Dadas n citas, busque todas las citas en conflicto.
Ejemplos:
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]
Una cita es conflictiva si entra en conflicto con alguna de las citas anteriores de la array.
Recomendamos encarecidamente minimizar el navegador y probarlo usted mismo primero.
Una solución simple es procesar una por una todas las citas desde la segunda cita hasta la última. Para cada cita i, compruebe si entra en conflicto con i-1, i-2, … 0. La complejidad temporal de este método es O(n 2 ).
Podemos usar Interval Tree para resolver este problema en tiempo O(nLogn). A continuación se muestra un algoritmo detallado.
- Cree un árbol de intervalos, inicialmente con la primera cita.
- Haga lo siguiente para todas las demás citas a partir de la segunda.
- Compruebe si la cita actual entra en conflicto con alguna de las citas existentes en el árbol de intervalos. Si hay conflicto, imprima la cita actual. Este paso se puede hacer O (Iniciar sesión) tiempo.
- Inserte la cita actual en el árbol de intervalos. Este paso también se puede realizar O(Log) time.
A continuación se muestra la implementación de la idea anterior.
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>
Producción
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]
Tenga en cuenta que la implementación anterior utiliza una simple operación de inserción de árbol de búsqueda binaria. Por lo tanto, la complejidad temporal de la implementación anterior es mayor que O(nLogn). Podemos usar técnicas de balanceo Red-Black Tree o AVL Tree para hacer la implementación anterior O(nLogn).
Este artículo es una contribución de Anmol . Escriba comentarios si encuentra algo incorrecto o si desea compartir más información sobre el tema tratado anteriormente.
Publicación traducida automáticamente
Artículo escrito por GeeksforGeeks-1 y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA