Check if all elements of the array are palindrome or not

Try it on GfG Practice

Given an array arr[]. The task is to check if the array is PalinArray or not i.e., if all elements of array are palindrome or not. 

Examples: 

Input: arr[] = [21, 131, 20] 
Output: false
Explanation: For the given array, element 20 is not a palindrome so false is the output.

Input: arr[] = [111, 121, 222, 333, 444] 
Output: true 
Explanation: For the given array, all the elements of the array are palindromes.

By Checking Each Element - O(n*k) Time and O(1) Space

The basic idea is to check each number in the array is a palindrome by converting each number to a string and comparing its characters symmetrically from both ends. If any non-palindrome is found then returns false otherwise true.

#include<bits/stdc++.h>
using namespace std;

    bool isPalindrome(int n) {
        string s = "" + n;
        int len = s.length();
        for (int i = 0; i < len / 2; i++) {
            if (s[i] != s[len - 1 - i])
                return false;
        }
        return true;
    }

    bool isPalinArray(vector<int>& arr)   {
      
        // Traversing each element of the array
        // and check if it is palindrome or not
        for (int i = 0; i < arr.size(); i++) {
            if (! isPalindrome(arr[i]))
                return false;
        }
        return true;
    }

    int main() {
        vector<int> arr = { 121, 131, 20 };
        bool res = isPalinArray(arr);
        if (res == true)
            cout<<"true";
        else
            cout<<"false";
    }

class GfG {
    static boolean isPalindrome(int n) {
        String s = Integer.toString(n);
        int len = s.length();
        for (int i = 0; i < len / 2; i++) {
            if (s.charAt(i) != s.charAt(len - 1 - i))
                return false;
        }
        return true;
    }

    public static boolean isPalinArray(int[] arr) {
      
        // Traversing each element of the array
        // and check if it is palindrome or not
        for (int i = 0; i < arr.length; i++) {
            if (! isPalindrome(arr[i]))
                return false;
        }
        return true;
    }

    public static void main(String[] args) {
        int[] arr = { 121, 131, 20 };
        boolean res = isPalinArray(arr);
        System.out.println(res);
    }
}

def isPalindrome(n):
    s = str(n)
    return s == s[::-1]

def isPalinArray(arr):
  
    # Traversing each element of the array
    # and check if it is palindrome or not
    for num in arr:
        if not isPalindrome(num):
            return False
    return True


if __name__ == "__main__":
  arr = [121, 131, 20]
  res = isPalinArray(arr)
  print("true" if res else "false")

using System;
using System.Linq;

class GfG {
    static bool IsPalindrome(int n) {
        string s = n.ToString();
        return s.SequenceEqual(s.Reverse());
    }

    static bool IsPalinArray(int[] arr) {
      
        // Traversing each element of the array
        // and check if it is palindrome or not
        foreach (var num in arr) {
            if (! IsPalindrome(num))
                return false;
        }
        return true;
    }

    static void Main() {
        int[] arr = { 121, 131, 20 };
        bool res = IsPalinArray(arr);
        Console.WriteLine(res ? "true" : "false");
    }
}

function isPalindrome(n) {
    let s = n.toString();
    return s === s.split('').reverse().join('');
}

function isPalinArray(arr) {

    // Traversing each element of the array
    // and check if it is palindrome or not
    for (let num of arr) {
        if (! isPalindrome(num))
            return false;
    }
    return true;
}

// Driver Code
const arr = [121, 131, 20];
const res = isPalinArray(arr);
console.log(res ? 'true' : 'false');

true

Complexity Analysis:

• Time Complexity: O(n*k) n is for traversing the array and k is for checking each element.
• Auxiliary Space: O(1)