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// Source : https://oj.leetcode.com/problems/word-ladder-ii/
// Author : Hao Chen
// Date : 2014-10-13
/**********************************************************************************
*
* Given two words (start and end), and a dictionary, find all shortest transformation
* sequence(s) from start to end, such that:
*
* Only one letter can be changed at a time
* Each intermediate word must exist in the dictionary
*
* For example,
*
* Given:
* start = "hit"
* end = "cog"
* dict = ["hot","dot","dog","lot","log"]
*
* Return
*
* [
* ["hit","hot","dot","dog","cog"],
* ["hit","hot","lot","log","cog"]
* ]
*
* Note:
*
* All words have the same length.
* All words contain only lowercase alphabetic characters.
*
*
**********************************************************************************/
#include <iostream>
#include <vector>
#include <map>
#include <queue>
#include <unordered_set>
using namespace std;
// Solution
//
// 1) Using BSF algorithm build a tree like below
// 2) Using DSF to parse the tree to the transformation path.
//
// For example:
//
// start = "hit"
// end = "cog"
// dict = ["hot","dot","dog","lot","log","dit","hig", "dig"]
//
// +-----+
// +-------------+ hit +--------------+
// | +--+--+ |
// | | |
// +--v--+ +--v--+ +--v--+
// | dit | +-----+ hot +---+ | hig |
// +--+--+ | +-----+ | +--+--+
// | | | |
// | +--v--+ +--v--+ +--v--+
// +----> dot | | lot | | dig |
// +--+--+ +--+--+ +--+--+
// | | |
// +--v--+ +--v--+ |
// +----> dog | | log | |
// | +--+--+ +--+--+ |
// | | | |
// | | +--v--+ | |
// | +--->| cog |<-- + |
// | +-----+ |
// | |
// | |
// +----------------------------------+
map< string, unordered_set<string> >&
buildTree(string& start, string& end, unordered_set<string> &dict) {
static map< string, unordered_set<string> > parents;
parents.clear();
unordered_set<string> level[3];
unordered_set<string> *previousLevel = &level[0];
unordered_set<string> *currentLevel = &level[1];
unordered_set<string> *newLevel = &level[2];
unordered_set<string> *p =NULL;
currentLevel->insert(start);
bool reachEnd = false;
while( !reachEnd ) {
newLevel->clear();
for(auto it=currentLevel->begin(); it!=currentLevel->end(); it++) {
for (int i=0; i<it->size(); i++) {
string newWord = *it;
for(char c='a'; c<='z'; c++){
newWord[i] = c;
if (newWord == end){
reachEnd = true;
parents[*it].insert(end);
continue;
}
if ( dict.count(newWord)==0 || currentLevel->count(newWord)>0 || previousLevel->count(newWord)>0 ) {
continue;
}
newLevel->insert(newWord);
//parents[newWord].insert(*it);
parents[*it].insert(newWord);
}
}
}
if (newLevel->empty()) break;
p = previousLevel;
previousLevel = currentLevel;
currentLevel = newLevel;
newLevel = p;
}
if ( !reachEnd ) {
parents.clear();
}
return parents;
}
void generatePath( string start, string end,
map< string, unordered_set<string> > &parents,
vector<string> path,
vector< vector<string> > &paths) {
if (parents.find(start) == parents.end()){
if (start == end){
paths.push_back(path);
}
return;
}
for(auto it=parents[start].begin(); it!=parents[start].end(); it++){
path.push_back(*it);
generatePath(*it, end, parents, path, paths);
path.pop_back();
}
}
vector< vector<string> >
findLadders(string start, string end, unordered_set<string> &dict) {
vector< vector<string> > ladders;
vector<string> ladder;
ladder.push_back(start);
if (start == end){
ladder.push_back(end);
ladders.push_back(ladder);
return ladders;
}
map< string, unordered_set<string> >& parents = buildTree(start, end, dict);
if (parents.size()<=0) {
return ladders;
}
generatePath(start, end, parents, ladder, ladders);
return ladders;
}
void printLadders(vector< vector<string> > &ladders){
int i, j;
for (i=0; i<ladders.size(); i++){
for (j=0; j<ladders[i].size()-1; j++){
cout << ladders[i][j] << " -> ";
}
cout << ladders[i][j] << endl;
}
}
int main(int argc, char** argv)
{
string start = "hit";
string end = "cog";
//unordered_set<string> dict ({"hot","dot","dog","lot","log"});
unordered_set<string> dict ({"bot","cig", "cog", "dit", "dut", "hot", "hit" ,"dot","dog","lot","log"});
vector< vector<string> > ladders;
ladders = findLadders(start, end, dict);
printLadders(ladders);
return 0;
}