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
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<bits/stdc++.h>using namespace std;// Function to return the reverse of nint 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 palindromebool isPalin(int n){ return (n == reverse(n));}// Function to return the// count of digits of nint 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 codeint main(){ int arr[] = { 121, 56, 434 }; int n = sizeof(arr) / sizeof(arr[0]); cout << (countPalinDigits(arr, n)); return 0;}// This code is contributed by mits |
Java
// Java implementation of the approachimport java.util.*;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 */ |
Javascript
<script>// Javascript implementation of the approach// Function to return the reverse of nfunction reverse(n){ let rev = 0; while (n > 0) { let d = n % 10; rev = rev * 10 + d; n = parseInt(n / 10); } return rev;}// Function that returns true// if n is a palindromefunction isPalin(n){ return (n == reverse(n));}// Function to return the// count of digits of nfunction countDigits(n){ let c = 0; while (n > 0) { n = parseInt(n / 10); c++; } return c;}// Function to return the count of digits// in all the palindromic numbers of arr[]function countPalinDigits(arr, n){ let s = 0; for (let 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 let arr = [ 121, 56, 434 ]; let n = arr.length; document.write(countPalinDigits(arr, n));</script> |
Output:
6
Time Complexity : O(n)
Auxiliary Space: O(1)
Shorter Python Implementation
C++
// C++ code to implement the approach#include <bits/stdc++.h>using namespace std;// Function to return the count of digits // in all the palindromic numbers of arr[] int countPalinDigits(vector<int> arr){ int sum = 0; for (int n: arr) { string n_str = to_string(n); int l = n_str.length(); string rev = to_string(n); reverse(rev.begin(), rev.end()); if (rev == n_str) // if palindrome sum += l; } return sum; }// Driver code int main(){ vector <int> arr = { 121, 56, 434 }; cout << countPalinDigits(arr) << endl; }// This code is contributed by phasing17 |
Java
// Java code to implement the approachimport java.util.*;class GFG{ // Function to return the count of digits // in all the palindromic numbers of arr[] static int countPalinDigits(int[] arr) { int sum = 0; for (int n : arr) { String n_str = String.valueOf(n); int l = n_str.length(); String rev = new StringBuilder(new String(n_str)).reverse().toString(); if (rev.equals(n_str)) // if palindrome sum += l; } return sum; } // Driver code public static void main(String[] args) { int[] arr = { 121, 56, 434 }; System.out.println(countPalinDigits(arr)); }}// This code is contributed by phasing17 |
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)); |
C#
// C# code to implement the approachusing System;using System.Collections.Generic;class GFG{ // Function to return the count of digits // in all the palindromic numbers of arr[] static int countPalinDigits(int[] arr) { int sum = 0; foreach (int n in arr) { string n_str = Convert.ToString(n); int l = n_str.Length; string rev = Convert.ToString(n); char[] revs = rev.ToCharArray(); Array.Reverse(revs); rev = new string(revs); if (rev.Equals(n_str)) // if palindrome sum += l; } return sum; } // Driver code public static void Main(string[] args) { int[] arr = { 121, 56, 434 }; Console.WriteLine(countPalinDigits(arr)); }}// This code is contributed by phasing17 |
Javascript
// Function to return the count of digits // in all the palindromic numbers of arr[] function countPalinDigits(arr){ let sum = 0 for (let n of arr) { let n_str = "" + n; let l = n_str.length let rev = n_str.split("").reverse().join("") if (rev.localeCompare(n_str) == 0) // if palindrome sum += l } return sum }// Driver code let arr = [ 121, 56, 434 ]; console.log(countPalinDigits(arr)); // This code is contributed by phasing17 |
Output:
6
Time Complexity : O(n)
Auxiliary Space: O(n)
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