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Find the nearest power of 2 for every array element
  • Last Updated : 11 Jan, 2021

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++

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// C++ program to implement
// the above approach
#include <bits/stdc++.h>
using namespace std;
  
// Function to find nearest power of two
// for every element in the given array
void nearestPowerOfTwo(int arr[], int N)
{
  
    // Traverse the array
    for (int i = 0; i < N; i++) {
  
        // Calculate log of the
        // current array element
        int lg = log2(arr[i]);
        int a = pow(2, lg);
        int b = pow(2, lg + 1);
  
        // Find the nearest
        if ((arr[i] - a) < (b - arr[i]))
            cout << a << " ";
        else
            cout << b << " ";
    }
}
  
// Driver Code
int main()
{
    int arr[] = { 5, 2, 7, 12 };
    int N = sizeof(arr) / sizeof(arr[0]);
    nearestPowerOfTwo(arr, N);
    return 0;
}

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Java

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// Java program to implement the above approach
import java.io.*;
  
class GFG {
  
    // Function to find the nearest power of two
    // for every element of the given array
    static void nearestPowerOfTwo(int[] arr, int N)
    {
  
        // Traverse the array
        for (int i = 0; i < N; i++) {
  
            // Calculate log of the
            // current array element
            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));
  
            // Find the nearest
            if ((arr[i] - a) < (b - arr[i]))
                System.out.print(a + " ");
            else
                System.out.print(b + " ");
        }
    }
  
    // Driver Code
    public static void main(String[] args)
    {
        int[] arr = { 5, 2, 7, 12 };
        int N = arr.length;
        nearestPowerOfTwo(arr, N);
    }
}

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Python3

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# Python program to implement the above approach
import math
  
# Function to find the nearest power
# of two for every array element
def nearestPowerOfTwo(arr, N):
  
    # Traverse the array
    for i in range(N):
  
        # Calculate log of current array element
        lg = (int)(math.log2(arr[i]))
  
        a = (int)(math.pow(2, lg))
        b = (int)(math.pow(2, lg + 1))
  
        # Find the nearest
        if ((arr[i] - a) < (b - arr[i])):
            print(a, end = " ")
        else:
            print(b, end = " ")
  
  
# Driver Code
arr = [5, 2, 7, 12]
N = len(arr)
nearestPowerOfTwo(arr, N)

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C#

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// C# program to implement the above approach
using System;
  
class GFG {
  
    // Function to find nearest power of two
    // for every array element
    static void nearestPowerOfTwo(int[] arr, int N)
    {
  
        // Traverse the array
        for (int i = 0; i < N; i++) {
  
            // Calculate log of the
            // current array element
            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));
  
            // Find the nearest
            if ((arr[i] - a) < (b - arr[i]))
                Console.Write(a + " ");
            else
                Console.Write(b + " ");
        }
    }
  
    // Driver Code
    public static void Main(String[] args)
    {
        int[] arr = { 5, 2, 7, 12 };
        int N = arr.Length;
        nearestPowerOfTwo(arr, N);
    }
}

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Output:

4 2 8 16

Time Complexity: O(N)
Auxiliary Space: O(1)

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