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 sorted fashion. The task is to reduce the array to a single element by performing the following operation:
All the elements at 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 print that element in the end.

Examples:

Input: N = 3
Output: 2
Initially the array will be arr[] = {1, 2, 3}
After 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 first iteration, array becomes {2, 4, 6}
After second iteration, 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 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

Input(N) Output
3 2
4 4
6 4
8 8
12 8
20 16

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

Input(N) Output
2-3 2
4-7 4
8-15 8
16-31 16
32-63 32
2i – 2i + 1 – 1 2i

Below is the implementation of the above approach:

C++

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// 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

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Java

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// 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));
    }
}

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Python3

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# 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

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C#

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// 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));
    }
}

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PHP

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<?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
?>

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Output:

8


My Personal Notes arrow_drop_up

LayCoder

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