139. Word Break

1. Description

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.

2. Example

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

3. 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.

4. Solutions

Dynamic Programming

n = s.size()
Time complexity: O($n^2$)
Space complexity: O(n)

class Solution {
public:
    bool wordBreak(const string &s, const vector<string> &wordDict) {
        unordered_set<string> words;
        int max_word_length = 0;
        for (const auto & word : wordDict) {
            words.emplace(word);
            max_word_length = max((int)word.size(), max_word_length);
        }

        vector<bool> dp(s.size() + 1); // dp[i] means if we can construct word whose length is i with the dict
        dp[0] = true;
        for (int i = 1; i <= s.size(); ++i) {
            for (int j = max(0, i - max_word_length); j < i; ++j) {
                if (dp[j] && words.find(s.substr(j, i - j)) != words.end()) {
                    dp[i] = true;
                    break;
                }
            }
        }

        return dp.back();
    }
};