return a new node root = node if (node = = None ): return newNode(data) # If key is smaller than root's key, go to left # subtree and set successor as current node # print(node) if (data[ 0 ] < node.i[ 0 ]): # print(node) root.left = insert(node.left, data) # Go to right subtree else : root.right = insert(node.right, data) if root. max < data[ 1 ]: root. max = data[ 1 ] return root # The main function that searches a given # interval i in a given Interval Tree. def overlapSearch(root, i): # Base Case, tree is empty if (root = = None ): return None # 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 ! = None and root.left. max > = i[ 0 ]): 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 # appointments. def printConflicting(appt, n): # Create an empty Interval Search Tree, # add first appointment root = None root = insert(root, appt[ 0 ]) # Process rest of the intervals for i in range ( 1 , n): # If current appointment conflicts # with any of the existing intervals, # print it res = overlapSearch(root, appt[i]) if (res ! = None ): print ( "[" , appt[i][ 0 ], "," , appt[i][ 1 ], "] Conflicts with [" , res[ 0 ], "," , res[ 1 ], "]" ) # Insert this appointment root = insert(root, appt[i]) # Driver code if __name__ = = '__main__' : # Let us create interval tree # shown in above figure appt = [ [ 1 , 5 ], [ 3 , 7 ], [ 2 , 6 ], [ 10 , 15 ], [ 5 , 6 ], [ 4 , 100 ] ] n = len (appt) print ( "Following are conflicting intervals" ) printConflicting(appt, n) # This code is contributed by mohit kumar 29 |
C#
// C# program to print all conflicting // appointments in a given set of appointments using System; public class GfG { // Structure to represent an interval public class Interval { public int low, high; } public class ITNode { // 'i' could also be a normal variable public Interval i; public int max; public ITNode left, right; } // A utility function to create a new node static Interval newNode( int l, int h) { Interval temp = new Interval(); temp.low = l; temp.high = h; return temp; } // A utility function to create a new node static ITNode newNode(Interval i) { ITNode temp = new ITNode(); temp.i = i; temp.max = i.high; temp.left = temp.right = null ; return temp; } // 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 to // maintain BST property static 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 static 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. static 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 appointments. static 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 ) Console.Write( "[" + appt[i].low + "," + appt[i].high + "] Conflicts with [" + res.low + "," + res.high + "]\n" ); // Insert this appointment root = insert(root, appt[i]); } } // Driver code public static void Main(String[] args) { Interval []appt = new Interval[6]; appt[0] = newNode(1, 5); appt[1] = newNode(3, 7); appt[2] = newNode(2, 6); appt[3] = newNode(10, 15); appt[4] = newNode(5, 6); appt[5] = newNode(4, 100); int n = appt.Length; Console.Write( "Following are conflicting intervals\n" ); printConflicting(appt, n); } } // This code is contributed by gauravrajput1 |
Javascript
<script> // Javascript program to print all conflicting // appointments in a given set of appointments // Structure to represent an interval class Interval { constructor() { this .low = 0; this .high = 0; } } class ITNode { // 'i' could also be a normal variable constructor() { this .max = 0; this .left = null ; this .right = null ; this .i = null ; } } // A utility function to create a new node function newNodeDouble(l, h) { var temp = new Interval(); temp.low = l; temp.high = h; return temp; } // A utility function to create a new node function newNodeSingle(i) { var temp = new ITNode(); temp.i = i; temp.max = i.high; temp.left = temp.right = null ; return temp; } // 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 to // maintain BST property function insert(root, i) { // Base case: Tree is empty, // new node becomes root if (root == null ) return newNodeSingle(i); // Get low value of interval at root var 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 function doOVerlap(i1, 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. function overlapSearch(root, 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 appointments. function printConflicting(appt, n) { // Create an empty Interval Search // Tree, add first appointment var root = null ; root = insert(root, appt[0]); // Process rest of the intervals for ( var i = 1; i < n; i++) { // If current appointment conflicts // with any of the existing intervals, // print it var res = overlapSearch(root, appt[i]); if (res != null ) document.write( "[" + appt[i].low + "," + appt[i].high + "] Conflicts with [" + res.low + "," + res.high + "]<br>" ); // Insert this appointment root = insert(root, appt[i]); } } // Driver code var appt = Array(6); appt[0] = newNodeDouble(1, 5); appt[1] = newNodeDouble(3, 7); appt[2] = newNodeDouble(2, 6); appt[3] = newNodeDouble(10, 15); appt[4] = newNodeDouble(5, 6); appt[5] = newNodeDouble(4, 100); var n = appt.length; document.write( "Following are conflicting intervals<br>" ); printConflicting(appt, n); </script> |
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 a simple Binary Search Tree insert operations. Therefore, the 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