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// Source : https://oj.leetcode.com/problems/word-break-ii/
// Author : Hao Chen
// Date : 2014-07-03
/**********************************************************************************
*
* Given a string s and a dictionary of words dict, add spaces in s to construct a sentence
* where each word is a valid dictionary word.
*
* Return all such possible sentences.
*
* For example, given
* s = "catsanddog",
* dict = ["cat", "cats", "and", "sand", "dog"].
*
* A solution is ["cats and dog", "cat sand dog"].
*
*
**********************************************************************************/
#include <stdio.h>
#include <stdlib.h>
#include <time.h>
#include <iostream>
#include <string>
#include <vector>
#include <set>
#include <map>
using namespace std;
// ---------------
// Recursive Way
// ---------------
// The recursive method is quite straight forward.
//
// 1) if a substring from 0 to i is a word, then take the rest substring to evaluate.
// 2) during the recursion, keep tracking the result
//
// For example:
//
// s = "catsanddog",
// dict = ["cat", "cats", "and", "sand", "dog"].
//
//
// +---> sand / dog ---> dog
// |
// +-------> cat / sanddog
// |
// catsanddog
// |
// +------> cats / anddog
// |
// +----> and / dog ---> dog
//
//
// However, the recursive could produce a lot duplicated calculation, we need use a cache to avoid.
//
//To avoid time limit error, need to add cache
vector<string> wordBreak(string s, set<string> &dict, map<string, vector<string> >& cache) {
if (cache.find(s)!=cache.end()){
return cache[s];
}
vector<string> result;
for(int i=0; i<s.size(); i++){
string w = s.substr(0,i+1);
if (dict.find(w)!=dict.end()) {
if (i==s.size()-1){
result.push_back(w);
break;
}
vector<string> ret = wordBreak(s.substr(i+1, s.size()-i-1), dict, cache);
for(int j=0; j<ret.size(); j++){
result.push_back( w + " " + ret[j] );
}
}
}
cache[s] = result;
return result;
}
//Time limit error
void wordBreak(string s, set<string> &dict, string str, vector<string>& result) {
string org_str = str;
for(int i=0; i<s.size(); i++){
string w = s.substr(0,i+1);
// if the current substring is a word
if (dict.find(w)!=dict.end()) {
str = org_str;
if (str.size()>0){
str +=" ";
}
str = str + w;
// foud the solution, add it into the result
if (i==s.size()-1){
result.push_back(str);
return;
}
//recursively to solve the rest subarray
wordBreak(s.substr(i+1, s.size()-i-1), dict, str, result);
}
}
}
//---------------------
// Dynamic Programming
//---------------------
//
// Define substring[i, j] is the sub string from i to j.
//
// (substring[i,j] == word) : result[i] = substring[i,j] + {result[j]}
//
// So, it's better to evaluate it backword.
//
// For example:
//
// s = "catsanddog",
// dict = ["cat", "cats", "and", "sand", "dog"].
//
// 0 c "cat" -- word[0,2] + {result[3]} ==> "cat sand dog"
// "cats" -- word[0,3] + {result[4]} ==> "cats and dog"
// 1 a ""
// 2 t ""
// 3 s "sand" -- word[3,6] + {result[7]} ==> "sand dog"
// 4 a "and" -- word[4,6] + {result[7]} ==> "and dog"
// 5 n ""
// 6 d ""
// 7 d "dog"
// 8 o ""
// 9 g ""
vector<string> wordBreak_dp(string s, set<string> &dict) {
vector< vector<string> > result(s.size());
for(int i=s.size()-1; i>=0; i--) {
vector<string> v;
result[i] = v;
for(int j=i+1; j<=s.size(); j++) {
string word = s.substr(i, j-i);
if (dict.find(word) != dict.end()){
if (j==s.size()){
result[i].push_back(word);
}else{
for(int k=0; k<result[j].size(); k++){
result[i].push_back(word + " " + result[j][k]);
}
}
}
}
}
return result[0];
}
vector<string> wordBreak(string s, set<string> &dict) {
vector<string> result;
switch (random()%3)
{
case 0:
{
cout << "---------Recursive Solution--------" << endl;
string str;
wordBreak(s, dict, str, result);
}
break;
case 1:
{
cout << "----Memorized Recursive Solution----" << endl;
map<string, vector<string> > cache;
result = wordBreak(s, dict, cache);
}
break;
case 2:
cout << "----Dynamic Programming Solution----" << endl;
result = wordBreak_dp(s, dict);
break;
}
return result;
}
void printVector(vector<string>& v)
{
for(int i=0; i<v.size(); i++){
cout << v[i] <<endl;
}
}
int main()
{
srand(time(NULL));
string d[]={"cat", "cats", "and", "sand", "dog"};
set<string> dict(d, d+5);
string s = "catsanddog";
vector<string> v = wordBreak(s, dict);
printVector(v);
string d0[]={"apple","app","le","pie"};
dict.clear();
dict.insert(d0, d0+4);
s = "applepie";
v = wordBreak(s, dict);
printVector(v);
string d1[]={"aaaa","aa","a"};
dict.clear();
dict.insert(d1, d1+3);
s = "aaaaaaa";
v = wordBreak(s, dict);
printVector(v);
string d5[]={"a","aa","aaa","aaaa","aaaaa","aaaaaa","aaaaaaa","aaaaaaaa","aaaaaaaaa","aaaaaaaaaa"};
dict.clear();
dict.insert(d5, d5+10);
s="aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaab";
v = wordBreak(s, dict);
printVector(v);
return 0;
}