Perform given queries on Queue according to the given rules
Last Updated :
21 Feb, 2022
Given a queue consisting of the first N natural numbers and queries Query[][] of the type {E, X}, the task is to perform the given queries on the given queue according to the following rules:
- If the value of E is 1, then pop the front element from the queue.
- If the value of E is 2, then remove the value X from the queue if it exists.
- If the value of E is 3, then find the position of X in the queue if it exists. Otherwise, print “-1”.
Examples:
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
Approach: The given problem can be solved by using Greedy Approach and Binary Search. Follow the steps below to solve the given problem.
- Initialize a variable say countE1 as 0 to count the number of occurrences of event E1.
- Increment the value of countE1 by 1 when the query is of type E1.
- Maintain a set data structure that stores the elements which are removed while performing queries operations.
- For the query of type 3 perform the following steps:
- Initialize a variable, say position to store the initial position of X.
- Find the value of X in the set to check whether X is already removed or not and if X is present in the set, the print -1. Otherwise, find the position of X.
- Traverse the set with iterator say it and if the value of it > X, then break out of the loop. Otherwise, decrease the position by 1.
- Print the final position of X store in position variable.
Below is the implementation of the above approach:
C++
#include <bits/stdc++.h>
using namespace std;
void solve(int n, int m, int** queries)
{
int countE1 = 0;
set<int> removed;
for (int i = 0; i < m; i++) {
if (queries[i][0] == 1)
++countE1;
else if (queries[i][0] == 2)
removed.insert(queries[i][1]);
else {
int position = queries[i][1]
- countE1;
if (removed.find(queries[i][1])
!= removed.end()
|| position <= 0)
cout << "-1\n";
else {
for (int it : removed) {
if (it > queries[i][1])
break;
position--;
}
cout << position << "\n";
}
}
}
}
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
import java.util.*;
class GFG{
static void solve(int n, int m, int [][]queries)
{
int countE1 = 0;
HashSet<Integer> removed = new HashSet<Integer>();
for (int i = 0; i < m; i++) {
if (queries[i][0] == 1)
++countE1;
else if (queries[i][0] == 2)
removed.add(queries[i][1]);
else {
int position = queries[i][1]
- countE1;
if (removed.contains(queries[i][1])
|| position <= 0)
System.out.print("-1\n");
else {
for (int it : removed) {
if (it > queries[i][1])
break;
position--;
}
System.out.print(position+ "\n");
}
}
}
}
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);
}
}
|
Python3
def solve(n, m, queries):
countE1 = 0
removed = set()
for i in range(0, m):
if (queries[i][0] == 1):
countE1 += 1
elif(queries[i][0] == 2):
removed.add(queries[i][1])
else:
position = queries[i][1] - countE1
if (queries[i][1] in removed or position <= 0):
print("-1")
else:
for it in removed:
if (it > queries[i][1]):
break
position -= 1
print(position)
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)
|
C#
using System;
using System.Collections.Generic;
using System.Linq;
public class GFG {
static void solve(int n, int m, int [,]queries)
{
int countE1 = 0;
HashSet<int> removed = new HashSet<int>();
for (int i = 0; i < m; i++) {
if (queries[i, 0] == 1)
++countE1;
else if (queries[i, 0] == 2)
removed.Add(queries[i, 1]);
else {
int position = queries[i, 1]
- countE1;
if (removed.Contains(queries[i, 1])
|| position <= 0)
Console.WriteLine("-1\n");
else {
foreach (int it in removed) {
if (it > queries[i, 1])
break;
position--;
}
Console.WriteLine(position+ "\n");
}
}
}
}
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);
}
}
|
Javascript
<script>
function solve(n, m, queries)
{
let countE1 = 0;
let removed = new Set();
for (let i = 0; i < m; i++) {
if (queries[i][0] == 1)
++countE1;
else if (queries[i][0] == 2)
removed.add(queries[i][1]);
else {
let position = queries[i][1]
- countE1;
if (removed.has(queries[i][1])
!= false
|| position <= 0)
document.write("-1" + "<br>");
else {
for (let it of removed) {
if (it > queries[i][1])
break;
position--;
}
document.write(position + "<br>");
}
}
}
}
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);
</script>
|
Time Complexity:
- For Query of type 1: O(1)
- For Query of type 2: O(log N)
- For Query of type 3: O(N2 log N)
Auxiliary Space: O(N)
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