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 &str, const vector<string> &wordDict) {
const int n = str.size();
int min_word_length = numeric_limits<int>::max(), max_word_length = 0;
unordered_set<string> valid_words;
for (const string &word : wordDict) {
min_word_length = min(min_word_length, static_cast<int>(word.size()));
max_word_length = max(max_word_length, static_cast<int>(word.size()));
valid_words.insert(word);
}
vector<bool> formable(n + 1, false);
formable[0] = true;
for (int i = min_word_length; i <= n; ++i) {
for (int j = max(0, i - max_word_length); j <= i - min_word_length; ++j) {
if (formable[j]) {
string word = str.substr(j, i - j);
if (valid_words.contains(word)) {
formable[i] = true;
break;
}
}
}
}
return formable.back();
}
};