Given an integer N, the task is to find the smallest power of four greater than or equal to N.
Examples:
Input: N = 12
Output: 16
24 = 16 which is the next required
greater number after 12.Input: N = 81
Output: 81
Approach:
- Find the fourth root of the given n.
- Calculate its floor value using floor() function.
- If n is itself a power of four then return n.
- Else add 1 to the floor value.
- Return the fourth power of that number.
Below is the implementation of the above approach:
C++
// C++ implementation of above approach#include <bits/stdc++.h>using namespace std;// Function to return the smallest power// of 4 greater than or equal to nint nextPowerOfFour(int n){ int x = floor(sqrt(sqrt(n))); // If n is itself is a power of 4 then return n if (pow(x, 4) == n) return n; else { x = x + 1; return pow(x, 4); }}// Driver codeint main(){ int n = 122; printf("%d", nextPowerOfFour(n)); return 0;}// This code is contributed by Sania Kumari Gupta |
C
// C implementation of above approach#include <math.h>#include <stdio.h>// Function to return the smallest power// of 4 greater than or equal to nint nextPowerOfFour(int n){ int x = floor(sqrt(sqrt(n))); // If n is itself is a power of 4 then return n if (pow(x, 4) == n) return n; else { x = x + 1; return pow(x, 4); }}// Driver codeint main(){ int n = 122; printf("%d", nextPowerOfFour(n)); return 0;}// This code is contributed by Sania Kumari Gupta |
Java
// Java implementation of above approach import java.util.*;import java.lang.Math;import java.io.*;class GFG{ // Function to return the smallest power // of 4 greater than or equal to n static int nextPowerOfFour(int n) { int x = (int)Math.floor(Math.sqrt(Math.sqrt(n))); // If n is itself is a power of 4 // then return n if (Math.pow(x, 4) == n) return n; else { x = x + 1; return (int)Math.pow(x, 4); } } // Driver code public static void main (String[] args) throws java.lang.Exception{ int n = 122; System.out.println(nextPowerOfFour(n)); }}// This code is contributed by nidhiva |
Python3
# Python3 implementation of above approachimport math# Function to return the smallest power# of 4 greater than or equal to ndef nextPowerOfFour(n): x = math.floor((n**(1/2))**(1/2)); # If n is itself is a power of 4 # then return n if ((x**4) == n): return n; else: x = x + 1; return (x**4);# Driver coden = 122;print(nextPowerOfFour(n));# This code is contributed by Rajput-Ji |
C#
// C# implementation of above approach using System;class GFG{ // Function to return the smallest power // of 4 greater than or equal to n static int nextPowerOfFour(int n) { int x = (int)Math.Floor(Math.Sqrt(Math.Sqrt(n))); // If n is itself is a power of 4 // then return n if (Math.Pow(x, 4) == n) return n; else { x = x + 1; return (int)Math.Pow(x, 4); } } // Driver code public static void Main () { int n = 122; Console.WriteLine(nextPowerOfFour(n)); }}// This code is contributed by anuj_67.. |
Javascript
<script>// Javascript implementation of above approach// Function to return the smallest power// of 4 greater than or equal to nfunction nextPowerOfFour(n){ let x = Math.floor(Math.sqrt( Math.sqrt(n))); // If n is itself is a power of 4 // then return n if (Math.pow(x, 4) == n) return n; else { x = x + 1; return Math.pow(x, 4); }}// Driver codelet n = 122;document.write(nextPowerOfFour(n));// This code is contributed by mohit kumar 29</script> |
Output:
256
Time Complexity: O(logx) where x is sqrt(sqrt(n))
Auxiliary Space: O(1)
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