Compute the maximum power with a given condition

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

Examples:

Input :N = 1, K = 2, M = 5 
Output :2
Explanation: 1*2² = 5, which is <= to M(5), also for p = 3 the value would be 1*2³=9, this is not a feasible solution. Thus 2 is the answer.

Input: N = 5, K = 25, M = 100 
Output: 0  
Explanation: 5*25?=5, which is <= to M(100), also for p =1 the value would be 125, which is greater than the M(100), thus not a feasible solution. Hence 0 is the required answer.

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

Time Complexity:– O(1)
Auxiliary Space: O(1), as no extra space is used.