# Word Break II Problem

## Description

LeetCode Problem 140.

Given a string s and a dictionary of strings wordDict, add spaces in s to construct a sentence where each word is a valid dictionary word. Return all such possible sentences in any order.

Note that the same word in the dictionary may be reused multiple times in the segmentation.

Example 1:

``````1
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Input: s = "catsanddog", wordDict = ["cat","cats","and","sand","dog"]
Output: ["cats and dog","cat sand dog"]
``````

Example 2:

``````1
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Input: s = "pineapplepenapple", wordDict = ["apple","pen","applepen","pine","pineapple"]
Output: ["pine apple pen apple","pineapple pen apple","pine applepen apple"]
Explanation: Note that you are allowed to reuse a dictionary word.
``````

Example 3:

``````1
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Input: s = "catsandog", wordDict = ["cats","dog","sand","and","cat"]
Output: []
``````

Constraints:

• 1 <= s.length <= 20
• 1 <= wordDict.length <= 1000
• 1 <= wordDict[i].length <= 10
• s and wordDict[i] consist of only lowercase English letters.
• All the strings of wordDict are unique.

## Sample C++ Code

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class Solution {
public:
struct TrieNode {
TrieNode* arr[26];
bool is_end;
TrieNode() {
for (int i = 0; i < 26; i ++) arr[i] = nullptr;
is_end = false;
}
};

void insert(TrieNode* root, string s) {
TrieNode* curr = root;
for (int i = 0; i < s.size(); i ++) {
if (curr->arr[s[i]-'a'] == nullptr)
curr->arr[s[i]-'a'] = new TrieNode();
curr = curr->arr[s[i]-'a'];
}
curr->is_end = true;
return;
}

vector<string> ans;

void dfs(string& currS, string& s, int idx, TrieNode* root) {
if (idx == s.size()) {
ans.push_back(currS.substr(0, currS.size()-1));
}
if (idx >= s.size())
return;

TrieNode* curr = root;
string tempS;
for (int i = idx; i < s.size(); i ++) {
tempS += s[i];
if (curr->arr[s[i]-'a'] == nullptr)
return;
curr = curr->arr[s[i]-'a'];
if (curr->is_end == true) {
currS = currS + tempS + ' ';
dfs(currS, s, i+1, root);
currS = currS.substr(0, currS.size()-tempS.size()-1);
}
}
}

vector<string> wordBreak(string s, vector<string>& wordDict) {
TrieNode* root = new TrieNode();
for (int i = 0; i < wordDict.size(); i ++) {
insert(root, wordDict[i]);
}

string currS;
dfs(currS, s, 0, root);
return ans;
}
};
``````