Sum of all palindrome numbers present in an Array
Last Updated :
11 Jul, 2025
Given an array arr[] of N positive integers. The task is to find the sum of all palindrome numbers present in the array. Print the total sum.
A palindrome number is a number that when reversed is equal to the initial number. Example: 121 is palindrome(reverse(121) = 121), 123 is not palindrome(reverse(123) = 321).
Note: Consider palindrome numbers of length greater than 1 while calculating the sum.
Examples:
Input : arr[] ={12, 313, 11, 44, 9, 1}
Output : 368
Input : arr[] = {12, 11, 121}
Output : 132
Approach: The idea is to implement a reverse function that reverses a number from the right to left. Implement a function that checks for palindrome numbers and finally traverses the array and calculates the sum of all elements which are palindrome.
Below is the implementation of the above approach:
C++
// C++ program to calculate the sum of all
// palindromic numbers in array
#include<bits/stdc++.h>
using namespace std;
// Function to reverse a number n
int reverse(int n)
{
int d = 0, s = 0;
while (n > 0)
{
d = n % 10;
s = s * 10 + d;
n = n / 10;
}
return s;
}
// Function to check if a number n is
// palindrome
bool isPalin(int n)
{
// If n is equal to the reverse of n
// it is a palindrome
return n == reverse(n);
}
// Function to calculate sum of all array
// elements which are palindrome
int sumOfArray(int arr[], int n)
{
int s = 0;
for (int i = 0; i < n; i++)
{
if ((arr[i] > 10) && isPalin(arr[i]))
{
// summation of all palindrome numbers
// present in array
s += arr[i];
}
}
return s;
}
// Driver Code
int main()
{
int n = 6;
int arr[] = { 12, 313, 11, 44, 9, 1 };
cout << sumOfArray(arr, n);
return 0;
}
// This code is contributed by mits
// Java program to calculate the sum of all
// palindromic numbers in array
class GFG {
// Function to reverse a number n
static int reverse(int n)
{
int d = 0, s = 0;
while (n > 0) {
d = n % 10;
s = s * 10 + d;
n = n / 10;
}
return s;
}
// Function to check if a number n is
// palindrome
static boolean isPalin(int n)
{
// If n is equal to the reverse of n
// it is a palindrome
return n == reverse(n);
}
// Function to calculate sum of all array
// elements which are palindrome
static int sumOfArray(int[] arr, int n)
{
int s = 0;
for (int i = 0; i < n; i++) {
if ((arr[i] > 10) && isPalin(arr[i])) {
// summation of all palindrome numbers
// present in array
s += arr[i];
}
}
return s;
}
// Driver Code
public static void main(String[] args)
{
int n = 6;
int[] arr = { 12, 313, 11, 44, 9, 1 };
System.out.println(sumOfArray(arr, n));
}
}
// C# program to calculate the sum of all
// palindromic numbers in array
using System;
class GFG
{
// Function to reverse a number n
static int reverse(int n)
{
int d = 0, s = 0;
while (n > 0)
{
d = n % 10;
s = s * 10 + d;
n = n / 10;
}
return s;
}
// Function to check if a number n is
// palindrome
static bool isPalin(int n)
{
// If n is equal to the reverse of n
// it is a palindrome
return n == reverse(n);
}
// Function to calculate sum of all array
// elements which are palindrome
static int sumOfArray(int[] arr, int n)
{
int s = 0;
for (int i = 0; i < n; i++)
{
if ((arr[i] > 10) && isPalin(arr[i]))
{
// summation of all palindrome numbers
// present in array
s += arr[i];
}
}
return s;
}
// Driver Code
public static void Main(String[] args)
{
int n = 6;
int[] arr = { 12, 313, 11, 44, 9, 1 };
Console.WriteLine(sumOfArray(arr, n));
}
}
/* This code contributed by PrinciRaj1992 */
# Python3 program to calculate the sum of all
# palindromic numbers in array
# Function to reverse a number n
def reverse(n) :
d = 0; s = 0;
while (n > 0) :
d = n % 10;
s = s * 10 + d;
n = n // 10;
return s;
