Find the minimum number to be added to N to make it a power of K

Given two positive integers N and K, the task is to find the minimum number to be added to N to make it a power of K.
Examples:

Input: N = 9, K = 10 
Output: 1 
Explanation:
9 + 1 = 10 = 10¹
Input: N = 20, K = 5 
Output: 5 
Explanation:
20 + 5 = 25 = 5²

Approach: The idea to solve this problem is to observe that the minimum power of K which can be formed from N is the next greater power of K. So, the idea is to find the next greater power of K and find the difference between N and this number. The next greater power of K can be found by the formula, 

K^{int(log(N)/log(K)) + 1}

Therefore, the minimum number to be added can be computed by: 

Minimum Number = K^{int(log(N)/log(K)) + 1} – N 

Below is the implementation of the above approach:

C++

// C++ program to find the minimum number
// to be added to N to make it a power of K

#include <bits/stdc++.h>
#define ll long long int
using namespace std;

// Function to return the minimum number
// to be added to N to make it a power of K.
int minNum(int n, int k)
{
    int x = (int)(log(n) / log(k)) + 1;

    // Computing the difference between
    // then next greater power of K
    // and N
    int mn = pow(k, x) - n;
    return mn;
}

// Driver code
int main()
{
    int n = 20, k = 5;
    cout << minNum(n, k);
    return 0;
}

Java

// Java program to find the minimum number
// to be added to N to make it a power of K

class GFG{

// Function to return the minimum number
// to be added to N to make it a power of K.
static int minNum(int n, int k)
{
    int x = (int)(Math.log(n) / Math.log(k)) + 1;

    // Computing the difference between
    // then next greater power of K
    // and N
    int mn = (int) (Math.pow(k, x) - n);
    return mn;
}

// Driver code
public static void main(String[] args)
{
    int n = 20, k = 5;
    System.out.print(minNum(n, k));
}
}

// This code is contributed by Amit Katiyar

Python3

# Python3 program to find the minimum number
# to be added to N to make it a power of K
import math

# Function to return the minimum number
# to be added to N to make it a power of K.
def minNum(n, k):

    x = int((math.log(n) // math.log(k))) + 1

    # Computing the difference between
    # then next greater power of K
    # and N
    mn = pow(k, x) - n
    return mn

# Driver code
if __name__=='__main__':

    n = 20
    k = 5
    print(minNum(n, k))

# This code is contributed by rutvik_56

C#

// C# program to find the minimum number
// to be added to N to make it a power of K
using System;
class GFG{

// Function to return the minimum number
// to be added to N to make it a power of K.
static int minNum(int n, int k)
{
    int x = (int)(Math.Log(n) /
                  Math.Log(k)) + 1;

    // Computing the difference between
    // then next greater power of K
    // and N
    int mn = (int)(Math.Pow(k, x) - n);
    return mn;
}

// Driver code
public static void Main(string[] args)
{
    int n = 20, k = 5;
    Console.Write(minNum(n, k));
}
}

// This code is contributed by Ritik Bansal

Javascript

<script>

// Javascript program to find the minimum number
// to be added to N to make it a power of K

// Function to return the minimum number
// to be added to N to make it a power of K.
function minNum(n, k)
{
    var x = parseInt(Math.log(n) / Math.log(k)) + 1;

    // Computing the difference between
    // then next greater power of K
    // and N
    var mn = Math.pow(k, x) - n;
    return mn;
}

// Driver code
var n = 20, k = 5;
document.write( minNum(n, k));

</script>

Output:

5

Time Complexity: O(log n)

Auxiliary Space: O(1)