Noisy Palindrome - Facebook Top Interview Questions


Problem Statement :


You are given a string s containing lowercase and uppercase alphabet characters as well as digits from "0" to "9". 

Return whether s is a palindrome if we only consider the lowercase alphabet characters.

Constraints

0 ≤ n ≤ 100,000 where n is the length of s

Example 1

Input

s = "a92bcbXa"

Output

True

Explanation

If we only consider the lowercase characters, then s is "abcba" which is a palindrome.



Example 2

Input

s = "abZ"

Output

False



Solution :



                        Solution in C++ :

bool solve(string s) {
    int n = s.length();
    int i = 0, j = n - 1;
    while (i < j) {
        if (!isalpha(s[i]) || isupper(s[i])) {
            i++;
        }
        if (!isalpha(s[j]) || isupper(s[j])) {
            j--;
        }
        if (isalpha(s[i]) && islower(s[i]) && isalpha(s[j]) && islower(s[j])) {
            if (s[i] == s[j]) {
                i++, j--;
            } else {
                return false;
            }
        }
    }
    return true;
}
                    

                        Solution in Java :

import java.util.*;

class Solution {
    public boolean solve(String s) {
        int left = 0, right = s.length() - 1;
        while (left < right) {
            char l = s.charAt(left), r = s.charAt(right);
            if (l >= '0' && l <= '9' || l >= 'A' && l <= 'Z')
                left++;
            else if (r >= '0' && r <= '9' || r >= 'A' && r <= 'Z')
                right--;
            else if (l != r)
                return false;
            else {
                left++;
                right--;
            }
        }
        return true;
    }
}
                    

                        Solution in Python : 
                            
class Solution:
    def solve(self, s):
        s0 = (c for c in s if c in string.ascii_lowercase)
        s1 = (c for c in reversed(s) if c in string.ascii_lowercase)
        return all(c0 == c1 for c0, c1 in zip(s0, s1))
                    

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