Find the nearest power of 2 for every array element
Given an array arr[] of size N, the task is to print the nearest power of 2 for each array element.
Note: If there happens to be two nearest powers of 2, consider the larger one.
Examples:
Input: arr[] = {5, 2, 7, 12}
Output: 4 2 8 16
Explanation:
The nearest power of arr[0] ( = 5) is 4.
The nearest power of arr[1] ( = 2) is 2.
The nearest power of arr[2] ( = 7) is 8.
The nearest power of arr[3] ( = 12) are 8 and 16. Print 16, as it is the largest.
Input: arr[] = {31, 13, 64}
Output: 32 16 64
Approach: Follow the steps below to solve the problem:
- Traverse the array from left to right.
- For every array element, find the nearest powers of 2 greater and smaller than it, i.e. calculate pow(2, log2(arr[i])) and pow(2, log2(arr[i]) + 1).
- Calculate difference of these two values from the current array element and print the nearest as specified in the problem statement.
Below is the implementation of the above approach:
C++
#include <bits/stdc++.h>
using namespace std;
void nearestPowerOfTwo( int arr[], int N)
{
for ( int i = 0; i < N; i++) {
int lg = log2(arr[i]);
int a = pow (2, lg);
int b = pow (2, lg + 1);
if ((arr[i] - a) < (b - arr[i]))
cout << a << " " ;
else
cout << b << " " ;
}
}
int main()
{
int arr[] = { 5, 2, 7, 12 };
int N = sizeof (arr) / sizeof (arr[0]);
nearestPowerOfTwo(arr, N);
return 0;
}
|
Java
import java.io.*;
class GFG {
static void nearestPowerOfTwo( int [] arr, int N)
{
for ( int i = 0 ; i < N; i++) {
int lg = ( int )(Math.log(arr[i])
/ Math.log( 2 ));
int a = ( int )(Math.pow( 2 , lg));
int b = ( int )(Math.pow( 2 , lg + 1 ));
if ((arr[i] - a) < (b - arr[i]))
System.out.print(a + " " );
else
System.out.print(b + " " );
}
}
public static void main(String[] args)
{
int [] arr = { 5 , 2 , 7 , 12 };
int N = arr.length;
nearestPowerOfTwo(arr, N);
}
}
|
Python3
import math
def nearestPowerOfTwo(arr, N):
for i in range (N):
lg = ( int )(math.log2(arr[i]))
a = ( int )(math. pow ( 2 , lg))
b = ( int )(math. pow ( 2 , lg + 1 ))
if ((arr[i] - a) < (b - arr[i])):
print (a, end = " " )
else :
print (b, end = " " )
arr = [ 5 , 2 , 7 , 12 ]
N = len (arr)
nearestPowerOfTwo(arr, N)
|
C#
using System;
class GFG {
static void nearestPowerOfTwo( int [] arr, int N)
{
for ( int i = 0; i < N; i++) {
int lg = ( int )(Math.Log(arr[i])
/ Math.Log(2));
int a = ( int )(Math.Pow(2, lg));
int b = ( int )(Math.Pow(2, lg + 1));
if ((arr[i] - a) < (b - arr[i]))
Console.Write(a + " " );
else
Console.Write(b + " " );
}
}
public static void Main(String[] args)
{
int [] arr = { 5, 2, 7, 12 };
int N = arr.Length;
nearestPowerOfTwo(arr, N);
}
}
|
Javascript
<script>
function nearestPowerOfTwo(arr , N) {
for (i = 0; i < N; i++) {
var lg = parseInt( (Math.log(arr[i]) / Math.log(2)));
var a = parseInt( (Math.pow(2, lg)));
var b = parseInt( (Math.pow(2, lg + 1)));
if ((arr[i] - a) < (b - arr[i]))
document.write(a + " " );
else
document.write(b + " " );
}
}
var arr = [ 5, 2, 7, 12 ];
var N = arr.length;
nearestPowerOfTwo(arr, N);
</script>
|
Time Complexity: O(N)
Auxiliary Space: O(1)
Last Updated :
07 Apr, 2021
Like Article
Save Article
Share your thoughts in the comments
Please Login to comment...