# Function to check if a number n is
# palindrome
def isPalin(n) :
# If n is equal to the reverse of n
# it is a palindrome
return n == reverse(n);
# Function to calculate sum of all array
# elements which are palindrome
def sumOfArray(arr, n) :
s = 0;
for i in range(n) :
if ((arr[i] > 10) and isPalin(arr[i])) :
# summation of all palindrome numbers
# present in array
s += arr[i];
return s;
# Driver Code
if __name__ == "__main__" :
n = 6;
arr = [ 12, 313, 11, 44, 9, 1 ];
print(sumOfArray(arr, n));
# This code is contributed by AnkitRai01
<script>
// Javascript program to calculate the
// sum of all palindromic numbers in array
// Function to reverse a number n
function reverse( n)
{
let d = 0, s = 0;
while (n > 0)
{
d = n % 10;
s = s * 10 + d;
n = Math.floor(n / 10);
}
return s;
}
// Function to check if a number n is
// palindrome
function isPalin(n)
{
// If n is equal to the reverse of n
// it is a palindrome
return n == reverse(n);
}
// Function to calculate sum of all array
// elements which are palindrome
function sumOfArray( arr, n)
{
let s = 0;
for(let i = 0; i < n; i++)
{
if ((arr[i] > 10) && isPalin(arr[i]))
{
// Summation of all palindrome
// numbers present in array
s += arr[i];
}
}
return s;
}
// Driver Code
let n = 6;
let arr = [ 12, 313, 11, 44, 9, 1 ];
document.write(sumOfArray(arr, n));
// This code is contributed by jana_sayantan
</script>
Time Complexity: O(n * max(arr)), where max(arr) is the largest element of the array arr.
Auxiliary Space: O(1), since no extra space has been taken.
Another approach :
In java, we can easily implement it by using StringBuilder object and the reverse() method.
Step 1: Get the input from the user
Step 2: Initialize sum=0 and iterate through each element.
Step 3: Now, convert the integer element to string by using Integer.toString(array[i])
Step 4:Reverse it by StringBuilder object using reverse() method and toString() is used to convert the object to string.
Step 5: Equalize both the string values and check element greater than 9. If it satisfies the condition, sum the elements.
C++
#include <bits/stdc++.h>
using namespace std;
string reverse(string str,int l,int r)
{
while(l<r)
{
swap(str[l++],str[r--]);
}
return str;
}
int main()
{
int array[]={12, 313, 11, 44, 9, 1};
int n = sizeof(array)/sizeof(array[0]);
int sum = 0;
for(int i = 0; i < n; i++){
string str = to_string(array[i]);
string rev=reverse(str,0,str.size()-1);
if(str == rev && array[i] > 9){
sum=sum+array[i];
}
}
cout << sum;
return 0;
}
// this code is contributed by aditya942003patil
/*package whatever //do not write package name here */
import java.io.*;
class GFG {
public static void main (String[] args) {
int array[]={12, 313, 11, 44, 9, 1};
int n=array.length;
int sum=0;
for(int i=0;i<n;i++){
String str=Integer.toString(array[i]);
String rev=new StringBuilder(str).reverse().toString();
if(str.equals(rev) && array[i]>9){
sum=sum+array[i];
}
}
System.out.println(sum);
}
}
# Python3 program to implement the approach
# This method reverses the characters in a string
# in the range [l, r] by swapping characters
# from each end one by one
def reverse(str_, l, r):
str_ = list(str_)
while l < r:
str_[i], str_[j] = str_[j], str_[i]
l += 1
r -= 1
return "".join(str_)
# Driver Code
array = [12, 313, 11, 44, 9, 1]
n = len(array)
sum = 0
for i in range(n):
string = str(array[i])
rev = string[::-1]
if string == rev and array[i] > 9:
sum += array[i]
print(sum)
# This code is contributed by phasing17
// C# code to implement the approach
using System;
using System.Collections.Generic;
class GFG {
// Function to reverse the string
static string reverse(string str)
{
// Convert string to char array
char[] charArray = str.ToCharArray();
// Char array to store reversed array
