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  2. Algorithms
  3. Strings
  4. Palindrome Index
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Palindrome Index

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1106 Discussions

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  • jkyeusun
     + 1 comment

    Can't for the life of me figure out what's wrong with my solution - it passes 12/15 test cases. 

    def palindromeIndex(s):
        for i in range(len(s) // 2):
                j = len(s) - 1 - i
                if (s[i] == s[j]):
                        continue
                if (s[i] == s[j - 1]):
                        return j
                if (s[i + 1] == s[j]):
                        return i
        return -1

  • nicoleliu1016
     + 0 comments

    Python Complexity:O(n)

    def checkPalindrome(s):
    for i in range(len(s) // 2):
        if s[i] != s[len(s) - i - 1]:
            return False
    return True

def palindromeIndex(s):
    idx = -1
    # s is palindrome
    if checkPalindrome(s):
        return -1
        
    # find the first character index that does not match
    for i in range(len(s) // 2):
        if s[i] == s[len(s) - i - 1]:
            continue
        else:
            idx = i
            break
    
    # drop the first character
    sub_s_1 = s[(idx+1) : (len(s) - idx)]
    if checkPalindrome(sub_s_1):
        return idx
    
    # drop the last character
    sub_s_2 = s[idx : (len(s) - idx - 1)]
    if checkPalindrome(sub_s_2):
        return len(s) - idx - 1

    return -1

  • afzal_saiyed931
     + 0 comments
    def palindromeIndex(s):
    # Write your code here
    
    def isPalindrome(s, start, end):
        while start < end:
            if s[start] != s[end]:
                return False
            start += 1
            end -= 1
        return True
    
    n = len(s)
    start = 0
    end = n - 1
    
    while start < end:
        if s[start] != s[end]:
            
            if isPalindrome(s, start + 1, end):
                return start
            elif isPalindrome(s, start, end - 1):
                return end
            else:
                return -1
        start += 1
        end -= 1
            
    return -1

  • cass6896
     + 0 comments

    C++

    int palindromeIndex(string s)
{
    for (size_t i = 0; i < s.length() / 2; i++)
    {
        if (s[i] != s[s.length() - 1 - i])
        {
            string tmp1 = s, tmp2 = s;
            tmp1.erase(i, 1), tmp2.erase(tmp2.length() - 1 - i, 1);
            
            reverse(tmp1.begin(), tmp1.begin() + tmp1.length() / 2);
            reverse(tmp2.begin(), tmp2.begin() + tmp2.length() / 2);
            
            if ((s.length() - 1) % 2)
            {
                if (tmp1.substr(0, tmp1.length() / 2) == tmp1.substr(tmp1.length() / 2 + 1, tmp1.length() / 2))
                    return i;
                else if (tmp2.substr(0, tmp2.length() / 2) == tmp2.substr(tmp2.length() / 2 + 1, tmp2.length() / 2))
                    return s.length() - 1 - i;
                else
                    return -1;
            }
            else
            {
                if (tmp1.substr(0, tmp1.length() / 2) == tmp1.substr(tmp1.length() / 2, tmp1.length() / 2))
                    return i;
                else if (tmp2.substr(0, tmp2.length() / 2) == tmp2.substr(tmp2.length() / 2, tmp2.length() / 2))
                    return s.length() - 1 - i;
                else
                    return -1;
            }
        }
    }

    return -1;
}

  • pradeep_tarakar
     + 1 comment

    My C++ Solution: Time complexity = O(n)

    string canPalindrome(string s){
    bool op1 = false;
        int slow = 1;
        int fast = s.length()-1;
        cout << s <<endl;
        while(fast > slow){
            if(s[fast] != s[slow]){
                op1 = true;
            }
            fast --;
            slow++;
        } 
        if(op1)return "second";
        else return "first";
}

int palindromeIndex(string s) {
    int idx = -1;
    int fast = s.length()-1;
    int slow = 0;
    
    while(slow < fast){
        if(s[slow] != s[fast]){
            string sub = s.substr(slow,(fast-slow)+1);
            string result = canPalindrome(sub);
            result == "first" ? idx=slow : idx = fast;
            return idx;
        }
        slow++;
        fast--;
    }
    return idx;
}