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*ND 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*ND = 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*ND 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 functionusing namespace std;// Function to check if N is a super-d numberbool 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 Codeint 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 numberclass GFG{ // Function to check if N// is a super-d numberstatic 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 Codeprivate static String newString(int d){ String ans = ""; for (int i = 0; i < d; i++) { ans += String.valueOf(d); } return ans;}// Driver Codepublic 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 numberdef 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 = 261if isSuperdNum(n) == True: print("Yes")else : print("No") |
C#
// C# implementation to // check if N is a super-d numberusing System;class GFG{ // Function to check if N// is a super-d numberstatic 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 Codepublic 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 numberfunction 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 codelet n = 261;if (isSuperdNum(n)) { console.log("Yes");} else { console.log("No");} |
Yes
Time Complexity: O(1)
Auxiliary Space: O(1) as it is using constant space
References: OEIS
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