# Count the number of digits of palindrome numbers in an array

Given an array arr[] with N integers. The task is to count all the digits of all palindrome numbers present in the array.

Examples:

Input: arr[] = {121, 56, 434}
Output: 6
Only 121 and 434 are palindromes
and digitCount(121) + digitCount(434) = 3 + 3 = 6

Input: arr[] = {56, 455, 546, 234}
Output: 0

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach: For every element of the array, if it is a one digit number then add 1 to the answer for its digit else check if the number is a palindrome. If yes then find the count of its digits and add it to the answer.

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach ` `#include ` `using` `namespace` `std; ` ` `  `// Function to return the reverse of n ` `int` `reverse(``int` `n) ` `{ ` `    ``int` `rev = 0; ` `    ``while` `(n > 0)  ` `    ``{ ` `        ``int` `d = n % 10; ` `        ``rev = rev * 10 + d; ` `        ``n = n / 10; ` `    ``} ` `    ``return` `rev; ` `} ` ` `  `// Function that returns true ` `// if n is a palindrome ` `bool` `isPalin(``int` `n) ` `{ ` `    ``return` `(n == reverse(n)); ` `} ` ` `  `// Function to return the ` `// count of digits of n ` `int` `countDigits(``int` `n) ` `{ ` `    ``int` `c = 0; ` `    ``while` `(n > 0)  ` `    ``{ ` `        ``n = n / 10; ` `        ``c++; ` `    ``} ` `    ``return` `c; ` `} ` ` `  `// Function to return the count of digits ` `// in all the palindromic numbers of arr[] ` `int` `countPalinDigits(``int` `arr[], ``int` `n) ` `{ ` `    ``int` `s = 0; ` `    ``for` `(``int` `i = 0; i < n; i++)  ` `    ``{ ` ` `  `        ``// If arr[i] is a one digit number ` `        ``// or it is a palindrome ` `        ``if` `(arr[i] < 10 || isPalin(arr[i]))  ` `        ``{ ` `            ``s += countDigits(arr[i]); ` `        ``} ` `    ``} ` `    ``return` `s; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `arr[] = { 121, 56, 434 }; ` `    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr); ` `    ``cout << (countPalinDigits(arr, n)); ` `    ``return` `0; ` `} ` ` `  `// This code is contributed by mits `

## Java

 `// Java implementation of the approach ` `class` `GFG { ` ` `  `    ``// Function to return the reverse of n ` `    ``static` `int` `reverse(``int` `n) ` `    ``{ ` `        ``int` `rev = ``0``; ` `        ``while` `(n > ``0``) { ` `            ``int` `d = n % ``10``; ` `            ``rev = rev * ``10` `+ d; ` `            ``n = n / ``10``; ` `        ``} ` `        ``return` `rev; ` `    ``} ` ` `  `    ``// Function that returns true ` `    ``// if n is a palindrome ` `    ``static` `boolean` `isPalin(``int` `n) ` `    ``{ ` `        ``return` `(n == reverse(n)); ` `    ``} ` ` `  `    ``// Function to return the ` `    ``// count of digits of n ` `    ``static` `int` `countDigits(``int` `n) ` `    ``{ ` `        ``int` `c = ``0``; ` `        ``while` `(n > ``0``) { ` `            ``n = n / ``10``; ` `            ``c++; ` `        ``} ` `        ``return` `c; ` `    ``} ` ` `  `    ``// Function to return the count of digits ` `    ``// in all the palindromic numbers of arr[] ` `    ``static` `int` `countPalinDigits(``int``[] arr, ``int` `n) ` `    ``{ ` `        ``int` `s = ``0``; ` `        ``for` `(``int` `i = ``0``; i < n; i++) { ` ` `  `            ``// If arr[i] is a one digit number ` `            ``// or it is a palindrome ` `            ``if` `(arr[i] < ``10` `|| isPalin(arr[i])) { ` `                ``s += countDigits(arr[i]); ` `            ``} ` `        ``} ` `        ``return` `s; ` `    ``} ` ` `  `    ``// Driver code ` `    ``public` `static` `void` `main(String[] args) ` `    ``{ ` `        ``int``[] arr = { ``121``, ``56``, ``434` `}; ` `        ``int` `n = arr.length; ` `        ``System.out.println(countPalinDigits(arr, n)); ` `    ``} ` `} `

