139. Word Break

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139. Word Break

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.

Example 1:

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Input: s = "leetcode", wordDict = ["leet","code"]
Output: true
Explanation: Return true because "leetcode" can be segmented as "leet code".

Example 2:

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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:

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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 .

Hints/Notes

Solution

Language: C++

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class Solution {
public:
    string s_;
    int minLen = INT_MAX, maxLen = 0;
    unordered_set<string> words;
    map<int, bool> m;

    bool wordBreak(string s, vector<string>& wordDict) {
        for (auto& word : wordDict) {
            words.insert(word);
            maxLen = max(maxLen, (int)word.size());
            minLen = min(minLen, (int)word.size());
        }
        s_ = s;
        bool found = traverse(0);
        return found;
    }

    bool traverse(int index) {
        if (index == s_.size()) {
            return true;
        }

        if (m.contains(index)) {
            return m[index];
        }

        bool res = false;
        for (int i = maxLen; i >= minLen; i--) {
            if (i + index > s_.size()) {
                continue;
            }
            string tmp = s_.substr(index, i);
            if (words.contains(tmp)) {
                res |= traverse(index + i);
                if (res) {
                    break;
                }
            }
        }
        m[index] = res;
        return res;
    }
};