Dada una cola que consta de los primeros N números naturales y consultas Query[][] del tipo {E, X}, la tarea es realizar las consultas dadas en la cola dada de acuerdo con las siguientes reglas:
- Si el valor de E es 1 , extraiga el elemento frontal de la cola .
- Si el valor de E es 2 , elimine el valor X de la cola si existe.
- Si el valor de E es 3 , busque la posición de X en la cola, si existe. De lo contrario, imprima «-1» .
Ejemplos:
Entrada: N = 5, Query[][] = {{1, 0}, {3, 3}, {2, 2}}
Salida: 2
Explicación:
Inicialmente, la cola parece { 1, 2, 3, 4, 5 }. Las siguientes son las operaciones realizadas de acuerdo con las consultas dadas:
1ra consulta E = 1, X = 0 -> 1 aparece cambiando la cola a { 2, 3, 4, 5 }.
2ª consulta E = 3, X = 3 -> Se imprime la posición de 3.
La tercera consulta E = 2, X = 2 -> 2 se elimina de la cola cambiando la cola a { 3, 4, 5 }.Entrada: N = 5, Consulta[][] = {{1, 0}, {3, 1}}
Salida: -1
Enfoque: el problema dado se puede resolver utilizando el enfoque codicioso y la búsqueda binaria . Siga los pasos a continuación para resolver el problema dado.
- Inicialice una variable, digamos countE1 como 0 para contar el número de ocurrencias del evento E1 .
- Incremente el valor de countE1 en 1 cuando la consulta sea de tipo E1 .
- Mantener una estructura de datos establecida que almacene los elementos que se eliminan al realizar operaciones de consulta.
- Para la consulta de tipo 3 realizar los siguientes pasos:
- Inicializa una variable, digamos position para almacenar la posición inicial de X .
- Encuentre el valor de X en el conjunto para verificar si X ya se eliminó o no y si X está presente en el conjunto , la impresión -1 . De lo contrario, encuentre la posición de X .
- Atraviese el conjunto con el iterador, dígalo y , si su valor > X , salga del bucle . De lo contrario, disminuya la posición en 1 .
- Imprime la posición final de la tienda X en la variable de posición.
A continuación se muestra la implementación del enfoque anterior:
C++
// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
// Function to perform all the queries
// operations on the given queue
void solve(int n, int m, int** queries)
{
// Stores the count query of type 1
int countE1 = 0;
set<int> removed;
for (int i = 0; i < m; i++) {
// Event E1: increase countE1
if (queries[i][0] == 1)
++countE1;
// Event E2: add the X in set
else if (queries[i][0] == 2)
removed.insert(queries[i][1]);
// Event E3: Find position of X
else {
// Initial position is
// (position - countE1)
int position = queries[i][1]
- countE1;
// If X is already removed or
// popped out
if (removed.find(queries[i][1])
!= removed.end()
|| position <= 0)
cout << "-1\n";
// Finding the position of
// X in queue
else {
// Traverse set to decrease
// position of X for all the
// number removed in front
for (int it : removed) {
if (it > queries[i][1])
break;
position--;
}
// Print the position of X
cout << position << "\n";
}
}
}
}
// Driver Code
int main()
{
int N = 5, Q = 3;
int** queries = new int*[Q];
for (int i = 0; i < Q; i++) {
queries[i] = new int[2];
}
queries[0][0] = 1;
queries[0][1] = 0;
queries[1][0] = 3;
queries[1][1] = 3;
queries[2][0] = 2;
queries[2][1] = 2;
solve(N, Q, queries);
return 0;
}
Java
// Java program for the above approach
import java.util.*;
class GFG{
// Function to perform all the queries
// operations on the given queue
static void solve(int n, int m, int [][]queries)
{
// Stores the count query of type 1
int countE1 = 0;
HashSet<Integer> removed = new HashSet<Integer>();
for (int i = 0; i < m; i++) {
// Event E1: increase countE1
if (queries[i][0] == 1)
++countE1;
// Event E2: add the X in set
else if (queries[i][0] == 2)
removed.add(queries[i][1]);
// Event E3: Find position of X
else {
// Initial position is
// (position - countE1)
int position = queries[i][1]
- countE1;
// If X is already removed or
// popped out
if (removed.contains(queries[i][1])
|| position <= 0)
System.out.print("-1\n");
// Finding the position of
// X in queue
else {
// Traverse set to decrease
// position of X for all the
// number removed in front
for (int it : removed) {
if (it > queries[i][1])
break;
position--;
}
// Print the position of X
System.out.print(position+ "\n");
}
}
}
}
// Driver Code
public static void main(String[] args)
