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.
- 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 <bits/stdc++.h>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 notimport 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 notdef 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 Coden = "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 notusing 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
<?php// PHP program to check whether a // number is undulating or notfunction isUndulating($n){ // Considering the definition // with restriction that there // should be at least 3 digits if (strlen($n) <= 2) return false; // Check if all alternate // digits are same or not. for ($i = 2; $i < strlen($n); $i++) if ($n[$i - 2] != $n[$i]) false; return true;}// Driver code$n = "1212121";if (isUndulating($n)) echo("Yes");else echo("No");// This code is contributed by Ajit.?> |
Javascript
<script>// JavaScript program to check whether a number// is undulating or not function isUndulating(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 (let i = 2; i < n.length; i++) if (n[i-2] != n[i]) return false; return true; }// Driver Code let n = "1212121"; if (isUndulating(n)==true) document.write("Yes"); else document.write("No"); </script> |
Output:
Yes
Time complexity: O(N) where N is no of digits of given number
Auxiliary Space: O(1)
Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!