Undulating numbers

An undulating number is a number that has only two types of digits and alternate digits are same, i.e., it is of the form “ababab….”. It is sometimes restricted to non-trivial undulating numbers which are required to have at least 3 digits and a is not equal to b.

The first few such numbers are: 101, 121, 131, 141, 151, 161, 171, 181, 191, 202, 212, 232, 242, 252, 262, 272, 282, 292, 303, 313, 323, 343, 353, 363, 373, 383, 393, 404, 414, 424, 434, 454, 464, 474, 484, 494, …
Some higher undulating numbers are: 6363, 80808, 1717171.

1. For any n >= 3, there are 9 × 9 = 81 non-trivial n-digit undulating numbers, since the first digit can have 9 values (it cannot be 0), and the second digit can have 9 values when it must be different from the first.

Given a number, check if it is Undulating numbers considering the definition of alternating digits, at least 3 digits and adjacent digits not same.

Examples :

Input : n = 121
Output : Yes

Input : n = 1991
Output : No

C++

 // C++ program to check whether a number // is undulating or not #include using namespace std;    bool isUndulating(string n) {     // Considering the definition     // with restriction that there     // should be at least 3 digits     if (n.length() <= 2)        return false;        // Check if all alternate digits are     // same or not.     for (int i = 2; i < n.length(); i++)          if (n[i - 2] != n[i])             false;        return true; }    int main() {     string n = "1212121";     if (isUndulating(n))         cout << "Yes";     else         cout << "No"; }

Java

 // Java program to check whether a number // is undulating or not import java.util.*;    class GFG {            public static boolean isUndulating(String n)     {                    // Considering the definition         // with restriction that there         // should be at least 3 digits             if (n.length() <= 2)                 return false;                // Check if all alternate digits are         // same or not.         for (int i = 2; i < n.length(); i++)              if (n.charAt(i-2) != n.charAt(i))                  return false;                return true;     }                   // Driver code     public static void main (String[] args)     {                    String n = "1212121";                    if (isUndulating(n)==true)             System.out.println("yes");         else             System.out.println("no");     } }    // This code is contributed by akash1295.

Python3

 # Python3 program to check whether a  # number is undulating or not    def isUndulating(n):        # Considering the definition     # with restriction that there     # should be at least 3 digits     if (len(n) <= 2):         return False        # Check if all alternate digits      # are same or not.     for i in range(2, len(n)):          if (n[i - 2] != n[i]):             return False        return True    # Driver Code n = "1212121" if (isUndulating(n)):     print("Yes") else:     print("No")    # This code is contributed by Smitha Dinesh Semwal.

C#

 // C# program to check whether a number // is undulating or not using System;    class GFG {            public static bool isUndulating(string n)     {                    // Considering the definition         // with restriction that there         // should be at least 3 digits             if (n.Length <= 2)                 return false;                // Check if all alternate digits are         // same or not.         for (int i = 2; i < n.Length; i++)              if (n[i-2] != n[i])                  return false;                return true;     }            // Driver code     public static void Main ()     {                    string n = "1212121";                    if (isUndulating(n)==true)             Console.WriteLine("yes");         else             Console.WriteLine("no");     } }    // This code is contributed by Vt_m.

PHP



Output:

Yes

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