Un número de Katadrome es un número cuyos dígitos están en orden decreciente.
Pocos números de Katadrome son:
0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 20, 21….
Comprobar si N es un Katadromes
Dado un número N , la tarea es comprobar si es Katadromes o no.
Ejemplos:
Entrada: 4321
Salida: Sí
Entrada: 1243
Salida: No
Planteamiento: La idea es recorrer los dígitos del número y verificar si el dígito actual es menor que el último dígito. Si todos los dígitos cumplen las condiciones, entonces el número es un número de Katadrome.
A continuación se muestra la implementación del enfoque anterior:
C++
// C++ implementation to check if
// a number is Katadrome or not.
#include <iostream>
using namespace std;
// Function to check if a number
// is a Katadrome number or not
bool isKatadrome(int num)
{
// To store previous digit (Assigning
// initial value which is less than any
// digit)
int prev = -1;
// Traverse all digits from right to
// left and check if any digit is
// smaller than previous.
while (num) {
int digit = num % 10;
num /= 10;
if (digit < prev)
return false;
prev = digit;
}
return true;
}
// Driver code
int main()
{
int num = 4321;
isKatadrome(num) ? cout << "Yes"
: cout << "No";
return 0;
}
Java
// Java implementation to check if
// a number is Katadrome or not.
class GFG{
// Function to check if a number
// is a Katadrome number or not
static boolean isKatadrome(int num)
{
// To store previous digit
// (Assigning initial value
// which is less than any digit)
int prev = -1;
// Traverse all digits from right
// to left and check if any digit
// is smaller than previous.
while (num > 0)
{
int digit = num % 10;
num /= 10;
if (digit < prev)
return false;
prev = digit;
}
return true;
}
// Driver Code
public static void main(String[] args)
{
int N = 4321;
// Function Call
if (isKatadrome(N))
System.out.print("Yes");
else
System.out.print("No");
}
}
// This code is contributed by Pratima Pandey
Python3
# Python3 program to print count of values such
# that n+i = n^i
def isKatadrome(num):
# To store previous digit (Assigning
# initial value which is less than any
# digit)
prev = -1
# Traverse all digits from right to
# left and check if any digit is
# smaller than previous.
while num:
digit = num % 10
num //= 10
if digit < prev:
return False
prev = digit
return True
# Driver code
if __name__=='__main__':
num = 4321
if isKatadrome(num):
print('Yes')
else:
print('No')
# This code is contributed by rutvik
C#
// C# implementation to check if
// a number is Katadrome or not.
using System;
class GFG{
// Function to check if a number
// is a Katadrome number or not
static bool isKatadrome(int num)
{
// To store previous digit
// (Assigning initial value
// which is less than any digit)
int prev = -1;
// Traverse all digits from right
// to left and check if any digit
// is smaller than previous.
while (num > 0)
{
int digit = num % 10;
num /= 10;
if (digit < prev)
return false;
prev = digit;
}
return true;
}
// Driver Code
public static void Main()
{
int N = 4321;
// Function Call
if (isKatadrome(N))
Console.Write("Yes");
else
Console.Write("No");
}
}
// This code is contributed by Code_Mech
Javascript
<script>
// Javascript implementation to check if
// a number is Katadrome or not.
// Function to check if a number
// is a Katadrome number or not
function isKatadrome( num) {
// To store previous digit
// (Assigning initial value
// which is less than any digit)
let prev = -1;
// Traverse all digits from right
// to left and check if any digit
// is smaller than previous.
while (num > 0) {
let digit = num % 10;
num = parseInt(num/10);
if (digit < prev)
return false;
prev = digit;
}
return true;
}
// Driver Code
let N = 4321;
// Function Call
if (isKatadrome(N))
document.write("Yes");
else
document.write("No");
// This code contributed by Rajput-Ji
</script>
Producción:
Yes
Complejidad de tiempo: O(d) donde d es el número de dígitos en un número dado.
Referencia: http://www.numbersaplenty.com/set/katadrome/