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> &word_dict) {
const int n = str.size();
vector<bool> formable(n + 1, false);
formable[0] = true;
unordered_set<string> words;
unordered_set<int> lengths;
size_t min_len = numeric_limits<size_t>::max(), max_len = 0;
for (const string &word : word_dict) {
min_len = min(min_len, word.size());
max_len = max(max_len, word.size());
lengths.insert(word.size());
words.insert(word);
}
for (int i = min_len; i <= n; ++i) {
for (auto j = lengths.begin(); j != lengths.end(); ++j) {
int start = i - *j;
if (start < 0) {
continue;
}
if (formable[start]) {
string s = str.substr(start, *j);
if (words.contains(s)) {
formable[i] = true;
break;
}
}
}
}
return formable.back();
}
};