[NeetCode 150] Word Break

Word Break

Given a string s and a dictionary of strings wordDict, return true if s can be segmented into a space-separated sequence of dictionary words.

You are allowed to reuse words in the dictionary an unlimited number of times. You may assume all dictionary words are unique.

Example 1:

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Input: s = "neetcode", wordDict = ["neet","code"]

Output: true

Explanation: Return true because "neetcode" can be split into "neet" and "code".

Example 2:

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Input: s = "applepenapple", wordDict = ["apple","pen","ape"]

Output: true

Explanation: Return true because "applepenapple" can be split into "apple", "pen" and "apple". Notice that we can reuse words and also not use all the words.

Example 3:

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Input: s = "catsincars", wordDict = ["cats","cat","sin","in","car"]

Output: false

Constraints:

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1 <= s.length <= 200
1 <= wordDict.length <= 100
1 <= wordDict[i].length <= 20

s and wordDict[i] consist of only lowercase English letters.

Solution

To break the string, we can break it step by step. If s 0 : n s_{0:n} s0:n can be broken into words, then we just need to consider whether there exists a position m m m that s n : m s_{n:m} sn:m is a word in dictionary. So, we can record the positions that the substrings before these positions are broken, and check whether we can find a substring after these position that appears in dictionary.

One plausible way is to enumerate the next position and check the substring between two positions, whose time complexity is O ( Len 2 ( s ) ) O(\text{Len}^2(s)) O(Len2(s)). Another way is to traverse the dictionary and find one can be put after the last position, whose time complexity is O ( Len ( s ) × ∑ w Len ( w ) ) O(\text{Len}(s)\times\sum_w\text{Len}(w)) O(Len(s)×∑wLen(w))

Code

Enumerate break position:

py 复制代码

class Solution:
    def wordBreak(self, s: str, wordDict: List[str]) -> bool:
        word_set = set(wordDict)
        brk = [False]*(len(s)+1)
        brk[-1] = True
        for i in range(len(s)):
            subword = ''
            for j in range(i, -1, -1):
                subword = s[j]+subword
                if brk[j-1] and subword in word_set:
                    brk[i] = True
                    break
        print(brk)
        return brk[len(s)-1]

Enumerate dictionary words:

py 复制代码

class Solution:
    def wordBreak(self, s: str, wordDict: List[str]) -> bool:
        dp = [False] * (len(s) + 1)
        dp[len(s)] = True

        for i in range(len(s) - 1, -1, -1):
            for w in wordDict:
                if (i + len(w)) <= len(s) and s[i : i + len(w)] == w:
                    dp[i] = dp[i + len(w)]
                if dp[i]:
                    break

        return dp[0]