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

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 = 2
Smallest power of 3 greater than 6 = 9

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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 `

Output:

```9 27
```

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