## Python3

 `# Python3 implementation of the approach  ` ` `  `# Function to return the reverse of n  ` `def` `reverse(n):  ` `    ``rev ``=` `0``;  ` `    ``while` `(n > ``0``): ` `        ``d ``=` `n ``%` `10``;  ` `        ``rev ``=` `rev ``*` `10` `+` `d;  ` `        ``n ``=` `n ``/``/` `10``;  ` `    ``return` `rev;  ` ` `  `# Function that returns true  ` `# if n is a palindrome  ` `def` `isPalin(n):  ` `    ``return` `(n ``=``=` `reverse(n));  ` ` `  ` `  `# Function to return the  ` `# count of digits of n  ` `def` `countDigits(n):  ` `    ``c ``=` `0``;  ` `    ``while` `(n > ``0``):  ` `        ``n ``=` `n ``/``/` `10``;  ` `        ``c ``+``=` `1``; ` `    ``return` `c;  ` ` `  `# Function to return the count of digits  ` `# in all the palindromic numbers of arr[]  ` `def` `countPalinDigits(arr, n):  ` `    ``s ``=` `0``;  ` `    ``for` `i ``in` `range``(n): ` ` `  `        ``# If arr[i] is a one digit number  ` `        ``# or it is a palindrome  ` `        ``if` `(arr[i] < ``10` `or` `isPalin(arr[i])): ` `            ``s ``+``=` `countDigits(arr[i]);  ` ` `  `    ``return` `s;  ` ` `  ` `  `# Driver code  ` `arr ``=` `[ ``121``, ``56``, ``434` `];  ` `n ``=` `len``(arr);  ` `print``(countPalinDigits(arr, n));  ` ` `  `# This code contributed by Rajput-Ji `

## C#

 `// C# implementation of the approach  ` `using` `System; ` `     `  `class` `GFG  ` `{ ` ` `  `    ``// Function to return the reverse of n ` `    ``static` `int` `reverse(``int` `n) ` `    ``{ ` `        ``int` `rev = 0; ` `        ``while` `(n > 0) ` `        ``{ ` `            ``int` `d = n % 10; ` `            ``rev = rev * 10 + d; ` `            ``n = n / 10; ` `        ``} ` `        ``return` `rev; ` `    ``} ` ` `  `    ``// Function that returns true ` `    ``// if n is a palindrome ` `    ``static` `bool` `isPalin(``int` `n) ` `    ``{ ` `        ``return` `(n == reverse(n)); ` `    ``} ` ` `  `    ``// Function to return the ` `    ``// count of digits of n ` `    ``static` `int` `countDigits(``int` `n) ` `    ``{ ` `        ``int` `c = 0; ` `        ``while` `(n > 0)  ` `        ``{ ` `            ``n = n / 10; ` `            ``c++; ` `        ``} ` `        ``return` `c; ` `    ``} ` ` `  `    ``// Function to return the count of digits ` `    ``// in all the palindromic numbers of arr[] ` `    ``static` `int` `countPalinDigits(``int``[] arr, ``int` `n) ` `    ``{ ` `        ``int` `s = 0; ` `        ``for` `(``int` `i = 0; i < n; i++) ` `        ``{ ` ` `  `            ``// If arr[i] is a one digit number ` `            ``// or it is a palindrome ` `            ``if` `(arr[i] < 10 || isPalin(arr[i]))  ` `            ``{ ` `                ``s += countDigits(arr[i]); ` `            ``} ` `        ``} ` `        ``return` `s; ` `    ``} ` ` `  `    ``// Driver code ` `    ``public` `static` `void` `Main() ` `    ``{ ` `        ``int``[] arr = { 121, 56, 434 }; ` `        ``int` `n = arr.Length; ` `        ``Console.WriteLine(countPalinDigits(arr, n)); ` `    ``} ` `} ` ` `  `/* This code contributed by PrinciRaj1992 */`

Output:

```6
```

Shorter Python Implementation

## Python3

 `# Function to return the count of digits  ` `# in all the palindromic numbers of arr[]  ` `def` `countPalinDigits(arr):  ` `   ``sum` `=` `0` `  `  `   ``for` `n ``in` `arr: ` `      ``n_str ``=` `str``(n) ` `      ``l ``=` `len``(n_str) ` `      ``if` `n_str[l::``-``1``] ``=``=` `n_str: ``# if palindrome ` `         ``sum` `+``=` `l  ` `   ``return` `sum` ` `  `# Driver code  ` `arr ``=` `[ ``121``, ``56``, ``434` `];  ` `print``(countPalinDigits(arr));  `

Output:

```6
```

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