vector<vector<string>> findLadders(string start, string end, unordered_set<string> &dict){
dict.emplace(start);
dict.emplace(end);
unordered_map<string, vector<string>> from;
queue<string> que;
que.push(start);
dict.erase(start);
int endLevel = -1;
int level = 0;
while(!que.empty()){
if(endLevel != -1) break;
int size = que.size();
unordered_set<string> toPush;//this is acctually what the next level is
for(int ttt = 0; ttt < size; ttt++){
string front = que.front();
if(front == end){
endLevel = level;
break;
}
for(unsigned int i = 0; i < front.size(); i++){
string neighbor = front;
for(int j = 0; j < 26; j++){
neighbor[i] = (char)((int)'a' + j);
if(neighbor == front) continue;
if((dict.count(neighbor) != 0)){
from[neighbor].push_back(front);
toPush.emplace(neighbor);
}
}
}
que.pop();
}
level++;
for(unordered_set<string>::iterator itr = toPush.begin(); itr != toPush.end(); itr++){
string temp = *itr;
que.push(temp);
dict.erase(temp);
//every time after next level is determined, erase them from dict
//since they won't be at the next next level(we need to exclude them from the future check)
}
}
vector<vector<string>> result;
if(from[end].size() == 0) return result;
vector<string> tempRes;
tempRes.push_back(end);
help(result, tempRes, endLevel, start, end, from);
return result;
}
shared_ptr<HashTables::BNode> HashTables::buildCannonicTree_helper(shared_ptr<BSTNode<int> > &n,
unordered_set<shared_ptr<BNode>, HashBNode, EqualBNode> &table) {
if (!n) return nullptr;
shared_ptr<BNode> left = buildCannonicTree_helper(n->left, table);
shared_ptr<BNode> right = buildCannonicTree_helper(n->right, table);
shared_ptr<BNode> newNode(new BNode(n->data, left, right));
table.emplace(newNode);
return newNode;
}
void reverseDFS(size_t idx) {
if (visit[idx]) return;
visit[idx] = true;
for (auto next : revGraph[idx]) {
reverseDFS(next);
}
selectedWord.emplace(sccId[idx]);
sum++;
}
void addClockReceiver(shared_ptr<Clocked> receiver)
{
clock_receivers.emplace(receiver);
}
/**
* @param start, a string
* @param end, a string
* @param dict, a set of string
* @return a list of lists of string
*/
vector<vector<string>> findLadders(string start, string end, unordered_set<string> &dict) {
vector<vector<string>> ans;
/*
in the pair, the vector contains all immediate and legal children of the node, int is the order of the node
*/
unordered_map<string, std::pair<vector<string>, int>> g;
/*
Put end on the dict to make sure we have an ending point on the graph
*/
dict.emplace(end);
size_t len = start.length();
char old_char = 0;
string str, key;
queue<string> q;
q.push(start);
while (false == q.empty()) {
key = std::move(q.front()); q.pop();
if (key == end) break;
str = key;
/*
Since we use a huge structure to store graph info, g[key] should be created already when we set up its order during BFS.
This is only for the string start as the key.
BTW, for an existing key-value in unordered_map, try to emplace again will simply be ignored. In other words, newly given
value will NOT replace the existing one.
*/
if (start == key)
g.emplace(key, std::make_pair(vector<string>(), 0));
for (size_t i = 0; i < len; ++i) {
for (char j = 'a'; j <= 'z'; ++j) {
if (j == str[i])continue;
old_char = str[i];
str[i] = j;
if (dict.end() != dict.find(str)) {
/*
We will only process for the following two conditions:
1. Node str has not been visited yet during a BFS. This can be confirmed by g.find(str) == g.end();
2. Node str has been visited. However, the current key's order is one level above node str. In other
words, according to the transform rule and common sense of a "shortest path", node key to node str
is legit and str should be considered as key's legal child even though str has other parent(s).
*/
if (g.end() == g.find(str) || g[str].second == g[key].second + 1) {
/*
If g.find on str returns end, it means node str has not been visited yet
*/
if (g.end() == g.find(str)) {
/*
we should only push str to the queue if str hasn't be on the queue before.
otherwise, we will introduce duplications onto the neighor list of str.
*/
q.push(str);
g.emplace(str, std::make_pair(vector<string>(), g[key].second + 1));
}
/*
however, we will put str onto key's neighor list even str is accessed before for an obvious reason:
we need to find all possible routes even they may overlap in between (not completely though). In such
scenario, str has multiple parents.
*/
g[key].first.push_back(str);
}
}
str[i] = old_char;
}
}
}
vector<string> vec(dict.size() + 2);
this->aux(ans, vec, g, start, end, 0);
return ans;
}
vector<vector<string>> findLadders(string start, string end, unordered_set<string> &dict){
dict.emplace(start);
dict.emplace(end);
unordered_map<string, vector<string>> from;
queue<string> que;
que.push(start);
dict.erase(start);
int endLevel = -1;
int level = 0;
while(!que.empty()){
if(endLevel != -1) break;
int size = que.size();
unordered_set<string> toPush;//this is acctually what the next level is
for(int ttt = 0; ttt < size; ttt++){
string front = que.front();
if(front == end){
endLevel = level;
break;
}
for(unsigned int i = 0; i < front.size(); i++){
string neighbor = front;
for(int j = 0; j < 26; j++){
neighbor[i] = (char)((int)'a' + j);
if(neighbor == front) continue;
if((dict.count(neighbor) != 0)){
from[neighbor].push_back(front);
//very important trick here: keep track of where the neighbor
//is from. This will improve the backtrack process since all
//the string we trace back from end will lead us back to nothing but start,
//we can avoid a lot of unnecessary check here
toPush.emplace(neighbor);
}
}
}
que.pop();
}
/*
* Here is what I don't understand. I firstly use following codes to find neighbors the current string, "front",
* can "jump" to, by iterating the whole dict. I thought this should be faster than iterating chars in
* the current string since the iterating the whole dict will take O(dict.size() * each level's size * length of string)
* while above code takes O(26^length fo string) in time, which should be much greater than the first one.
* However, I failed with the following case. Can anyone explain the reason, please?
string front = que.front();
dict.erase(front);
if(front == end){
endLevel = level;
break;
}
else{
for(unordered_set<string>::iterator itr = dict.begin(); itr != dict.end(); itr++){
string neighbor = *itr;
if(help2(neighbor, front)){
from[neighbor].push_back(front);
toPush.emplace(neighbor);
}
}
}
que.pop();
}
*/
level++;
for(unordered_set<string>::iterator itr = toPush.begin(); itr != toPush.end(); itr++){
string temp = *itr;
que.push(temp);
dict.erase(temp);
//every time after next level is determined, erase them from dict
//since they won't be at the next next level(we need to exclude them from the future check)
}
}
vector<vector<string>> result;
if(from[end].size() == 0) return result;
vector<string> tempRes;
tempRes.push_back(end);
help(result, tempRes, endLevel, start, end, from);
return result;
}