char[] result = new char[charArray.Length];
for (int i = 0, j = str.Length - 1; i < str.Length;
i++, j--) {
result[i] = charArray[j];
}
return new string(result);
}
public static void Main(string[] args)
{
int[] array = { 12, 313, 11, 44, 9, 1 };
int n = array.Length;
int sum = 0;
for (int i = 0; i < n; i++) {
// Convert to string
string str = Convert.ToString(array[i]);
// Reverse the string
string rev = reverse(str);
// If string is palindromic
// Update sum
if (str.Equals(rev) && array[i] > 9) {
sum = sum + array[i];
}
}
// Display result
Console.WriteLine(sum);
}
}
// This code is contributed by phasing17
// JavaScript code to implement the approach
function reverse(str, l, r)
{
str = str.split("");
while (l < r) {
let temp = str[r];
str[r--] = str[l];
str[l++] = temp;
}
return str.join("");
}
let array = [ 12, 313, 11, 44, 9, 1 ];
let n = array.length;
let sum = 0;
for (let i = 0; i < n; i++) {
let str = "" + (array[i]);
let rev = reverse(str, 0, str.length - 1);
if (str == rev && array[i] > 9) {
sum = sum + array[i];
}
}
console.log(sum);
// this code is contributed by phasing17
Time Complexity: O(n), where n is the size of the given array.
Auxiliary Space: O(d), where d is the maximum digits in number in the given array.
In python, we can execute it by implementing the below approach.
Step 1 : Initialize the list or array and sum=0.
Step 2 : Iterate the list using for loop and convert the integer element to string using str().
Step 3 : Reverse the string using string_element[ : : -1].
Step 4 :Equalize both the string values and check element greater than 9. If it satisfies the condition, sum the elements.
C++
// C++ program to implement the approach
#include <bits/stdc++.h>
using namespace std;
int main()
{
int lis[] = {12, 313, 11, 44, 9, 1};
int sum = 0;
for (int i : lis)
{
string string_conversion = to_string(i);
string rev_string = "" + string_conversion;
reverse(rev_string.begin(), rev_string.end());
if(string_conversion == (rev_string) && i>9)
sum = sum + i;
}
cout << sum;
}
// This code is contributed by phasing17
// Java program to implement the approach
import java.util.*;
class GFG {
public static void main(String[] args)
{
int lis[] = { 12, 313, 11, 44, 9, 1 };
int sum = 0;
for (int i : lis) {
String string_conversion = String.valueOf(i);
StringBuilder rev
= new StringBuilder(string_conversion);
rev.reverse();
String rev_string = rev.toString();
if (string_conversion.equals(rev_string)
&& i > 9)
sum = sum + i;
}
System.out.println(sum);
}
}
// This code is contributed by phasing17
# code
lis=[12, 313, 11, 44, 9, 1]
sum=0;
for i in lis:
string_conversion=str(i)
rev_string=string_conversion[ : : -1]
if(string_conversion==rev_string and i>9):
sum=sum+i
print(sum)
// C# program to implement the approach
using System;
using System.Collections.Generic;
class GFG {
public static void Main(string[] args)
{
int[] lis = { 12, 313, 11, 44, 9, 1 };
int sum = 0;
foreach(int i in lis)
{
string string_conversion = Convert.ToString(i);
char[] rev = string_conversion.ToCharArray();
Array.Reverse(rev);
string rev_string = new string(rev);
if (string_conversion.Equals(rev_string)
&& i > 9)
sum = sum + i;
}
Console.WriteLine(sum);
}
}
// This code is contributed by phasing17
// JS program to implement the approach
let lis = [12, 313, 11, 44, 9, 1]
let sum = 0;
for (var i of lis)
{
let string_conversion = "" + i
let rev_string = string_conversion.split("").reverse().join("");
if((string_conversion.localeCompare(rev_string) == 0) && i>9)
sum = sum + i
}
console.log(sum)
// This code is contributed by phasing17
Time Complexity: O(n)
Auxiliary Space: O(d), where d is the maximum digits in the number.
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