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  1. Prepare
  2. Algorithms
  3. Strings
  4. Palindrome Index
  5. Discussions

Palindrome Index

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1084 Discussions

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  • janvenyong
     + 0 comments
    function palindromeIndex(s) {
    let left = 0, right = s.length - 1
    while (left <= right) {
        if (s[left] !== s[right]) {
            if (isPalindrome(s.substring(left, right))) return right
            else if (isPalindrome(s.substring(left + 1, right + 1))) return left
            return -1
        } else {
            left++
            right--
        }
    }
    return -1
}

function isPalindrome(s) {
    for (let i = 0; i < Math.floor(s.length / 2); i++) {
        if (s[i] !== s[s.length - 1 - i]) return false
    }
    return true
}

  • manyatrivedi60
     + 1 comment

    my code returning right answer but still showing wrong submission.

  • alban_tyrex
     + 0 comments

    Here is my c++ solution : explanation here https://youtu.be/QjHpvMdfnqs,

     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;
}

  • bojan_milojkovi2
     + 0 comments

    // java solution

    private static int findIndexOfMismatch(String s) { for(int i=0 ; i 

    public static int palindromeIndex(String s) {
    int mismatchIndex = findIndexOfMismatch(s);
    if(mismatchIndex==-1) {
        // is already a palindrome
        return -1;
    }

    // option 1 - remove mismatching char at the left and check
    String maybePalindrome = s.substring(mismatchIndex+1, s.length()-mismatchIndex);
    if(findIndexOfMismatch(maybePalindrome)==-1) {
        return mismatchIndex;
    }

    // option 2 - remove mismatching char at the right and check
    maybePalindrome = s.substring(mismatchIndex, s.length()-mismatchIndex-1);
    if(findIndexOfMismatch(maybePalindrome)==-1) {
        return s.length()-mismatchIndex-1;
    }

    // option 3 - removing one char is not enough to make this into a palindrome
    return -1;
}

  • jmstr8261
     + 0 comments

    C# failed first tests using if(s == s,Reverse()) It was too slow for the tests - I'd still use it in production for readability though. Here's the final:

    public static int palindromeIndex(string s)
{
    int stringLength = s.Length;            
    int returnPos = 0;

    if (IsReversible(s))
        return -1;

    for (int i = 0; i < stringLength/2; i++)
    {
        if (s[i] != s[stringLength-i-1])
        {
            returnPos = RemoveCharacterAndTest(s, i);
            if (returnPos>-1)
                return returnPos;
        }
    }
    return -1;
}

private static int RemoveCharacterAndTest(string s, int removeAt)
{
    string testString = s.Remove(removeAt,1);
    int returnPos = 0;
    if (IsReversible(testString))
        return removeAt;

    testString = s.Remove(s.Length - removeAt - 1, 1);
    if (IsReversible(testString))
    {
        returnPos = (s.Length - removeAt - 1);
        return returnPos;
    }                
    return -1;
}

private static bool IsReversible(string s)
{
    int strLength = s.Length;
    for (int i = 0; i < strLength / 2; i++)
    {
        if (s[i] != s[strLength - i - 1])
            return false;
    }
    return true;
}