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concatenatedWords.cpp
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concatenatedWords.cpp
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// TRIE
class TrieNode
{
public:
TrieNode *link[26];
bool is_end;
TrieNode()
{
for (int i = 0; i < 26; i++)
{
link[i] = NULL;
}
is_end = false;
}
static TrieNode *insert(TrieNode *root, string str)
{
if (root == NULL)
{
root = new TrieNode;
}
TrieNode *temp = root;
int i = 0;
while (i < str.size())
{
if (temp->link[str[i] - 'a'] == NULL)
{
temp->link[str[i] - 'a'] = new TrieNode;
}
temp = temp->link[str[i] - 'a'];
i++;
}
temp->is_end = true;
return root;
}
static void mark(TrieNode *root, string str, vector<bool> &dp, int l, int r)
{
if (root == NULL)
{
return;
}
if (root->is_end)
{
dp[l] = true;
}
if (l == r)
{
return;
}
if (root->link[str[l] - 'a'] != NULL)
{
mark(root->link[str[l] - 'a'], str, dp, l + 1, r);
}
}
};
class Solution
{
public:
// sort on the basis of string length
static bool sorter(string a, string b)
{
return a.size() < b.size();
}
vector<string> findAllConcatenatedWordsInADict(vector<string> &words)
{
vector<string> ans;
sort(words.begin(), words.end(), sorter);
TrieNode *root = NULL;
vector<bool> singles(26, false); // this is for the hack
for (int i = 0; i < words.size(); i++)
{
if (words[i] == "")
{
continue;
}
if (words[i].size() == 1)
{
// this can never be the answer, but is a good building block
singles[words[i][0] - 'a'] = true;
root = TrieNode::insert(root, words[i]);
continue;
}
// check if the current string can be made using our single building blocks
bool check = true;
for (int j = 0; j < words[i].size(); j++)
{
if (!singles[words[i][j] - 'a'])
{
check = false;
break;
}
}
if (check)
{
ans.push_back(words[i]);
continue;
}
// our vanilla trie with dp solution
vector<bool> dp(words[i].size() + 1, false);
dp[0] = true;
TrieNode::mark(root, words[i], dp, 0, words[i].size());
for (int j = 1; j < words[i].size(); j++)
{
if (dp[j])
{
TrieNode::mark(root, words[i], dp, j, words[i].size());
}
}
if (dp[words[i].size()])
{
ans.push_back(words[i]);
}
else
{
root = TrieNode::insert(root, words[i]);
}
}
return ans;
}
};
// unordered_sets
class Solution
{
public:
vector<string> findAllConcatenatedWordsInADict(vector<string> &words)
{
unordered_set<string> s(words.begin(), words.end());
vector<string> res;
for (auto w : words)
{
int n = w.size();
vector<int> dp(n + 1);
dp[0] = 1;
for (int i = 0; i < n; i++)
{
if (dp[i] == 0)
continue;
for (int j = i + 1; j <= n; j++)
{
if (j - i < n && s.count(w.substr(i, j - i)))
dp[j] = 1;
}
if (dp[n])
{
res.push_back(w);
break;
}
}
}
return res;
}
};