Reduce the array to a single element with the given operation

Given an integer N and an array arr containing integers from 1 to N in a sorted fashion. The task is to reduce the array to a single element by performing the following operation: 
All the elements in the odd positions will be removed after a single operation. This operation will be performed until only a single element is left int the array and it prints that element at the end.

Examples:  

Input: N = 3 
Output: 2 
Initially the array will be arr[] = {1, 2, 3} 
After the 1st operation, ‘1’ and ‘3’ will be removed and the array becomes arr[] = {2} 
So 2 is the only element left at the end.

Input: N = 6 
Output: 4 
arr[] = {1, 2, 3, 4, 5, 6} 
After the first iteration, the array becomes {2, 4, 6} 
After the second iteration, the array becomes {4} 
So 4 is the last element.  

Approach: For this kind of problem:  

• Write multiple test cases and the respective output.
• Analyze the output for the given input and the relation between them.
• Once we find the relation we will try to express it in the form of a mathematical expression if possible.
• Write the code/algorithm for the above expression.

So let’s create a table for the given input N and its respective output. 

Analyzed Relation: The output is at 2ⁱ. Using the above table, we can create the output table for the range of inputs. 

Below is the implementation of the above approach:  

// C++ implementation of the approach 

#include<bits/stdc++.h>
using namespace std;

// Function to return the final element 
long getFinalElement(long n) 
{ 
    long finalNum; 
    for (finalNum = 2; finalNum * 2 <= n; finalNum *= 2) 
            ; 
    return finalNum; 
} 


// Driver code
int main()
{
       int N = 12; 
    cout << getFinalElement(N) ;
   
   return 0;
}
// This code is contributed by Ryuga

// Java implementation of the approach
class OddPosition {

    // Function to return the final element
    public static long getFinalElement(long n)
    {
        long finalNum;
        for (finalNum = 2; finalNum * 2 <= n; finalNum *= 2)
            ;
        return finalNum;
    }

    // Driver code
    public static void main(String[] args)
    {
        int N = 12;
        System.out.println(getFinalElement(N));
    }
}

# Python 3 implementation of the approach 

# Function to return the final element 
def getFinalElement(n):

    finalNum = 2
    while finalNum * 2 <= n: 
        finalNum *= 2
    return finalNum

# Driver code
if __name__ =="__main__":

    N = 12
    print( getFinalElement(N))

# This code is contributed 
# by ChitraNayal

 // C# implementation of the approach
using System;

public class GFG{
    
    // Function to return the final element
    public static long getFinalElement(long n)
    {
        long finalNum;
        for (finalNum = 2; finalNum * 2 <= n; finalNum *= 2)
            ;
        return finalNum;
    }

    // Driver code
    static public void Main (){
        int N = 12;
        Console.WriteLine(getFinalElement(N));
    }
}

<script>
// Javascript implementation of the approach
    
    // Function to return the final element
    function getFinalElement(n)
    {
        let finalNum;
        for (finalNum = 2; finalNum * 2 <= n; finalNum *= 2)
            ;
        return finalNum;
    }
    
    // Driver code
    let N = 12;
    document.write(getFinalElement(N));
    
// This code is contributed by avanitrachhadiya2155
</script>

    

<?php
//PHP implementation of the approach 

// Function to return the final element 
function  getFinalElement($n) 
{ 
    $finalNum=0; 
    for ($finalNum = 2; ($finalNum * 2) <= $n; $finalNum *= 2) ;
        
    return $finalNum; 
} 


// Driver code
    $N = 12; 
    echo  getFinalElement($N) ;
    
    
// This code is contributed by akt_mit
?>

8

Time Complexity: O(logN), since every time the finalNum value is becoming twice its current value.
Auxiliary Space: O(1), since no extra space has been taken.