Dado un número entero N , la tarea es verificar si la suma de los dígitos de N es palíndromo o no.
Ejemplo:
Entrada: N = 56
Salida: Sí
Explicación: La suma de dígitos es (5 + 6) = 11, que es un palíndromo.Entrada: N = 51241
Salida: No
Enfoque: Encuentre la suma de los dígitos de N y guárdela en una suma variable . Ahora verifique si sum es palíndromo o no usando el enfoque discutido en este artículo.
A continuación se muestra la implementación del enfoque anterior:
C++
// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
// Function to return the
// sum of digits of n
int digitSum(int n)
{
int sum = 0;
while (n > 0) {
sum += (n % 10);
n /= 10;
}
return sum;
}
// Function that returns true
// if n is palindrome
bool isPalindrome(int n)
{
// Find the appropriate divisor
// to extract the leading digit
int divisor = 1;
while (n / divisor >= 10)
divisor *= 10;
while (n != 0) {
int leading = n / divisor;
int trailing = n % 10;
// If first and last digit
// not same return false
if (leading != trailing)
return false;
// Removing the leading and trailing
// digit from number
n = (n % divisor) / 10;
// Reducing divisor by a factor
// of 2 as 2 digits are dropped
divisor = divisor / 100;
}
return true;
}
// Function that returns true if
// the digit sum of n is palindrome
bool isDigitSumPalindrome(int n)
{
// Sum of the digits of n
int sum = digitSum(n);
// If the digit sum is palindrome
if (isPalindrome(sum))
return true;
return false;
}
// Driver code
int main()
{
int n = 56;
if (isDigitSumPalindrome(n))
cout << "Yes";
else
cout << "No";
return 0;
}
Java
// Java implementation of the approach
import java.util.*;
class GFG
{
// Function to return the
// sum of digits of n
static int digitSum(int n)
{
int sum = 0;
while (n > 0)
{
sum += (n % 10);
n /= 10;
}
return sum;
}
// Function that returns true
// if n is palindrome
static boolean isPalindrome(int n)
{
// Find the appropriate divisor
// to extract the leading digit
int divisor = 1;
while (n / divisor >= 10)
divisor *= 10;
while (n != 0)
{
int leading = n / divisor;
int trailing = n % 10;
// If first and last digit
// not same return false
if (leading != trailing)
return false;
// Removing the leading and trailing
// digit from number
n = (n % divisor) / 10;
// Reducing divisor by a factor
// of 2 as 2 digits are dropped
divisor = divisor / 100;
}
return true;
}
// Function that returns true if
// the digit sum of n is palindrome
static boolean isDigitSumPalindrome(int n)
{
// Sum of the digits of n
int sum = digitSum(n);
// If the digit sum is palindrome
if (isPalindrome(sum))
return true;
return false;
}
// Driver code
public static void main(String []args)
{
int n = 56;
if (isDigitSumPalindrome(n))
System.out.println("Yes");
else
System.out.println("No");
}
}
// This code is contributed by Surendra_Gangwar
Python3
# Python3 implementation of the approach
# Function to return the
# sum of digits of n
def digitSum(n) :
sum = 0;
while (n > 0) :
sum += (n % 10);
n //= 10;
return sum;
# Function that returns true
# if n is palindrome
def isPalindrome(n) :
# Find the appropriate divisor
# to extract the leading digit
divisor = 1;
while (n // divisor >= 10) :
divisor *= 10;
while (n != 0) :
leading = n // divisor;
trailing = n % 10;
# If first and last digit
# not same return false
if (leading != trailing) :
return False;
# Removing the leading and trailing
# digit from number
n = (n % divisor) // 10;
# Reducing divisor by a factor
# of 2 as 2 digits are dropped
divisor = divisor // 100;
return True;
# Function that returns true if
# the digit sum of n is palindrome
def isDigitSumPalindrome(n) :
# Sum of the digits of n
sum = digitSum(n);
# If the digit sum is palindrome
if (isPalindrome(sum)) :
return True;
return False;
# Driver code
if __name__ == "__main__" :
n = 56;
