In this Palindrome Index HackerRank solution, Given a string of lowercase letters in the range ascii[a-z], determine the index of a character that can be removed to make the string a palindrome. There may be more than one solution, but any will do. If the word is already a palindrome or there is no solution, return -1. Otherwise, return the index of a character to remove.
Example
Either remove ‘b’ at index or ‘c’ at index .
Function Description
Complete the palindromeIndex function in the editor below.
palindromeIndex has the following parameter(s):
- string s: a string to analyze
Returns
- int: the index of the character to remove or
Input Format
The first line contains an integer , the number of queries.
Each of the next lines contains a query string .
Constraints
- All characters are in the range ascii[a-z].
Sample Input
STDIN Function
----- --------
3 q = 3
aaab s = 'aaab' (first query)
baa s = 'baa' (second query)
aaa s = 'aaa' (third query)
Sample Output
3
0
-1
Explanation
Query 1: “aaab”
Removing ‘b’ at index results in a palindrome, so return .
Query 2: “baa”
Removing ‘b’ at index results in a palindrome, so return .
Query 3: “aaa”
This string is already a palindrome, so return . Removing any one of the characters would result in a palindrome, but this test comes first.
Note: The custom checker logic for this challenge is available here.
Palindrome Index HackerRank solution
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Problem Solution in C
int string_length (char* str)
{
int len;
for(len = 0; str[len]; len++);
return len;
}
int palindromeIndex(char* s)
{
int len = string_length(s);
for (int counter = 0; counter < len/2; counter++)
{
if (s[counter] != s[len - 1 - counter])
{
if (s[counter] == s[len - 2 - counter])
{
if (s[counter + 1] == s[len - 3 - counter])
{
return len - 1 - counter;
}
}
return counter;
}
}
return -1;
}Problem Solution in Python
def palindromeIndex(s):
if s == s[::-1]:
return -1
for i in range(len(s)//2):
if s[i] != s[len(s) - 1 - i]:
newstr = s[:i] + s[i+1:]
if newstr == newstr[::-1]:
return i
newstr = s[:len(s) - i - 1]+ s[len(s)-i:]
if newstr == newstr[::-1]:
return len(s) - i - 1
return -1Problem Solution in C++
bool isPalindrome(string s){
string r = s;
reverse(r.begin(), r.end());
return s == r;
}
int palindromeIndex(string input) {
int s, e;
for(s = 0, e = input.size() - 1; s < e; s++, e--){
if(input[s] != input[e]) break;
}
if(s >= e) return -1;
string s1 = input, s2 = input;
s1.erase(s1.begin() + s);
if(isPalindrome(s1)) return s;
s2.erase(s2.begin() + e);
if(isPalindrome(s2)) return e;
return -1;
}