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++
// 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 arrayvoid 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 Codeint main(){ int arr[] = { 5, 2, 7, 12 }; int N = sizeof(arr) / sizeof(arr[0]); nearestPowerOfTwo(arr, N); return 0;} |
Java
// Java program to implement the above approachimport 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); }} |
Python3
# Python program to implement the above approachimport math# Function to find the nearest power# of two for every array elementdef 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 Codearr = [5, 2, 7, 12]N = len(arr)nearestPowerOfTwo(arr, N) |
C#
// C# program to implement the above approachusing 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); }} |
Javascript
<script>// JavaScript program to implement // the above approach // Function to find the nearest power of two // for every element of the given array function nearestPowerOfTwo(arr , N) { // Traverse the array for (i = 0; i < N; i++) { // Calculate log of the // current array element 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))); // Find the nearest if ((arr[i] - a) < (b - arr[i])) document.write(a + " "); else document.write(b + " "); } } // Driver Code var arr = [ 5, 2, 7, 12 ]; var N = arr.length; nearestPowerOfTwo(arr, N);// This code is contributed by todaysgaurav </script> |
Output:
4 2 8 16
Time Complexity: O(N)
Auxiliary Space: O(1)
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