Largest set of numbers upto N such that either i or i/2 is present

Given a positive integer N, the task is to find the length of largest set that can be generated such that if i is present, then i/2 will not be present and 1<=i<=N.

Note: For multiple solutions, print anyone satisfying the condition.

Examples:

Input: N = 2
Output: 1
Explanation: There are two possible values 1 and 2. If 2 is present, 1 can not be present since 2/2=1. So only possibilities are 1 or 2. Both 1 and 2 separately can be the answers.

Input: N = 10
Output: 1 3 4 5 7 9
Explanation: In the output, there are no i for which i/2 exists.

Approach: Create a map for easy storing and searching and a vector to store the final set. Run a loop from 1 to N (say i) If i is odd then add i to the answer vector and as a key in the map. Else if i is even, search i/2 as key in the map. If i/2 is absent in the map then add i to the answer vector and as a key in the map. Follow the steps below to solve the problem:

• Initialize the map<int, bool> mp[].
• Initialize the vector<int> ans[] to store the result.
• Iterate over the range [, N] using the variable i and perform the following tasks:
  • If i is odd, then push i into the vector ans[] and insert {i, 1} into the map mp[].
  • Else if i/2 does not exist in the map mp[] then push i into the vector ans[] and insert {i, 1} into the map mp[].
• After performing the above steps, print the values of the vector ans[] as the answer.

Below is the implementation of the above approach.

C++14

// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;

// Function to get the largest set
vector<int> reqsubarray(int& n)
{

    // Initialize the map
    unordered_map<int, bool> mp;

    // Vector to store the answer
    vector<int> ans;
    int i;

    // Traverse the range
    for (i = 1; i <= n; i++) {
        if (i & 1) {
            ans.push_back(i);
            mp.insert({ i, 1 });
        }
        else if (!mp[i/2]) {
            ans.push_back(i);
            mp.insert({ i, 1 });
        }
    }

    // Return the answer
    return ans;
}

// Driver Code
int main()
{
    int n = 10, i;

    vector<int> ans;

    ans = reqsubarray(n);

    for (i = 0; i < ans.size(); i++)
        cout << ans[i] << " ";

    return 0;
}

Java

// Java program for the above approach
import java.util.ArrayList;
import java.util.HashMap;

class GFG
{

  // Function to get the largest set
  static ArrayList<Integer> reqsubarray(int n)
  {

    // Initialize the map
    HashMap<Integer, Integer> mp = new HashMap<Integer, Integer>();

    // Vector to store the answer
    ArrayList<Integer> ans = new ArrayList<Integer>();
    int i;

    // Traverse the range
    for (i = 1; i <= n; i++) {
      if ((i & 1) == 1) {
        ans.add(i);
        mp.put(i, 1);
      }
      else if (!mp.containsKey(i / 2)) {
        ans.add(i);
        mp.put(i, 1);
      }
    }

    // Return the answer
    return ans;
  }

  // Driver Code
  public static void main(String args[])
  {
    int n = 10, i;

    ArrayList<Integer> ans = reqsubarray(n);

    for (i = 0; i < ans.size(); i++)
      System.out.print(ans.get(i) + " ");

  }
}

// This code is contributed by Saurabh Jaiswal

Python3

# Python 3 program for the above approach

# Function to get the largest set
def reqsubarray(n):

    # Initialize the map
    mp = {}

    # Vector to store the answer
    ans = []

    # Traverse the range
    for i in range(1, n + 1):
        if (i & 1):
            ans.append(i)
            mp[i]= 1

        elif ((i // 2) not in mp):
            ans.append(i)
            mp[i] = 1

    # Return the answer
    return ans

# Driver Code
if __name__ == "__main__":

    n = 10

    ans = reqsubarray(n)

    for i in range(len(ans)):
        print(ans[i], end=" ")

        # This code is contributed by ukasp.

C#

// C# program for the above approach
using System;
using System.Collections;
using System.Collections.Generic;

class GFG
{

  // Function to get the largest set
  static ArrayList reqsubarray(int n)
  {

    // Initialize the map
    Dictionary<int, int> mp =
      new Dictionary<int, int>();

    // Vector to store the answer
    ArrayList ans = new ArrayList();
    int i;

    // Traverse the range
    for (i = 1; i <= n; i++) {
      if ((i & 1) == 1) {
        ans.Add(i);
        mp.Add(i, 1);
      }
      else if (!mp.ContainsKey(i / 2)) {
        ans.Add(i);
        mp.Add(i, 1);
      }
    }

    // Return the answer
    return ans;
  }

  // Driver Code
  public static void Main()
  {
    int n = 10, i;

    ArrayList ans = reqsubarray(n);

    for (i = 0; i < ans.Count; i++)
      Console.Write(ans[i] + " ");

  }
}

// This code is contributed by Samim Hossain Mondal.

Javascript

<script>
       // JavaScript code for the above approach

       // Function to get the largest set
       function reqsubarray(n) {

           // Initialize the map
           let mp = new Map();

           // Vector to store the answer
           let ans = [];
           let i;

           // Traverse the range
           for (i = 1; i <= n; i++) {
               if (i & 1) {
                   ans.push(i);
                   mp.set(i, 1);
               }
               else if (!mp.has(Math.floor(i / 2))) {
                   ans.push(i);
                   mp.set(i, 1);
               }
           }

           // Return the answer
           return ans;
       }

       // Driver Code
       let n = 10;
       let ans;
       ans = reqsubarray(n);
       for (let i = 0; i < ans.length; i++)
           document.write(ans[i] + " ");

 // This code is contributed by Potta Lokesh
   </script>

Output

1 3 4 5 7 9 

Time Complexity: O(N)
Auxiliary Space: O(N)