[LeetCode] 135: Word Break

[Problem]

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.

For example, given

s = “leetcode”,

dict = [“leet”, “code”].

Return true because “leetcode” can be segmented as “leet code”.

[Solution]

class Solution {
public:
bool wordBreak(string s, unordered_set<string> &dict) {
// Note: The Solution object is instantiated only once and is reused by each test case.
if(s.size() == 0)return true;

// dp
int len = s.size();
bool dp[len];
memset(dp, 0, sizeof(dp));
vector<pair<int, int> > next;
string word;
for(int i = len - 1; i >= 0; --i){
// no word is matched right now
if(next.size() == 0){
word = s.substr(i, len - i);
if(dict.find(word) != dict.end()){
dp[i] = true;
next.insert(next.begin(), pair<int, int>(i, len-1));
}
}
// some words are matched
else{
pair<int, int> first(-1, -1), second(-1, -1);
for(int j = 0; j < next.size(); ++j){
// match the word [i : next[j].first-1]
word = s.substr(i, next[j].first - i);
if(dict.find(word) != dict.end()){
dp[i] = true;
first = pair<int, int>(i, next[j].first - 1);
}

// match the word [i : next[j].second]
word = s.substr(i, next[j].second + 1 - i);
if(dict.find(word) != dict.end()){
dp[i] = true;
second = pair<int, int>(i, next[j].second);
}
}

// update next
if(first.first != -1){
next.insert(next.begin(), first);
}
if(second.first != -1){
next.insert(next.begin(), second);
}
}
}

return dp[0];
}
};