**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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