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[dp] [dfs] [word break] [string] Given a string s and a dictionary of words dict, determine if s can be segmented into a space-separated sequence of one or more dictionary words
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| // dp solution | |
| bool word_break(const string& s, unordered_set<string>& dict) { | |
| if(s.empty() || dict.empty()) return true; | |
| vector<bool> dp(s.size()+1, false); // <---- why define a 'dp' of length N+1? because dp[i] means whether a string of length i can be segmented using dict | |
| dp[0] = true; // cannot forget! | |
| for(int i=1; i<=s.size(); i++) { | |
| for(int j=0; j<i; j++) { // looping j from 0 to i-1 seems better | |
| dp[i] = dp[j] && dict.find(s.substr(j+1, i-j)) != dict.end(); | |
| if(dp[i] == true) break; | |
| } | |
| } | |
| return dp[s.size()]; | |
| } | |
| // dfs | |
| bool dfs(string s, unordered_set<string> &dict, int start_idx) { | |
| if(start_idx >= s.size()) return true; | |
| for(int j=start_idx; j<s.size(); j++) { | |
| string t = s.substr(start_idx, j-start_idx+1); | |
| if(dict.find(t) == dict.end()) continue; | |
| if(dfs(s, dict, j+1)) return true; | |
| } | |
| return false; | |
| } | |
| bool word_break(string s, unordered_set<string> &dict) { | |
| return dfs(s, dict, 0); | |
| } |
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