0139-word-break.py

# Tries
# time complexity: O(n^2 + m*k)
# space complexity: O(n + m*k)
from collections import deque
from typing import List


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


class Solution:
    def wordBreak(self, s: str, wordDict: List[str]) -> bool:
        root = TrieNode()
        for word in wordDict:
            curr = root
            for c in word:
                if c not in curr.children:
                    curr.children[c] = TrieNode()
                curr = curr.children[c]
            curr.isWord = True

        dp = [False] * len(s)
        for i in range(len(s)):
            if i == 0 or dp[i-1]:
                curr = root
                for j in range(i, len(s)):
                    c = s[j]
                    if c not in curr.children:
                        break
                    curr = curr.children[c]
                    if curr.isWord:
                        dp[j] = True
        return dp[-1]

# Bottom Up
# time complexity: O(n^3 +m*k)
# space complexity: O(n+m*k)
# from typing import List


class Solution:
    def wordBreak(self, s: str, wordDict: List[str]) -> bool:
        n = len(s)
        words = set(wordDict)
        dp = [False] * (n + 1)
        dp[0] = True
        for i in range(1, n + 1):
            for j in range(i):
                if dp[j] and s[j:i] in words:
                    dp[i] = True
                    break
        return dp[-1]

s = "leetcode"
wordDict = ["leet", "code"]
print(Solution().wordBreak(s, wordDict))
s = "applepenapple"
wordDict = ["apple", "pen"]
print(Solution().wordBreak(s, wordDict))
s = "catsandog"
wordDict = ["cats", "dog", "sand", "and", "cat"]
print(Solution().wordBreak(s, wordDict))

# DP 1
# time complexity: O(n*m*k)
# space complexity: O(n)


class Solution:
    def wordBreak(self, s: str, wordDict: List[str]) -> bool:
        sLen = len(s)
        dp = [False] * sLen
        for i in range(sLen):
            for word in wordDict:
                wordLen = len(word)
                if i < wordLen - 1:
                    continue
                if i == wordLen - 1 or dp[i-wordLen]:
                    if s[i-wordLen+1:i + 1] == word:
                        dp[i] = True
                        break
        return dp[sLen-1]


# BFS
# time complexity: O(n^3 + m*k)
# space complexity: O(n+m*k)


class Solution:
    def wordBreak(self, s: str, wordDict: List[str]) -> bool:
        words = set(wordDict)
        queue = deque([0])
        seen = set()
        while queue:
            start = queue.popleft()
            if start == len(s):
                return True
            for end in range(start + 1, len(s) + 1):
                if end in seen:
                    continue
                if s[start:end] in words:
                    queue.append(end)
                    seen.add(end)
        return False


s = "applepenapple"
wordDict = ["apple", "pen"]
print(Solution().wordBreak(s, wordDict))