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

m = s.size(), n = wordDict.size()
Time complexity: O($m^2$)
Space complexity: O(m + n)

class Solution {
public:
    bool wordBreak(const string &s, const vector<string> &wordDict) {
        unordered_set<string> dictionary;
        int min_word_length = numeric_limits<int>::max(), max_word_length = 0;
        for (const string &word : wordDict) {
            dictionary.insert(word);

            min_word_length = min(int(word.size()), min_word_length);
            max_word_length = max(int(word.size()), max_word_length);
        }

        const int n = s.size();
        vector<bool> valid_break(n + 1, false);
        valid_break.front() = true;

        for (int i = min_word_length; i <= n; ++i) {
            int left = max(0, i - max_word_length);
            int right = i - min_word_length;

            for (int j = right; j >= left; --j) {
                if (valid_break[j] && dictionary.find(s.substr(j, i - j)) != dictionary.end()) {
                    valid_break[i] = true;
                    break;
                }
            }
        }

        return valid_break.back();
    }
};