Números ondulantes no triviales

Dado un número entero N , la tarea es comprobar si N es un número ondulante no trivial.

Número ondulante no trivial son números en base 10 > 100 que son de la forma aba, abab, ababa, …, donde a != b.

Ejemplos:

Entrada: N = 121 
Salida: Sí 
Explicación: 
121 tiene la forma aba

Entrada: N = 123 
Salida: No

Enfoque: La idea es convertir el número en la string. Si la longitud de la string es par, comprobaremos si la primera mitad de la string es igual a la segunda mitad o no. Si la longitud es impar, agregaremos un segundo carácter de la string en el último de la string para que su longitud sea uniforme. Finalmente, verifique que la primera mitad de la string sea igual a la segunda mitad o no.

A continuación se muestra la implementación del enfoque anterior:

// C++ implementation to check if N
// is a Nontrivial undulant number
 
#include<bits/stdc++.h>
using namespace std;
 
 
// Function to check if a string
// is double string or not
bool isDouble(int num)
{
    string s = to_string(num);
    int l = s.length();
     
    // a and b should not be equal
    if(s[0] == s[1])
       return false;
         
    // Condition to check
    // if length is odd
    // make length even
    if(l % 2 == 1)
    {
        s = s + s[1];
        l++;
    }
         
    // first half of s
    string s1 = s.substr(0, l/2);
    // second half of s
    string s2 = s.substr(l/2);
         
    // Double string if first
    // and last half are equal
    return s1 == s2;
}
 
// Function to check if N is an
// Nontrivial undulant number
bool isNontrivialUndulant(int N)
{   
    return N > 100 && isDouble(N);
}
 
// Driver Code
int main()
{
    int n = 121;
    if (isNontrivialUndulant(n))
        cout << "Yes";
    else
        cout << "No";
    return 0;
}

// Java implementation to check if N
// is a Nontrivial undulant number
class GFG{
 
// Function to check if a string
// is double string or not
static boolean isDouble(int num)
{
    String s = Integer.toString(num);
    int l = s.length();
     
    // a and b should not be equal
    if(s.charAt(0) == s.charAt(1))
       return false;
         
    // Condition to check if length
    // is odd make length even
    if(l % 2 == 1)
    {
        s = s + s.charAt(1);
        l++;
    }
         
    // First half of s
    String s1 = s.substring(0, l / 2);
     
    // Second half of s
    String s2 = s.substring(l / 2);
         
    // Double string if first
    // and last half are equal
    return s1.equals(s2);
}
 
// Function to check if N is an
// Nontrivial undulant number
static boolean isNontrivialUndulant(int N)
{
    return N > 100 && isDouble(N);
}
 
// Driver code
public static void main(String[] args)
{
    int n = 121;
     
    if (isNontrivialUndulant(n))
    {
        System.out.println("Yes");
    }
    else
    {
        System.out.println("No");
    }
}
}
 
// This code is contributed by shubham

# Python3 implementation to check if N
# is a Nontrivial undulant number
 
# Function to check if a string
# is double string or not
def isDouble(num):
 
    s = str(num)
    l = len(s)
 
    # a and b should not be equal
    if(s[0] == s[1]):
        return False
 
    # Condition to check
    # if length is odd
    # make length even
    if(l % 2 == 1):
 
        s = s + s[1]
        l += 1
 
    # First half of s
    s1 = s[:l // 2]
 
    # Second half of s
    s2 = s[l // 2:]
 
    # Double string if first
    # and last half are equal
    return s1 == s2
 
# Function to check if N is an
# Nontrivial undulant number
def isNontrivialUndulant(N):
 
    return N > 100 and isDouble(N)
 
# Driver Code
n = 121
 
if (isNontrivialUndulant(n)):
    print("Yes")
else:
    print("No")
     
# This code is contributed by vishu2908

// C# implementation to check if N
// is a Nontrivial undulant number
using System;
class GFG{
  
// Function to check if a string
// is double string or not
static bool isDouble(int num)
{
    String s = num.ToString();
    int l = s.Length;
      
    // a and b should not be equal
    if(s[0] == s[1])
       return false;
          
    // Condition to check if length
    // is odd make length even
    if(l % 2 == 1)
    {
        s = s + s[1];
        l++;
    }
          
    // First half of s
    String s1 = s.Substring(0, l / 2);
      
    // Second half of s
    String s2 = s.Substring(l / 2);
          
    // Double string if first
    // and last half are equal
    return s1.Equals(s2);
}
  
// Function to check if N is an
// Nontrivial undulant number
static bool isNontrivialUndulant(int N)
{
    return N > 100 && isDouble(N);
}
  
// Driver code
public static void Main(String[] args)
{
    int n = 121;
      
    if (isNontrivialUndulant(n))
    {
        Console.WriteLine("Yes");
    }
    else
    {
        Console.WriteLine("No");
    }
}
}
 
// This code is contributed by gauravrajput1

<script>
 
// JavaScript implementation to check if N
// is a Nontrivial undulant number
 
// Function to check if a string
// is double string or not
function isDouble(num)
{
    let s = num.toString();
    let l = s.length;
     
    // a and b should not be equal
    if (s[0] == s.charAt[1])
       return false;
         
    // Condition to check if length
    // is odd make length even
    if (l % 2 == 1)
    {
        s = s + s[1];
        l++;
    }
         
    // First half of s
    let s1 = s.substr(0, l / 2);
     
    // Second half of s
    let s2 = s.substr(l / 2);
         
    // Double string if first
    // and last half are equal
    return (s1 == s2);
}
 
// Function to check if N is an
// Nontrivial undulant number
function isNontrivialUndulant(N)
{
    return N > 100 && isDouble(N);
}
   
// Driver Code
let n = 121;
 
if (isNontrivialUndulant(n))
{
    document.write("Yes");
}
else
{
    document.write("No");
}
 
// This code is contributed by susmitakundugoaldanga
       
</script>

Yes

Complejidad de tiempo: O(|N|)

Referencias: OEIS