根据给定的规则对 Queue 执行给定的查询
给定一个由前N个自然数组成的队列和类型为 {E, X} 的查询Query[][] ，任务是根据以下规则对给定队列执行给定查询：
- 如果E的值为1 ，则从队列中弹出最前面的元素。
- 如果E的值为2 ，则将值X从队列中删除（如果存在）。
- 如果E的值为3 ，则在队列中找到X的位置（如果存在）。否则，打印“-1” 。
例子：
Input: N = 5, Query[][] = {{1, 0}, {3, 3}, {2, 2}}
Output: 2
Explanation:
Initially queue looks like { 1, 2, 3, 4, 5 }. Following are the operations performed according to the queries given:
1st query E = 1, X = 0 -> 1 is popped out changing queue to { 2, 3, 4, 5 }.
2nd query E = 3, X = 3 -> The position of 3 is printed.
3rd query E = 2, X = 2 -> 2 is removed from queue changing queue to { 3, 4, 5 }.
Input: N = 5, Query[][] = {{1, 0}, {3, 1}}
Output: -1
方法：给定的问题可以通过使用贪婪方法和二分搜索来解决。请按照以下步骤解决给定的问题。
- 将变量countE1初始化为0以计算事件E1的发生次数。
- 当查询类型为E1时，将countE1的值增加1 。
- 维护一个集合数据结构，该结构存储在执行查询操作时删除的元素。
- 对于类型 3的查询，请执行以下步骤：
- 初始化一个变量，比如position来存储X的初始位置。
- 在集合中查找X的值以检查X是否已被删除，如果X存在于集合中，则打印-1 。否则，求X的位置。
- 用迭代器遍历集合，如果它的值> X ，则跳出循环。否则，将位置减少1 。
- 在位置变量中打印X存储的最终位置。
下面是上述方法的实现：
C++
// C++ program for the above approach
#include
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 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 removed = new HashSet();
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 rakeshsahniC#
// 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 removed = new HashSet();
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
输出：
2时间复杂度：
- 对于类型 1 的查询： O(1)
- 对于类型 2 的查询： O(log N)
- 对于类型 3 的查询： O(N 2 log N)
辅助空间： O(N)