Largest N digit Octal number which is a Perfect square

Given a natural number N, the task is to find the largest N digit Octal number which is a perfect square.
Examples:

Input: N = 1 
Output: 4 
Explanation:
4 is the largest 1 digit Octal number which is also perfect square
Input: N = 2 
Output: 61 
Explanation:
49 is the largest number which is a 2-Digit Octal Number and also a perfect square. 
Therefore 49 in Octal = 61 

Brute Force Approach:

Check each octal number starting from the largest one until it finds a perfect square. It uses the isPerfectSquare function to check if a number is a perfect square, and the decimalToOctal function to convert decimal numbers to octal.

C++

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

// Function to check if a number is a perfect square
bool isPerfectSquare(int n) {
    int root = sqrt(n);
    return (root * root == n);
}

// Function to convert decimal to octal
int decimalToOctal(int n) {
    int octalNum = 0, count = 0;
    while (n != 0) {
        int rem = n % 8;
        octalNum += rem * pow(10, count);
        count++;
        n /= 8;
    }
    return octalNum;
}

int nDigitPerfectSquares(int n){
     // Largest N digit octal number is 777...N times
    int largestOctalNum = 0;
    for (int i = 1; i <= n; i++) {
        largestOctalNum = largestOctalNum * 10 + 7;
    }

    // Check perfect squares starting from the largest octal number
    for (int i = largestOctalNum; i >= 0; i--) {
        int decimalNum = 0, count = 0;
        int num = i;
        while (num != 0) {
            int rem = num % 10;
            decimalNum += rem * pow(8, count);
            count++;
            num /= 10;
        }
        if (isPerfectSquare(decimalNum)) {
            cout << decimalToOctal(decimalNum) << endl;
            break;
        }
    }
}
int main() {
    int n = 2;
    nDigitPerfectSquares(n);
    return 0;
}

Java

import java.util.*;

public class Main {
    // Function to check if a number is a perfect square
    public static boolean isPerfectSquare(int n) {
        int root = (int)Math.sqrt(n);
        return (root * root == n);
    }
    // Function to convert decimal to octal
    public static int decimalToOctal(int n) {
        int octalNum = 0, count = 0;
        while (n != 0) {
            int rem = n % 8;
            octalNum += rem * Math.pow(10, count);
            count++;
            n /= 8;
        }
        return octalNum;
    }
    public static void nDigitPerfectSquares(int n) {
        // Largest N digit octal number is 777...N times
        int largestOctalNum = 0;
        for (int i = 1; i <= n; i++) {
            largestOctalNum = largestOctalNum * 10 + 7;
        }

        // Check perfect squares starting from the largest octal number
        for (int i = largestOctalNum; i >= 0; i--) {
            int decimalNum = 0, count = 0;
            int num = i;
            while (num != 0) {
                int rem = num % 10;
                decimalNum += rem * Math.pow(8, count);
                count++;
                num /= 10;
            }
            if (isPerfectSquare(decimalNum)) {
                System.out.println(decimalToOctal(decimalNum));
                break;
            }
        }
    }
    public static void main(String[] args) {
        int n = 2;
        nDigitPerfectSquares(n);
    }
}

Python3

import math

# Function to check if a number is a perfect square
def isPerfectSquare(n):
    root = int(math.sqrt(n))
    return (root * root == n)

# Function to convert decimal to octal
def decimalToOctal(n):
    octalNum = 0
    count = 0
    while (n != 0):
        rem = n % 8
        octalNum += rem * pow(10, count)
        count += 1
        n //= 8
    return octalNum

def nDigitPerfectSquares(n):
    # Largest N digit octal number is 777...N times
    largestOctalNum = 0
    for i in range(1, n+1):
        largestOctalNum = largestOctalNum * 10 + 7

    # Check perfect squares starting from the largest octal number
    for i in range(largestOctalNum, -1, -1):
        decimalNum = 0
        count = 0
        num = i
        while (num != 0):
            rem = num % 10
            decimalNum += rem * pow(8, count)
            count += 1
            num //= 10
        if (isPerfectSquare(decimalNum)):
            print(decimalToOctal(decimalNum))
            break

n = 2
nDigitPerfectSquares(n)
# This code is contributed by Prajwal Kandekar

Output

61

Time Complexity: O(N^2*log(N)), where N is the input variable from the problem statement

Space Complexity: O(1)

Approach:
It can be observed that the series of largest numbers which is also a perfect square in Octal is: 

4, 61, 744, 7601, 77771, 776001 ….. 

