Largest value of x such that axx is N-digit number of base b

Given the integers a, b, N, the task is to find the largest number x such that is an N-digit number of base b.

Examples:

Input: a = 2, b = 10, N = 2 
Output: 3 
Explanation:
Here 2 * 3³ = 54, which has the number of digits = 2, 
but 2 * 4⁴ = 512 which has a number of digits = 3, which is not equal to N. 
Therefore, the largest value of x is 2.

Input: a = 1, b = 2, N = 3 
Output: 2 
Explanation:
1 * 2² = 4 whose binary representation is 100 and it has 3 digits.  

Approach: This problem can be solved using binary search. 

• The number of digits of in base is .
• Binary search is used to find the largest such that the number of digits of in base is exactly .
• In binary search, we will check the number of digits , where , and change the pointer according to that. 
  

Below is the implementation of the above approach:

C++

// C++ implementation of the above approach

#include <bits/stdc++.h>
using namespace std;

// Function to find log_b(a)
double log(int a, int b)
{
    return log10(a) / log10(b);
}

int get(int a, int b, int n)
{

    // Set two pointer for binary search
    int lo = 0, hi = 1e6;

    int ans = 0;

    while (lo <= hi) {
        int mid = (lo + hi) / 2;

        // Calculating number of digits
        // of a*mid^mid in base b
        int dig = ceil((mid * log(mid, b)
                        + log(a, b)));

        if (dig > n) {

            // If number of digits > n
            // we can simply ignore it
            // and decrease our pointer
            hi = mid - 1;
        }
        else {

            // if number of digits <= n,
            // we can go higher to
            // reach value exactly equal to n
            ans = mid;
            lo = mid + 1;
        }
    }

    // return the largest value of x
    return ans;
}

// Driver Code
int main()
{

    int a = 2, b = 2, n = 6;

    cout << get(a, b, n)
         << "\n";

    return 0;
}

Java

// Java implementation of the above approach
import java.util.*;

class GFG{

// Function to find log_b(a)
static int log(int a, int b)
{
    return (int)(Math.log10(a) /
                 Math.log10(b));
}

static int get(int a, int b, int n)
{

    // Set two pointer for binary search
    int lo = 0, hi = (int) 1e6;

    int ans = 0;

    while (lo <= hi)
    {
        int mid = (lo + hi) / 2;

        // Calculating number of digits
        // of a*mid^mid in base b
        int dig = (int) Math.ceil((mid * log(mid, b) +
                                         log(a, b)));

        if (dig > n)
        {

            // If number of digits > n
            // we can simply ignore it
            // and decrease our pointer
            hi = mid - 1;
        }
        else
        {

            // If number of digits <= n,
            // we can go higher to reach
            // value exactly equal to n
            ans = mid;
            lo = mid + 1;
        }
    }

    // Return the largest value of x
    return ans;
}

// Driver Code
public static void main(String[] args)
{
    int a = 2, b = 2, n = 6;

    System.out.print(get(a, b, n) + "\n");
}
}

// This code is contributed by amal kumar choubey

Python3

# Python3 implementation of the above approach

from math import log10,ceil,log

# Function to find log_b(a)
def log1(a,b):
    return log10(a)//log10(b)

def get(a,b,n):
    # Set two pointer for binary search
    lo = 0
    hi = 1e6

    ans = 0

    while (lo <= hi):
        mid = (lo + hi) // 2

        # Calculating number of digits
        # of a*mid^mid in base b
        dig = ceil((mid * log(mid, b) + log(a, b)))

        if (dig > n):
            # If number of digits > n
            # we can simply ignore it
            # and decrease our pointer
            hi = mid - 1

        else:
            # if number of digits <= n,
            # we can go higher to
            # reach value exactly equal to n
            ans = mid
            lo = mid + 1

    # return the largest value of x
    return ans

# Driver Code
if __name__ == '__main__':
    a = 2
    b = 2
    n = 6

    print(int(get(a, b, n)))

# This code is contributed by Surendra_Gangwar

C#

// C# implementation of the above approach
using System;
class GFG{

// Function to find log_b(a)
static int log(int a, int b)
{
    return (int)(Math.Log10(a) /
                 Math.Log10(b));
}

static int get(int a, int b, int n)
{

    // Set two pointer for binary search
    int lo = 0, hi = (int) 1e6;

    int ans = 0;

    while (lo <= hi)
    {
        int mid = (lo + hi) / 2;

        // Calculating number of digits
        // of a*mid^mid in base b
        int dig = (int)Math.Ceiling((double)(mid *
                                         log(mid, b) +
                                         log(a, b)));

        if (dig > n)
        {

            // If number of digits > n
            // we can simply ignore it
            // and decrease our pointer
            hi = mid - 1;
        }
        else
        {

            // If number of digits <= n,
            // we can go higher to reach
            // value exactly equal to n
            ans = mid;
            lo = mid + 1;
        }
    }

    // Return the largest value of x
    return ans;
}

// Driver Code
public static void Main(String[] args)
{
    int a = 2, b = 2, n = 6;

    Console.Write(get(a, b, n) + "\n");
}
}

// This code is contributed by amal kumar choubey

Javascript

<script>

// Javascript implementation of the above approach

// Function to find log_b(a)
function log(a, b)
{
    return (Math.log10(a) /
                 Math.log10(b));
}

function get(a, b, n)
{

    // Set two pointer for binary search
    let lo = 0, hi = 1e6;

    let ans = 0;

    while (lo <= hi)
    {
        let mid = Math.floor((lo + hi) / 2);

        // Calculating number of digits
        // of a*mid^mid in base b
        let dig =  Math.ceil((mid * log(mid, b) +
                                         log(a, b)));

        if (dig > n)
        {

            // If number of digits > n
            // we can simply ignore it
            // and decrease our pointer
            hi = mid - 1;
        }
        else
        {

            // If number of digits <= n,
            // we can go higher to reach
            // value exactly equal to n
            ans = mid;
            lo = mid + 1;
        }
    }

    // Return the largest value of x
    return ans;
}
// Driver Code

    let a = 2, b = 2, n = 6;

    document.write(get(a, b, n) + "\n");

</script>

Output:

3

Time Complexity: O(log(k)) where k=1e6.
Auxiliary Space: O(1)