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

Palindrome Index

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1087 Discussions

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  • johalun0
     + 0 comments

    Has anyone managed to pass all the test using Rust? I'm hitting the time limit on some tests and I'm pretty sure I got a really optimized O(n), zero copy solution (with help from solutions below in other languages).

  • 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;
}

  • zaidnsour12
     + 0 comments

    Java solution:

     public static boolean isPalindrome(String s){
    int start = 0, end = s.length() - 1;
     while(start < end){
      if(s.charAt(start) != s.charAt(end))
        return false;
      start++;
      end--;
      } 
    return true;
    }
  
  public static int palindromeIndex(String s) {
    int start = 0, end = s.length() - 1;

    if(isPalindrome(s))
      return -1;
    
    while(start < end){
      int start_elem = s.charAt(start);
      int end_elem = s.charAt(end);
      
      if(start_elem != end_elem){
        String s_without_start = s.substring(start + 1, end + 1);
        if(isPalindrome(s_without_start))
          return start;
      
        String s_without_end = s.substring(start, end);
        if(isPalindrome(s_without_end))
          return end;  
        }
      start++;
      end--;
      }
    
    return -1; 
  }

  • v_trclam
     + 0 comments

    Below is my C# solution with explaination in comments. Hope you find it useful.

    public static int palindromeIndex(string s)
{
    for (int i = 0, j = s.Length - 1; i < j; i++, j--)
    {
        if (s[i] != s[j])
        {
            // remove 1 char at i or j
            // then check for condition
            // if false -> do not need to check the remain positions
            // because it can not be palindrome anymore
            // return -1
            if (CheckPalindrome(s, i + 1, j)) return i; //remove at position i & check
            if (CheckPalindrome(s, i, j - 1)) return j; //remove at position j & check
            return -1; // there is no solution
        }
    }
    return -1; // it's palindrome already  
}
public static bool CheckPalindrome(string s, int left, int right)
{
    for (int i = left, j = right; i < j; i++, j--)
    {
        if (s[i] != s[j]) return false;
    }
    return true;

}

  • 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
}