LeetCode-in-All

139. Word Break

Medium

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

Note that the same word in the dictionary may be reused multiple times in the segmentation.

Example 1:

Input: s = “leetcode”, wordDict = [“leet”,”code”]

Output: true

Explanation: Return true because “leetcode” can be segmented as “leet code”.

Example 2:

Input: s = “applepenapple”, wordDict = [“apple”,”pen”]

Output: true

Explanation: Return true because “applepenapple” can be segmented as “apple pen apple”. Note that you are allowed to reuse a dictionary word.

Example 3:

Input: s = “catsandog”, wordDict = [“cats”,”dog”,”sand”,”and”,”cat”]

Output: false

Constraints:

• 1 <= s.length <= 300
• 1 <= wordDict.length <= 1000
• 1 <= wordDict[i].length <= 20
• s and wordDict[i] consist of only lowercase English letters.
• All the strings of wordDict are unique.

To solve this problem with the Solution class, we can use dynamic programming to efficiently determine whether the string s can be segmented into words from the dictionary wordDict. Here are the steps:

1. Define the Solution class with a method wordBreak that takes a string s and a list of strings wordDict as input and returns a boolean indicating whether s can be segmented into words from wordDict.
2. Initialize a boolean list dp of size len(s) + 1, where dp[i] represents whether the substring s[:i] can be segmented into words from wordDict.
3. Set dp[0] to True since an empty string can always be segmented.
4. Iterate i from 1 to len(s), and for each index i, iterate j from 0 to i.
5. Check if the substring s[j:i] exists in wordDict and if dp[j] is True. If both conditions are met, set dp[i] to True.
6. After completing the iterations, return dp[len(s)], which represents whether the entire string s can be segmented into words from wordDict.

Here’s how the Solution class would look like in Python:

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

# Example usage:
solution = Solution()
print(solution.wordBreak("leetcode", ["leet", "code"]))         # Output: True
print(solution.wordBreak("applepenapple", ["apple", "pen"]))    # Output: True
print(solution.wordBreak("catsandog", ["cats", "dog", "sand", "and", "cat"]))  # Output: False

This solution effectively utilizes dynamic programming to determine whether the string s can be segmented into words from the dictionary wordDict, satisfying the constraints provided in the problem statement.

Solution

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

        def solve(i, s):
            if i < 0:
                return True
            if dp[i] != -1:
                return dp[i] == 1
            for w in wordDict:
                sz = len(w)
                if i - sz + 1 >= 0 and s[i - sz + 1 : i + 1] == w:
                    if solve(i - sz, s):
                        dp[i] = 1
                        return True
            dp[i] = 0
            return False

        return solve(n - 1, s)

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