{
int N = 5, Q = 3;
int [][]queries = new int[Q][2];
queries[0][0] = 1;
queries[0][1] = 0;
queries[1][0] = 3;
queries[1][1] = 3;
queries[2][0] = 2;
queries[2][1] = 2;
solve(N, Q, queries);
}
}
// This code is contributed by shikhasingrajput
Python3
# python program for the above approach
# Function to perform all the queries
# operations on the given queue
def solve(n, m, queries):
# Stores the count query of type 1
countE1 = 0
removed = set()
for i in range(0, m):
# Event E1: increase countE1
if (queries[i][0] == 1):
countE1 += 1
# Event E2: add the X in set
elif(queries[i][0] == 2):
removed.add(queries[i][1])
# Event E3: Find position of X
else:
# Initial position is
# (position - countE1)
position = queries[i][1] - countE1
# If X is already removed or
# popped out
if (queries[i][1] in removed or position <= 0):
print("-1")
# Finding the position of
# X in queue
else:
# Traverse set to decrease
# position of X for all the
# number removed in front
for it in removed:
if (it > queries[i][1]):
break
position -= 1
# Print the position of X
print(position)
# Driver Code
if __name__ == "__main__":
N = 5
Q = 3
queries = [[0 for _ in range(2)] for _ in range(Q)]
queries[0][0] = 1
queries[0][1] = 0
queries[1][0] = 3
queries[1][1] = 3
queries[2][0] = 2
queries[2][1] = 2
solve(N, Q, queries)
# This code is contributed by rakeshsahni
C#
// C# program for the above approach
using System;
using System.Collections.Generic;
using System.Linq;
public class GFG {
// Function to perform all the queries
// operations on the given queue
static void solve(int n, int m, int [,]queries)
{
// Stores the count query of type 1
int countE1 = 0;
HashSet<int> removed = new HashSet<int>();
for (int i = 0; i < m; i++) {
// Event E1: increase countE1
if (queries[i, 0] == 1)
++countE1;
// Event E2: add the X in set
else if (queries[i, 0] == 2)
removed.Add(queries[i, 1]);
// Event E3: Find position of X
else {
// Initial position is
// (position - countE1)
int position = queries[i, 1]
- countE1;
// If X is already removed or
// popped out
if (removed.Contains(queries[i, 1])
|| position <= 0)
Console.WriteLine("-1\n");
// Finding the position of
// X in queue
else {
// Traverse set to decrease
// position of X for all the
// number removed in front
foreach (int it in removed) {
if (it > queries[i, 1])
break;
position--;
}
// Print the position of X
Console.WriteLine(position+ "\n");
}
}
}
}
// Driver Code
public static void Main (string[] args) {
int N = 5, Q = 3;
int [,]queries = new int[Q, 2];
queries[0, 0] = 1;
queries[0, 1] = 0;
queries[1, 0] = 3;
queries[1, 1] = 3;
queries[2, 0] = 2;
queries[2, 1] = 2;
solve(N, Q, queries);
}
}
// This code is contributed by code_hunt.
Javascript
<script>
// JavaScript Program to implement
// the above approach
// Function to perform all the queries
// operations on the given queue
function solve(n, m, queries)
{
// Stores the count query of type 1
let countE1 = 0;
let removed = new Set();
for (let i = 0; i < m; i++) {
// Event E1: increase countE1
if (queries[i][0] == 1)
++countE1;
// Event E2: add the X in set
else if (queries[i][0] == 2)
removed.add(queries[i][1]);
// Event E3: Find position of X
else {
// Initial position is
// (position - countE1)
let position = queries[i][1]
- countE1;
// If X is already removed or
// popped out
if (removed.has(queries[i][1])
!= false
|| position <= 0)
document.write("-1" + "<br>");
// Finding the position of
// X in queue
else {
// Traverse set to decrease
// position of X for all the
// number removed in front
for (let it of removed) {
if (it > queries[i][1])
break;
position--;
}
// Print the position of X
document.write(position + "<br>");
}
}
}
}
// Driver Code
let N = 5, Q = 3;
let queries = new Array(Q);
for (let i = 0; i < Q; i++) {
queries[i] = new Array(2);
}
queries[0][0] = 1;
queries[0][1] = 0;
queries[1][0] = 3;
queries[1][1] = 3;
queries[2][0] = 2;
queries[2][1] = 2;
solve(N, Q, queries);
// This code is contributed by Potta Lokesh
</script>
Producción:
2
Complejidad del tiempo:
- Para consulta de tipo 1: O(1)
- Para Consulta de tipo 2: O(log N)
- Para consulta de tipo 3: O(N 2 log N)
Espacio Auxiliar: O(N)
Publicación traducida automáticamente
Artículo escrito por tewatia5355 y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA