Find first undeleted integer from K to N in given unconnected Graph after performing Q queries

Given a positive integer N representing the set of integers [1, N] and an array queries[] of length Q of type {L, K}, the task is to perform the given queries according to the following rules and print the result:

• If the value of L is 1, then delete the given integer K.
• If the value of L is 2, then print the first integer from K to N which is not deleted.

Note: When all the elements to the right are deleted, the answer will be -1.

Examples:

Input: N = 3, queries[] = {{2, 1}, {1, 1}, {2, 1}, {2, 3}}
Output: 1 2 3
Explanation:
For the first query: the rightmost element for 1 is 1 only.
After the second query: 1 is deleted.
For the third query: the rightmost element for 1 is 2.
For the fourth query: the rightmost element for 3 is 3.

Input: N = 5, queries = {{2, 1}, {1, 3}, {2, 3}, {1, 2}, {2, 2}, {1, 5}, {2, 5}}
Output: 1 4 4 -1 

Approach: The given problem can be solved by using Disjoint Set Union Data Structure. Initially, all the elements are in different sets, for the first type query perform the Union Operation on the given integer. For the second type of query, get the parent of the given vertex and print the value stored in the array graph[]. Follow the steps below to solve the problem:

• Initialize the vector graph(N + 2).
• Iterate over the range [1, N+1] using the variable i and set the value of graph[i] as i.
• Iterate over the queries[] using the variable query and perform the following steps:
  • If query.first is 1 then call for function operation Union(graph, query.second, query.second + 1).
  • Otherwise, call for function operation Get(graph, query.second) and assign it to variable a. Now, if the value of a is (N + 1), then print -1. Otherwise, print the value of graph[a] as the result.

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 the Get operation
// of disjoint set union
int Get(vector<int>& graph, int a)
{
    return graph[a]
           = (graph[a] == a ? a
                            : Get(graph, graph[a]));
}

// Function to perform the union
// operation of disjoint set union
void Union(vector<int>& graph,
           int a, int b)
{

    a = Get(graph, a);
    b = Get(graph, b);

    // Update the graph[a] as b
    graph[a] = b;
}

// Function to perform given queries on
// set of vertices initially not connected
void Queries(vector<pair<int, int> >& queries,
             int N, int M)
{

    // Stores the vertices
    vector<int> graph(N + 2);

    // Mark every vertices rightmost
    // vertex as i
    for (int i = 1; i <= N + 1; i++) {

        graph[i] = i;
    }

    // Traverse the queries array
    for (auto query : queries) {

        // Check if it is first type
        // of the given query
        if (query.first == 1) {

            Union(graph, query.second,
                  query.second + 1);
        }
        else {

            // Get the parent of a
            int a = Get(graph, query.second);

            // Print the answer for
            // the second query
            if (a == N + 1)
                cout << -1 << " ";
            else
                cout << graph[a] << " ";
        }
    }
}

// Driver Code
int main()
{
    int N = 5;
    vector<pair<int, int> > queries{
        { 2, 1 }, { 1, 1 }, { 2, 1 }, { 2, 3 }
    };
    int Q = queries.size();

    Queries(queries, N, Q);

    return 0;
}

Java

// Java program for the above approach
class GFG{

// Function to perform the Get operation
// of disjoint set union
public static int Get(int[] graph, int a)
{
    return graph[a]
           = (graph[a] == a ? a
                            : Get(graph, graph[a]));
}

// Function to perform the union
// operation of disjoint set union
public static void Union(int[] graph,
           int a, int b)
{

    a = Get(graph, a);
    b = Get(graph, b);

    // Update the graph[a] as b
    graph[a] = b;
}

// Function to perform given queries on
// set of vertices initially not connected
public static void Queries(int[][] queries,
             int N, int M)
{

    // Stores the vertices
    int[] graph = new int[N + 2];

    // Mark every vertices rightmost
    // vertex as i
    for (int i = 1; i <= N + 1; i++) {

        graph[i] = i;
    }

    // Traverse the queries array
    for (int[] query : queries) {

        // Check if it is first type
        // of the given query
        if (query[0] == 1) {

            Union(graph, query[1],
                  query[1] + 1);
        }
        else {

            // Get the parent of a
            int a = Get(graph, query[1]);

            // Print the answer for
            // the second query
            if (a == N + 1)
                System.out.print(-1);
            else
                System.out.print(graph[a] + " ");
        }
    }
}

// Driver Code
public static void main(String args[])
{
    int N = 5;
    int[][] queries  = { { 2, 1 }, { 1, 1 }, { 2, 1 }, { 2, 3 }};
    int Q = queries.length;

    Queries(queries, N, Q);

}
}

// This code is contributed by gfgking.

Python3

# Python 3 program for the above approach

# Function to perform the Get operation
# of disjoint set union
def Get(graph, a):
    if graph[graph[a]]!= graph[a]:
        graph[a] = Get(graph, graph[a])
    else:
        return graph[a]

# Function to perform the union
# operation of disjoint set union
def Union(graph, a, b):
    a = Get(graph, a)
    b = Get(graph, b)

    # Update the graph[a] as b
    graph[a] = b

# Function to perform given queries on
# set of vertices initially not connected
def Queries(queries, N, M):

    # Stores the vertices
    graph = [0 for i in range(N + 2)]

    # Mark every vertices rightmost
    # vertex as i
    for i in range(1,N + 2,1):
        graph[i] = i

    # Traverse the queries array
    for query in queries:

        # Check if it is first type
        # of the given query
        if (query[0] == 1):
            Union(graph, query[1], query[1] + 1)

        else:

            # Get the parent of a
            a = Get(graph, query[1])

            # Print the answer for
            # the second query
            if (a == N + 1):
                print(-1)
            else:
                print(graph[a],end = " ")

# Driver Code
if __name__ == '__main__':
    N = 5
    queries = [[2, 1],[1, 1],[2, 1],[2, 3]]
    Q = len(queries)

    Queries(queries, N, Q)

    # This code is contributed by SURENDRA_GANGWAR.

C#

// C# program for the above approach

using System;

class GFG
{
    // Function to perform the Get operation
    // of disjoint set union
    public static int Get(int[] graph, int a)
    {
        return graph[a] = (graph[a] == a ? a : Get(graph, graph[a]));
    }

    // Function to perform the union
    // operation of disjoint set union
    public static void Union(int[] graph, int a, int b)
    {
        a = Get(graph, a);
        b = Get(graph, b);

        // Update the graph[a] as b
        graph[a] = b;
    }

    // Function to perform given queries on
    // set of vertices initially not connected
    public static void Queries(int[][] queries, int N, int M)
    {
        // Stores the vertices
        int[] graph = new int[N + 2];

        // Mark every vertices rightmost
        // vertex as i
        for (int i = 1; i <= N + 1; i++)
        {
            graph[i] = i;
        }

        // Traverse the queries array
        foreach (int[] query in queries)
        {
            // Check if it is first type
            // of the given query
            if (query[0] == 1)
            {
                Union(graph, query[1], query[1] + 1);
            }
            else
            {
                // Get the parent of a
                int a = Get(graph, query[1]);

                // Print the answer for
                // the second query
                if (a == N + 1)
                    Console.Write(-1);
                else
                    Console.Write(graph[a] + " ");
            }
        }
    }

    // Driver Code
    static void Main(string[] args)
    {
        int N = 5;
        int[][] queries = {
            new int[] { 2, 1 },
            new int[] { 1, 1 },
            new int[] { 2, 1 },
            new int[] { 2, 3 }
        };
        int Q = queries.Length;

        Queries(queries, N, Q);
    }
}

Javascript

<script>
// Javascript program for the above approach

// Function to perform the Get operation
// of disjoint set union
function Get(graph, a) {
  return (graph[a] = graph[a] == a ? a : Get(graph, graph[a]));
}

// Function to perform the union
// operation of disjoint set union
function Union(graph, a, b) {
  a = Get(graph, a);
  b = Get(graph, b);

  // Update the graph[a] as b
  graph[a] = b;
}

// Function to perform given queries on
// set of vertices initially not connected
function Queries(queries, N, M) {
  // Stores the vertices
  let graph = new Array(N).fill(2);

  // Mark every vertices rightmost
  // vertex as i
  for (let i = 1; i <= N + 1; i++) {
    graph[i] = i;
  }

  // Traverse the queries array
  for (let query of queries) {
    // Check if it is first type
    // of the given query
    if (query[0] == 1) {
      Union(graph, query[1], query[1] + 1);
    } else {
      // Get the parent of a
      let a = Get(graph, query[1]);

      // Print the answer for
      // the second query
      if (a == N + 1) document.write(-1 + " ");
      else document.write(graph[a] + " ");
    }
  }
}

// Driver Code
let N = 5;
let queries = [
  [2, 1],
  [1, 1],
  [2, 1],
  [2, 3],
];
let Q = queries.length;

Queries(queries, N, Q);

// This code is contributed by gfgking.
</script>

Output:

1 2 3

Time Complexity: O(M*log N)
Auxiliary Space: O(N)