Reduce the array to a single element with the given operation
Last Updated :
02 Jul, 2022
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.
Input(N) |
Output |
3 |
2 |
4 |
4 |
6 |
4 |
8 |
8 |
12 |
8 |
20 |
16 |
|
|
Analyzed Relation: The output is at 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++
#include<bits/stdc++.h>
using namespace std;
long getFinalElement( long n)
{
long finalNum;
for (finalNum = 2; finalNum * 2 <= n; finalNum *= 2)
;
return finalNum;
}
int main()
{
int N = 12;
cout << getFinalElement(N) ;
return 0;
}
|
Java
class OddPosition {
public static long getFinalElement( long n)
{
long finalNum;
for (finalNum = 2 ; finalNum * 2 <= n; finalNum *= 2 )
;
return finalNum;
}
public static void main(String[] args)
{
int N = 12 ;
System.out.println(getFinalElement(N));
}
}
|
Python3
def getFinalElement(n):
finalNum = 2
while finalNum * 2 < = n:
finalNum * = 2
return finalNum
if __name__ = = "__main__" :
N = 12
print ( getFinalElement(N))
|
C#
using System;
public class GFG{
public static long getFinalElement( long n)
{
long finalNum;
for (finalNum = 2; finalNum * 2 <= n; finalNum *= 2)
;
return finalNum;
}
static public void Main (){
int N = 12;
Console.WriteLine(getFinalElement(N));
}
}
|
PHP
<?php
function getFinalElement( $n )
{
$finalNum =0;
for ( $finalNum = 2; ( $finalNum * 2) <= $n ; $finalNum *= 2) ;
return $finalNum ;
}
$N = 12;
echo getFinalElement( $N ) ;
?>
|
Javascript
<script>
function getFinalElement(n)
{
let finalNum;
for (finalNum = 2; finalNum * 2 <= n; finalNum *= 2)
;
return finalNum;
}
let N = 12;
document.write(getFinalElement(N));
</script>
|
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.
Share your thoughts in the comments
Please Login to comment...