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

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  • Difficulty Level : Easy
  • Last Updated : 18 Jan, 2022
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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:

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) 


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