[LeetCode September Challange]Day29-Word Break

Given a non-empty string s and a dictionary wordDict containing a list of non-empty words, determine if s can be segmented into a space-separated sequence of one or more dictionary words.

Note:

• The same word in the dictionary may be reused multiple times in the segmentation.
• You may assume the dictionary does not contain duplicate words.

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

---

Solution

Time complexity : O(n^2)
Space complexity : O(n)

class Solution {
public:
    bool wordBreak(string s, vector<string>& wordDict) {
        for (string word: wordDict) {
            m_[word] = true;
        }
        return helper(s);
    }
private:
    unordered_map<string, bool> m_;
    bool helper(const string& s) {
        if (m_.count(s)) return m_[s];
        
        // for each break point in s
        for (int i=1; i<s.length(); ++i) {
            auto it = m_.find(s.substr(i));
            if (it != m_.end() && it->second && helper(s.substr(0, i)))
                return m_[s] = true;
        }
        return m_[s] = false;
    }
};

memorized recursive.
記憶已算過的答案，避免重覆計算，加快速度。