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,



Kint(log(N)/log(K)) + 1

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

Minimum Number = Kint(log(N)/log(K)) + 1 – N

Below is the implementation of the above approach:

C++

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// 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;
}

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Java

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// 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

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Python3

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# 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

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Output:

5

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Improved By : amit143katiyar, rutvik_56