Check Palindrome by Filtering Non-Letters
Difficulty: Easy
Given a string containing letters, digits, and symbols, determine if it reads the same forwards and backwards when considering only alphabetic characters (case-insensitive).
Example
Input
code = A1b2B!a
Output
1
Explanation
- Step 1: Extract only letters → ['A','b','B','a']
- Step 2: Convert to lowercase → ['a','b','b','a']
- Step 3: Compare sequence forward and backward: 'abba' == 'abba' → true
Input Format
- A string code containing letters (A–Z, a–z), digits (0–9), and symbols
Constraints
- 0 <= code.length <= 1000
- For all 0 <= i < code.length: 33 <= ASCII(code[i]) <= 126
- code contains only printable ASCII characters (letters, digits, symbols)
Output Format
- Return a boolean value: 1 if true & 0 if false.
Sample Input 0
Z
Sample Output 0
1
Sample Input 1
abc123cba
Sample Output 1
1
Consideration:
1. Two Pointers: Left index, Right index to compare the alphabetic characters.
2. If not alphabetic characters, jump to the next index. (Left Index ++, Right Index --)
3. case insensitive: compare the distance with 'A' or 'a'.
Code:
/*
* Complete the 'isAlphabeticPalindrome' function below.
*
* The function is expected to return a BOOLEAN.
* The function accepts STRING code as parameter.
*/
bool isAlphabeticPalindrome(string code) {
int n = code.size();
//cout << "n = " << n << "\n";
int i = 0, j = n-1;
if (n==0 || n==1) return 1;
while (i<j) {
//cout << "i = " << i << "\n";
//cout << "j = " << j << "\n";
if (((code[i]>='A') && (code[i] <= 'Z')) || ((code[i] >= 'a') && (code[i] <= 'z'))) {
if (((code[j]>='A') && (code[j] <= 'Z')) || ((code[j] >= 'a') && (code[j] <= 'z'))) {
if (code[i]==code[j]) {
i++;
j--;
}
else if (code[i]>code[j]) {
if (code[i]>='a') {
if ((code[j]-'A')==(code[i]-'a')){
i++;
j--;
}
else {
//cout << "i = " << i << "[" << code[i] << "]" << "\n";
//cout << "j = " << j << "[" << code[j] << "]" << "\n";
return 0;
}
}
else {
//cout << "i = " << i << "[" << code[i] << "]" << "\n";
//cout << "j = " << j << "[" << code[j] << "]"<< "\n";
return 0;
}
}
else if (code[i]<code[j]) {
if (code[j]>='a') {
if ((code[i]-'A')==(code[j]-'a')){
i++;
j--;
}
else {
//cout << "i = " << i << "[" << code[i] << "]" << "\n";
//cout << "j = " << j << "[" << code[j] << "]" << "\n";
return 0;
}
}
else {
//cout << "i = " << i << "[" << code[i] << "]"<< "\n";
//cout << "j = " << j << "[" << code[j] << "]"<< "\n";
return 0;
}
}
}
else j--;
}
else i++;
}
return 1;
}
