Skip to main content

Leetcode 211. Design Add and Search Words Data Structure. Python (Prefix Tree)

 

211. Design Add and Search Words Data Structure


Design a data structure that supports adding new words and finding if a string matches any previously added string.

Implement the WordDictionary class:

  • WordDictionary() Initializes the object.
  • void addWord(word) Adds word to the data structure, it can be matched later.
  • bool search(word) Returns true if there is any string in the data structure that matches word or false otherwise. word may contain dots '.' where dots can be matched with any letter.

 

Example:

Input
["WordDictionary","addWord","addWord","addWord","search","search","search","search"]
[[],["bad"],["dad"],["mad"],["pad"],["bad"],[".ad"],["b.."]]
Output
[null,null,null,null,false,true,true,true]

Explanation
WordDictionary wordDictionary = new WordDictionary();
wordDictionary.addWord("bad");
wordDictionary.addWord("dad");
wordDictionary.addWord("mad");
wordDictionary.search("pad"); // return False
wordDictionary.search("bad"); // return True
wordDictionary.search(".ad"); // return True
wordDictionary.search("b.."); // return True

 

Constraints:

  • 1 <= word.length <= 25
  • word in addWord consists of lowercase English letters.
  • word in search consist of '.' or lowercase English letters.
  • There will be at most 3 dots in word for search queries.
  • At most 104 calls will be made to addWord and search.


class TrieNode:
    def __init__(self):
        self.children = {}
        self.eow = False

class WordDictionary:
    def __init__(self):
        self.root = TrieNode()
        
    def addWord(self, word: str) -> None:
        curr = self.root
        for c in word: 
            if c not in curr.children:
                curr.children[c] = TrieNode()
            curr = curr.children[c]
        curr.eow = True

    def search(self, word: str) -> bool:
        def dfs(k, root):
            curr = root
            for i in range(k, len(word)):
                c = word[i]
                if c == ".":
                    for child in curr.children.values():
                        if dfs(i+1, child):
                            return True
                    return False
                else:
                    if c not in curr.children:
                        return False
                    curr = curr.children[c]
            return curr.eow
        
        return dfs(0, self.root)
        


# Your WordDictionary object will be instantiated and called as such:
# obj = WordDictionary()
# obj.addWord(word)
# param_2 = obj.search(word)


Explained :









Labels:

Comments

Post a Comment

Popular posts from this blog

Leetcode 371. Sum of Two Integers. C++ / Java

371 .  Sum of Two Integers   Given two integers  a  and  b , return  the sum of the two integers without using the operators   +   and   - .   Example 1: Input: a = 1, b = 2 Output: 3 Example 2: Input: a = 2, b = 3 Output: 5   Constraints: -1000 <= a, b <= 1000 Solution :  C++ : class Solution { public: int getSum(int a, int b) { if (b==0) return a; int sum = a ^ b; int cr = (unsigned int) (a & b) << 1; return getSum(sum, cr); } }; Java :  class Solution { public int getSum(int a, int b) { while(b != 0){ int tmp = (a & b) << 1; a = a ^ b; b = tmp; } return a; } } Explaination :
Post a Comment
Read more

Number of Connected Components in an Undirected Graph (Python)

66.  Number of Connected Components in an Undirected Graph Question Link :  check here Givennnodes labeled from0ton - 1and a list of undirected edges (each edge is a pair of nodes), write a function to find the number of connected components in an undirected graph. Example 1:      0          3      |          |      1 --- 2    4 Givenn = 5andedges = [[0, 1], [1, 2], [3, 4]], return2. Example 2:      0           4      |           |      1 --- 2 --- 3 Givenn = 5andedges = [[0, 1], [1, 2], [2, 3], [3, 4]], return1. Note: You can assume that no duplicate edges will appear inedges. Since all edges are undirected,[0, 1]is the same as[1, 0]and thus will not appear together inedges. Solution : class Solution: def counComponents(self, n: int, edges : List[List[int]]) -> i...
Post a Comment
Read more

Leetcode 347. Top K Frequent Elements. Python (Bubble Sort)

Top K Frequent Elements Given an integer array  nums  and an integer  k , return  the   k   most frequent elements . You may return the answer in  any order .   Example 1: Input: nums = [1,1,1,2,2,3], k = 2 Output: [1,2] Example 2: Input: nums = [1], k = 1 Output: [1]   Constraints: 1 <= nums.length <= 10 5 -10 4  <= nums[i] <= 10 4 k  is in the range  [1, the number of unique elements in the array] . It is  guaranteed  that the answer is  unique .   Follow up:  Your algorithm's time complexity must be better than  O(n log n) , where n is the array's size. Solution : class Solution:     def topKFrequent(self, n: List[int], k: int) -> List[int]:                  # [1,1,1,2,2,3] &  k = 2                  f = [[] for i in range(len(n) + 1)]         # f = [...
Post a Comment
Read more