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

Palindrome Index

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1099 Discussions

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  • contact1827
     + 0 comments

    O(n) c++ solution:

    int palindromeIndex(string s) {
    size_t left{}, right{s.size() - 1};
		// last occurrence 
    std::pair<int, int> index{};
    int cursor{-1};
		// track direction pointer is moving in case of char mismatch
		// 0: left++
		// 1: right--
		// 2: no solution found left-right 
    size_t counter{};
    while(left < right)
    {
        if(s[left] == s[right])
        {
            left++;
            right--;
        }
        else {
            if(counter >= 2)
                return -1;
            else if(counter == 1)
            {
                left = index.first;
                right = index.second;
                cursor = right;
                right--;        
            }
            else {
                index = {left, right};
                cursor = left;
                left++;
            }
            counter++;
        }
    }
        
    return cursor;
}

  • 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;
}

  • emhhacker
     + 0 comments

    My Java solution:

    public static int palindromeIndex(String s) {
        //goal: find idx of char that will make str a palindrome when its removed
        if(s.length() == 1) return -1; //already a palindrome
        
        //use two pointers to iterate from left and right of str
        int left = 0, right = s.length() - 1;
        while(left < right){
           char leftChar = s.charAt(left);
           char rightChar = s.charAt(right);
           //if left pointer != right pointer, check which idx will make a palindrome if removed
           if(leftChar != rightChar){
                if(isPalindrome(s, left + 1, right)) return left;
                else return right;
           }
            left++;
            right--;
        }
        return -1; //str is already a palindrome
    }
    
    //uses two pointers to make sure chars are mirrored
    public static boolean isPalindrome(String s, int left, int right){
        while(left < right){
            if(s.charAt(left) != s.charAt(right)) return false; //chars arent mirrored
            left++;
            right--;
        }
        return true;
    }

  • adithyarupesh81
     + 0 comments

    the solution is: def ispalindromes(s): return s==s[::-1]

    def palindromeIndex(s): n=len(s) if ispalindromes(s): return -1

    for i in range(n//2):
    if s[i]!=s[n-i-1]:
        if ispalindromes(s[:1]+s[i+1:]):
            return i
        else:
            return n-i-1
return -1

  • patrickgao2020
     + 2 comments

    My Python solution doesn't pass many of the testcases! Can anyone help?

    def ispalindrome(s):
    if s[::-1] == s:
        return True
    return False

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