# Highest and Smallest power of K less than and greater than equal to N respectively

• Last Updated : 19 Jan, 2022

Given positive integers N and K, the task is to find the highest and smallest power of K greater than equal to and less than equal to N respectively.

Examples:

Input: N = 3, K = 2
Output: 2 4
Highest power of 2 less than 3 = 2
Smallest power of 2 greater than 3 = 4

Input: N = 6, K = 3
Output: 3 9
Highest power of 3 less than 6 = 3
Smallest power of 3 greater than 6 = 9

Approach:

1. Compute the log of N in base K (logK N) to get the exponential power such that K raised to this exponent is the Highest power of K less than equal to N.
2. For the Smallest power of K less than equal to N, find the next power of K computed from the last step

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach` `#include ``using` `namespace` `std;` `// Function to return the highest power``// of k less than or equal to n``int` `prevPowerofK(``int` `n, ``int` `k)``{``    ``int` `p = (``int``)(``log``(n) / ``log``(k));``    ``return` `(``int``)``pow``(k, p);``}` `// Function to return the smallest power``// of k greater than or equal to n``int` `nextPowerOfK(``int` `n, ``int` `k)``{``    ``return` `prevPowerofK(n, k) * k;``}` `// Function to print the result``void` `printResult(``int` `n, ``int` `k)``{``    ``cout << prevPowerofK(n, k)``         ``<< ``" "` `<< nextPowerOfK(n, k)``         ``<< endl;``}` `// Driver code``int` `main()``{``    ``int` `n = 25, k = 3;` `    ``printResult(n, k);` `    ``return` `0;``}`

## Java

 `// Java implementation of the approach``import` `java.io.*;` `class` `GFG{` `// Function to return the highest power``// of k less than or equal to n``static` `int` `prevPowerofK(``int` `n, ``int` `k)``{``    ``int` `p = (``int``)(Math.log(n) / Math.log(k));``    ``return` `(``int``) Math.pow(k, p);``}` `// Function to return the smallest power``// of k greater than or equal to n``static` `int` `nextPowerOfK(``int` `n, ``int` `k)``{``    ``return` `prevPowerofK(n, k) * k;``}` `// Function to print the result``static` `void` `printResult(``int` `n, ``int` `k)``{``    ``System.out.println(prevPowerofK(n, k) + ``" "` `+``                       ``nextPowerOfK(n, k));``}` `// Driver Code``public` `static` `void` `main (String args[])``{``    ``int` `n = ``25``, k = ``3``;``    ``printResult(n, k);``}``}` `// This code is contributed by shivanisinghss2110`

## Python3

 `# Python3 implementation of the approach``import` `math` `# Function to return the highest power``# of k less than or equal to n``def` `prevPowerofK(n, k):` `    ``p ``=` `int``(math.log(n) ``/` `math.log(k))``    ``return` `int``(math.``pow``(k, p))` `# Function to return the smallest power``# of k greater than or equal to n``def` `nextPowerOfK(n, k):` `    ``return` `prevPowerofK(n, k) ``*` `k` `# Function to print the result``def` `printResult(n, k):` `    ``print``(prevPowerofK(n, k), nextPowerOfK(n, k))` `# Driver code``n ``=` `6``k ``=` `3` `printResult(n, k)` `# This code is contributed by divyamohan123`

## C#

 `// C# implementation of the approach``using` `System;``class` `GFG{` `// Function to return the highest power``// of k less than or equal to n``static` `int` `prevPowerofK(``int` `n, ``int` `k)``{``    ``int` `p = (``int``)(Math.Log(n) / Math.Log(k));``    ``return` `(``int``) Math.Pow(k, p);``}` `// Function to return the smallest power``// of k greater than or equal to n``static` `int` `nextPowerOfK(``int` `n, ``int` `k)``{``    ``return` `prevPowerofK(n, k) * k;``}` `// Function to print the result``static` `void` `printResult(``int` `n, ``int` `k)``{``    ``Console.WriteLine(prevPowerofK(n, k) + ``" "` `+``                      ``nextPowerOfK(n, k));``}` `// Driver Code``public` `static` `void` `Main(String []args)``{``    ``int` `n = 25, k = 3;``    ``printResult(n, k);``}``}` `// This code is contributed by gauravrajput1`

## Javascript

 ``
Output:
`9 27`

Time Complexity: O(logkn)

Auxiliary Space: O(1)

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