Palindrome Partitioning, Medium Leetcode

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Problem is this one: https://leetcode.com/problems/palindrome-partitioning/

Given a string s, partition s such that every substring of the partition is a palindrome.
Return all possible palindrome partitioning of s.
Example:
Input: "aab"
Output:
[
  ["aa","b"],
  ["a","a","b"]
]
Since the author wants all the solutions, chances are DP won't work here and a brute-force seems to be a better choice. Create a function to check whether a string is a palindrome. Then, do a brute-force recursively passing in the candidates that are palindrome. Add at the end, and return. That's it. Cheers, ACC.


public class Solution
{
 public IList> Partition(string s)
 {
  List> solution = new List>();
  Partition(s, new LinkedList(), solution);
  return solution;
 }

 private void Partition(string s,
       LinkedList pal,
       List> solution)
 {
  if (String.IsNullOrEmpty(s))
  {
   if (pal.Count > 0)
   {
    List palCopy = new List();
    foreach (string p in pal)
    {
     palCopy.Add(p);
    }
    solution.Add(palCopy);
   }
   return;
  }

  string current = "";
  for (int i = 0; i < s.Length; i++)
  {
   current += s[i].ToString();
   if (IsPalindrome(current))
   {
    pal.AddLast(current);
    Partition(s.Substring(i + 1), pal, solution);
    pal.RemoveLast();
   } 
  }
 }

 private bool IsPalindrome(string s)
 {
  if (String.IsNullOrEmpty(s)) return true;
  int left = 0;
  int right = s.Length - 1;
  while (left < right)
  {
   if (s[left] != s[right]) return false;
   left++;
   right--;
  }

  return true;
 }
}

Comments

  1. TarasMay 9, 2019 at 10:22 PM

    Python is a bit shorter :)

    import functools

    class Solution:
    def is_palindrome(self, s):
    left, right = 0, len(s)-1
    while left < right:
    if s[left] != s[right]:
    return False
    left += 1
    right -= 1
    return True

    @functools.lru_cache(maxsize=None)
    def partition(self, s):
    """
    :type s: str
    :rtype: List[List[str]]
    """
    if not s:
    return [[]]
    return [[s[:right+1]] + p for right in range(len(s)) for p in self.partition(s[right+1:]) if self.is_palindrome(s[:right+1])]

    ReplyDelete

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