Find the Largest N digit perfect square number in Base B

Given two integers N and B, the task is to find the largest N digit numbers of Base B which is a perfect square.
Examples:

Input: N = 2, B = 10 
Output: 81 
Explanation:
81 is the largest 2-digit perfect square in base 10.
Input: N = 1, B = 8 
Output: 4 
Explanation:
4 is the largest 1 digit Octal number which is also a perfect square. 

Approach: The largest number N in base B is given by . So if we find the square root of this number in integer form and then we have to again do its square then it will be the largest perfect square of N digits which is given by the formula: .
Below is the implementation of the above approach: 

C++

// C++ implementation to find Largest
// N digit perfect square number in Base B

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

// Function to find the
// largest N digit number
void nDigitPerfectSquares(int n, int b)
{
    // Largest n-digit perfect square
    int largest
        = pow(ceil(sqrt(pow(b, n))) - 1, 2);

    // Print the result
    cout << largest;
}

// Driver Code
int main()
{
    int N = 1, B = 8;

    nDigitPerfectSquares(N, B);

    return 0;
}

Java

// Java implementation to find largest N
// digit perfect square number in base B
import java.io.*;
import java.util.*;

class GFG {

// Function to find the
// largest N digit number
static double nDigitPerfectSquares(int n, int b)
{

    // Largest n-digit perfect square
    double largest = Math.pow(Math.ceil
                             (Math.sqrt
                             (Math.pow(b, n))) - 1, 2);

    // Print the result
    return largest;
}

// Driver code
public static void main(String[] args)
{
    int N = 1, B = 8;
    System.out.println(nDigitPerfectSquares(N, B));
}
}

// This code is contributed by coder001

Python3

# Python3 implementation to find the largest
# N digit perfect square number in base B
import math

# Function to find the
# largest N digit number
def nDigitPerfectSquares(n, b):

    # Largest n-digit perfect square
    largest = pow(math.ceil
                 (math.sqrt(pow(b, n))) - 1, 2)

    # Print the result
    print(largest)

# Driver Code
N = 1
B = 8

nDigitPerfectSquares(N, B)

# This code is contributed by divyamohan123

C#

// C# implementation to find largest N
// digit perfect square number in base B
using System;

class GFG {

// Function to find the
// largest N digit number
static double nDigitPerfectSquares(int n, int b)
{

    // Largest n-digit perfect square
    double largest = Math.Pow(Math.Ceiling
                             (Math.Sqrt
                             (Math.Pow(b, n))) - 1, 2);

    // Print the result
    return largest;
}

// Driver code
public static void Main(String[] args)
{
    int N = 1, B = 8;

    Console.WriteLine(nDigitPerfectSquares(N, B));
}
}

// This code is contributed by PrinciRaj1992

Javascript

<script>

// Javascript implementation to find Largest
// N digit perfect square number in Base B

// Function to find the
// largest N digit number
function nDigitPerfectSquares(n, b)
{
    // Largest n-digit perfect square
    var largest
        = Math.pow(Math.ceil(Math.sqrt(Math.pow(b, n))) - 1, 2);

    // Print the result
    document.write(largest);
}

// Driver Code
var N = 1, B = 8;
nDigitPerfectSquares(N, B);

</script>

Output:

4

Time Complexity: O(logn)

Auxiliary Space: O(1)