if (isDigitSumPalindrome(n)) :
print("Yes");
else :
print("No");
# This code is contributed by AnkitRai01
C#
// C# implementation of the approach
using System;
class GFG
{
// Function to return the
// sum of digits of n
static int digitSum(int n)
{
int sum = 0;
while (n > 0)
{
sum += (n % 10);
n /= 10;
}
return sum;
}
// Function that returns true
// if n is palindrome
static bool isPalindrome(int n)
{
// Find the appropriate divisor
// to extract the leading digit
int divisor = 1;
while (n / divisor >= 10)
divisor *= 10;
while (n != 0)
{
int leading = n / divisor;
int trailing = n % 10;
// If first and last digit
// not same return false
if (leading != trailing)
return false;
// Removing the leading and trailing
// digit from number
n = (n % divisor) / 10;
// Reducing divisor by a factor
// of 2 as 2 digits are dropped
divisor = divisor / 100;
}
return true;
}
// Function that returns true if
// the digit sum of n is palindrome
static bool isDigitSumPalindrome(int n)
{
// Sum of the digits of n
int sum = digitSum(n);
// If the digit sum is palindrome
if (isPalindrome(sum))
return true;
return false;
}
// Driver code
static public void Main ()
{
int n = 56;
if (isDigitSumPalindrome(n))
Console.Write("Yes");
else
Console.Write("No");
}
}
// This code is contributed by ajit
Javascript
<script>
// Javascript implementation of the approach
// Function to return the
// sum of digits of n
function digitSum(n)
{
let sum = 0;
while (n > 0)
{
sum += (n % 10);
n = parseInt(n / 10, 10);
}
return sum;
}
// Function that returns true
// if n is palindrome
function isPalindrome(n)
{
// Find the appropriate divisor
// to extract the leading digit
let divisor = 1;
while (parseInt(n / divisor, 10) >= 10)
divisor *= 10;
while (n != 0)
{
let leading = parseInt(n / divisor, 10);
let trailing = n % 10;
// If first and last digit
// not same return false
if (leading != trailing)
return false;
// Removing the leading and trailing
// digit from number
n = parseInt((n % divisor) / 10, 10);
// Reducing divisor by a factor
// of 2 as 2 digits are dropped
divisor = parseInt(divisor / 100, 10);
}
return true;
}
// Function that returns true if
// the digit sum of n is palindrome
function isDigitSumPalindrome(n)
{
// Sum of the digits of n
let sum = digitSum(n);
// If the digit sum is palindrome
if (isPalindrome(sum))
return true;
return false;
}
let n = 56;
if (isDigitSumPalindrome(n))
document.write("Yes");
else
document.write("No");
// This code is contributed by suresh07.
</script>
Producción
Yes
Complejidad temporal: O(logN)
Espacio auxiliar: O(1)
Otro enfoque: la idea es encontrar la suma de los dígitos de N y almacenarla en una suma variable y luego convertir la suma en una string, digamos s1 y verificar si invertir la string s1 es igual a s1. En caso afirmativo, entonces esta suma de el dígito del número dado N es palíndromo, de lo contrario no lo es.
Java
/*package whatever //do not write package name here */
import java.io.*;
class GFG {
public static void main(String[] args)
{
int n = 56;
int sum = 0;
while (n != 0) {
int temp = n % 10;
// Addition of digits
sum = sum + temp;
n = n / 10;
}
// Converting int sum to string
String str = String.valueOf(sum);
// Reverse the string
String rev
= new StringBuilder(str).reverse().toString();
if (str.equals(rev)) {
System.out.println("Yes");
}
else {
System.out.println("No");
}
}
}
// This code is contributed by keerthikarathan123
Python3
# code
num = int(56)
sum = int(0)
while num != 0:
temp = int(num % 10)
sum = sum+temp
num = num/10
# convert sum to string
string = str(sum)
# reverse the string
string_rev = string[:: -1]
# check the palindrome
if string == string_rev:
print("Yes")
else:
print("No")
# This code is contributed by keerthikarathan123
Producción
Yes
Complejidad temporal: O(N)
Espacio auxiliar: O(1)