Given n appointments, find all conflicting appointments

Difficulty Level : Hard
Last Updated : 11 Dec, 2020

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.

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 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;
};
 
// 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;
}

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 apointments.
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 
# apointments.
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

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 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

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