As we know the digits in the octal system increases when a number Greater than 8^k where k denotes the number of digits in the number. So for any N digit number in the octal number system must be less than the value of 8^N+1. So, the general term that can be derived using this observation is – 

N-Digit Octal Number = octal(pow(ceil(sqrt(pow(8, N))) -1, 2)) 

Below is the implementation of the above approach: 

CPP

// C++ implementation to find the maximum
// N-digit octal number which is perfect square

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

// Function to convert decimal number
// to a octal number
void decToOctal(int n)
{

    // Array to store octal number
    int octalNum[100];

    // Counter for octal number array
    int i = 0;
    while (n != 0) {

        // Store remainder in
        // octal array
        octalNum[i] = n % 8;
        n = n / 8;
        i++;
    }

    // Print octal number array
    // in reverse order
    for (int j = i - 1; j >= 0; j--)
        cout << octalNum[j];
    cout << "\n";
}

void nDigitPerfectSquares(int n)
{
    // Largest n-digit perfect square
    int decimal = pow(
        ceil(sqrt(pow(8, n))) - 1, 2
        );
    decToOctal(decimal);
}

// Driver Code
int main()
{
    int n = 2;
    nDigitPerfectSquares(n);

    return 0;
}

Java

// Java implementation to find the maximum
// N-digit octal number which is perfect square
import java.util.*;
import java.lang.*;
import java.io.*;

class GFG
{

    // Function to convert decimal number
    // to a octal number
    static void decToOctal(int n)
    {

        // Array to store octal number
        int octalNum[] = new int[100];

        // Counter for octal number array
        int i = 0;
        while (n != 0)
        {

            // Store remainder in
            // octal array
            octalNum[i] = n % 8;
            n = n / 8;
            i++;
        }

        // Print octal number array
        // in reverse order
        for (int j = i - 1; j >= 0; j--)
            System.out.print(octalNum[j]);
        System.out.println("\n");
    }

    static void nDigitPerfectSquares(int n)
    {
        // Largest n-digit perfect square
        int decimal = (int) Math.pow(Math.ceil(Math.sqrt(Math.pow(8, n))) - 1, 2);
        decToOctal(decimal);
    }

    // Driver code
    public static void main (String[] args)
    {
        int n = 2;
        nDigitPerfectSquares(n);
    }
}

// This code is contributed by nidhiva

Python3

# Python3 implementation to find the maximum
# N-digit octal number which is perfect square
from math import sqrt,ceil

# Function to convert decimal number
# to a octal number
def decToOctal(n) :

    # Array to store octal number
    octalNum = [0]*100;

    # Counter for octal number array
    i = 0;
    while (n != 0) :

        # Store remainder in
        # octal array
        octalNum[i] = n % 8;
        n = n // 8;
        i += 1;

    # Print octal number array
    # in reverse order
    for j in range(i - 1, -1, -1) :
        print(octalNum[j], end= "");
    print();

def nDigitPerfectSquares(n) :

    # Largest n-digit perfect square
    decimal = pow(ceil(sqrt(pow(8, n))) - 1, 2);
    decToOctal(decimal);

# Driver Code
if __name__ == "__main__" :

    n = 2;
    nDigitPerfectSquares(n);

# This code is contributed by AnkitRai01

C#

// C# implementation to find the maximum
// N-digit octal number which is perfect square
using System;

class GFG
{

    // Function to convert decimal number
    // to a octal number
    static void decToOctal(int n)
    {

        // Array to store octal number
        int []octalNum = new int[100];

        // Counter for octal number array
        int i = 0;
        while (n != 0)
        {

            // Store remainder in
            // octal array
            octalNum[i] = n % 8;
            n = n / 8;
            i++;
        }

        // Print octal number array
        // in reverse order
        for (int j = i - 1; j >= 0; j--)
            Console.Write(octalNum[j]);
    Console.WriteLine();
    }

    static void nDigitPerfectSquares(int n)
    {
        // Largest n-digit perfect square
        int _decimal = (int) Math.Pow(Math.Ceiling(Math.Sqrt(Math.Pow(8, n))) - 1, 2);
        decToOctal(_decimal);
    }

    // Driver code
    public static void Main()
    {
        int n = 2;
        nDigitPerfectSquares(n);
    }
}

// This code is contributed by AnkitRai01

Javascript

<script>

// Javascript implementation to find the maximum
// N-digit octal number which is perfect square

// Function to convert decimal number
// to a octal number
function decToOctal(n)
{

    // Array to store octal number
    var octalNum = Array(100).fill(0);

    // Counter for octal number array
    var i = 0;
    while (n != 0) {

        // Store remainder in
        // octal array
        octalNum[i] = n % 8;
        n = parseInt(n / 8);
        i++;
    }

    // Print octal number array
    // in reverse order
    for (var j = i - 1; j >= 0; j--)
        document.write(octalNum[j]);
    document.write("<br>");
}

function nDigitPerfectSquares(n)
{
    // Largest n-digit perfect square
    var decimal = Math.pow(
        Math.ceil(Math.sqrt(Math.pow(8, n))) - 1, 2
        );
    decToOctal(decimal);
}

// Driver Code
var n = 2;
nDigitPerfectSquares(n);


</script>

Output:

61

Time Complexity: O(log₈n)

Auxiliary Space: O(1)