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Perform given queries on Queue according to the given rules

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  • 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++




// 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>


Output: 

2

 

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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