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Word Break⚓︎

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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:

  • 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.

Solution⚓︎

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class Solution {
public:
    bool wordBreak(string s, vector<string>& wordDict) {
        unordered_set<string> wordSet(wordDict.begin(), wordDict.end());
        vector<bool> dp(s.size() + 1, false);
        dp[0] = true;
        for (int i = 1; i <= s.size(); i++) {
            for (int j = 0; j < i; j++) {
                if ((wordSet.find(s.substr(j, i - j)) != wordSet.end()) && dp[j])
                    dp[i] = true;
            }
        }
        return dp[s.size()];
    }
};