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Compute the maximum power with a given condition
  • Last Updated : 27 Apr, 2021

Given 3 integers N, M, and K. The task is to find maximum P such that N * KP ≤ M.
 

Examples:  

Input :N = 1, K = 2, M = 5 
Output :2
Input: N = 5, K = 25, M = 100 
Output: 0  

Approach: 
In this algorithm, simply multiply N with K and update the current value of N with the result and increase variable power (initially 0) by 1
To achieve this, a recursive function is defined which has 2 base cases.  

  1. Initially, N is greater than required, therefore, return 0.
  2. Otherwise, return power – 1.
  • If the current value of N is equal to the M then return power.
  • Recursive condition if current N < M
      Update N as (N * k) and power as current power + 1
     

Below is the implementation of the above approach: 
 



C++




// Compute maximum power to which K can be raised so
// that given condition remains true
#include <bits/stdc++.h>
using namespace std;
 
#define ll long long
 
// Function to return the largest
// power
int calculate(ll int n, ll int k,
              ll int m, ll int power)
{
 
    // If n is greater than given M
    if (n > m) {
        if (power == 0)
            return 0;
        else
            return power - 1;
    }
 
    // If n == m
    else if (n == m)
        return power;
 
    else
        // Checking for the next power
        return calculate(n * k, k, m, power + 1);
}
 
// Driver Code
int main()
{
    ll N = 1, K = 2, M = 5;
 
    cout << calculate(N, K, M, 0);
 
    return 0;
}

Java




// Java program for Compute maximum power
// to which K can be raised so that
// given condition remains true
class GFG
{
 
// Function to return the largest
// power
static int calculate(int n, int k,
                     int m, int power)
{
 
    // If n is greater than given M
    if (n > m)
    {
        if (power == 0)
            return 0;
        else
            return power - 1;
    }
 
    // If n == m
    else if (n == m)
        return power;
 
    else
        // Checking for the next power
        return calculate(n * k, k, m,
                          power + 1);
}
 
// Driver Code
public static void main (String[] args)
{
    int N = 1, K = 2, M = 5;
 
    System.out.println(calculate(N, K, M, 0));
}
}
 
// This code is contributed by AnkitRai01

Python




# Compute maximum power to
# which K can be raised so
# that given condition
# remains true
 
 
# Function to return the largest
# power
def calculate(n, k, m, power):    
     
    # If n is greater than given M                        
    if n > m:
        if power == 0:
            return 0
        else:
            return power-1
     
    # If n == m
    elif n == m:
        return power
    else:
        # Checking for the next power
        return calculate(n * k, k, m, power + 1)
 
# Driver's code    
if __name__=="__main__":
     
    N = 1
    K = 2
    M = 5
     
    print(calculate(N, K, M, 0))

C#




// C# program for Compute maximum power
// to which K can be raised so that
// given condition remains true
using System;
class GFG
{
 
// Function to return the largest
// power
static int calculate(int n, int k,
                     int m, int power)
{
 
    // If n is greater than given M
    if (n > m)
    {
        if (power == 0)
            return 0;
        else
            return power - 1;
    }
 
    // If n == m
    else if (n == m)
        return power;
 
    else
        // Checking for the next power
        return calculate(n * k, k, m,
                         power + 1);
}
 
// Driver Code
public static void Main (String[] args)
{
    int N = 1, K = 2, M = 5;
 
    Console.WriteLine(calculate(N, K, M, 0));
}
}
 
// This code is contributed by PrinciRaj1992

Javascript




<script>
// javascript program for Compute maximum power
// to which K can be raised so that
// given condition remains true    
// Function to return the largest
    // power
    function calculate(n , k , m , power) {
 
        // If n is greater than given M
        if (n > m) {
            if (power == 0)
                return 0;
            else
                return power - 1;
        }
 
        // If n == m
        else if (n == m)
            return power;
 
        else
            // Checking for the next power
            return calculate(n * k, k, m, power + 1);
    }
 
    // Driver Code
     
        var N = 1, K = 2, M = 5;
 
        document.write(calculate(N, K, M, 0));
 
// This code contributed by aashish1995
</script>
Output: 
2

 

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