Super-D Numbers

Given an integer N, the task is to check N is a super-d Number.

Super-D Number is a number N such that D*N^D contains a substring made of D digits containing D only, where D is more than 1 and less than 10 

Examples:

Input: N = 261 
Output: Yes 
Explanation:
It will be true for D = 3 
D*N^D = 3*261³ = 53338743 
which contains a substring made of 3 digits 333 containing 3 only.
Input: N = 10 
Output: No 

Approach: The idea is to create every possible string concatenating digit D, D number of times and then will check if the concatenation is present as a substring in D*N^D or not, where D will be in the range [2, 9].
Below is the implementation of the above approach: 

C++

#include <iostream>
#include <string>   // include string library for string manipulation
#include <cmath>    // include cmath library for pow function

using namespace std;

// Function to check if N is a super-d number
bool isSuperdNum(int n)
{
    for (int d = 2; d < 10; d++)
    {
        string subString = "";    // create an empty string
        for (int i = 0; i < d; i++)
        {
            subString += to_string(d);  // add d as string to the subString
        }
        if (to_string(d * pow(n, d)).find(subString) != string::npos)
        // check if subString is present in d * n^d as a substring
            return true;
    }
    return false;
}

// Driver Code
int main()
{
    int n = 261;
    if (isSuperdNum(n) == true)
        cout << "Yes" << endl;
    else
        cout << "No" << endl;
    return 0;
}

Java

// Java implementation to
// check if N is a super-d number
class GFG{

// Function to check if N
// is a super-d number
static boolean isSuperdNum(int n)
{
    for (int d = 2; d < 10; d++)
    {
        String subString = newString(d);
        if (String.valueOf(
           (d * Math.pow(n, d))).contains(subString))
            return true;
    }
    return false;
}

// Driver Code
private static String newString(int d)
{
    String ans = "";
    for (int i = 0; i < d; i++)
    {
        ans += String.valueOf(d);
    }
    return ans;
}

// Driver Code
public static void main(String[] args)
{
    int n = 261;
    if (isSuperdNum(n) == true)
        System.out.println("Yes");
    else
        System.out.println("No");
}
}

// This code is contributed by Rajput-Ji

Python3

# Python3 implementation to
# check if N is a super-d number

# Function to check if N
# is a super-d number
def isSuperdNum(n):
    for d in range (2, 10):
        substring = str(d) * d;
        if substring in str(d * pow(n, d)):
            return True
    return False

# Driver Code
n = 261
if isSuperdNum(n) == True:
    print("Yes")
else :
    print("No")

C#

// C# implementation to
// check if N is a super-d number
using System;

class GFG{

// Function to check if N
// is a super-d number
static bool isSuperdNum(int n)
{
    for(int d = 2; d < 10; d++)
    {
       String subString = newString(d);
       if (String.Join("",
          (d * Math.Pow(n, d))).Contains(subString))
           return true;
    }
    return false;
}

private static String newString(int d)
{
    String ans = "";

    for(int i = 0; i < d; i++)
    {
       ans += String.Join("", d);
    }
    return ans;
}

// Driver Code
public static void Main(String[] args)
{
    int n = 261;

    if (isSuperdNum(n) == true)
        Console.WriteLine("Yes");
    else
        Console.WriteLine("No");
}
}

// This code is contributed by Rajput-Ji

Javascript

// Function to check if N is a super-d number
function isSuperdNum(n) {
    for (let d = 2; d < 10; d++) {
        let substring = String(d).repeat(d);
        if (String(d * Math.pow(n, d)).includes(substring)) {
            return true;
        }
    }
    return false;
}

// Driver code
let n = 261;
if (isSuperdNum(n)) {
    console.log("Yes");
} else {
    console.log("No");
}

Output:

Yes

Time Complexity: O(1)

Auxiliary Space: O(1) as it is using constant space

References: OEIS