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  1. #include <vector>
  2. #include <list>
  3. #include <map>
  4. #include <set>
  5. #include <deque>
  6. #include <stack>
  7. #include <bitset>
  8. #include <algorithm>
  9. #include <functional>
  10. #include <numeric>
  11. #include <utility>
  12. #include <sstream>
  13. #include <iostream>
  14. #include <iomanip>
  15. #include <cstdio>
  16. #include <cmath>
  17. #include <cstdlib>
  18. #include <ctime>
  19. #include <unordered_set>
  20. #define debug(a) { cout<< #a << "  " << a << " " ;}
  21. using namespace std;
  22.  
  23. bool onedist(const string &a, const string &b);
  24. int dfs(string &begin, string &end, unordered_set<string>& words, bool visited[],int &counting);
  25. int ladderLength(string begin, string end, unordered_set<string>& words);
  26.  
  27. // Function that returns true if the two strings are at an edit distance of 1 from each other
  28. bool onedist(const string &a, const string &b) {
  29.         int mismatch = 0;
  30.         for(int i = 0;i< a.size(); i++) {
  31.             if(a[i] == b[i])
  32.             {
  33.                 // do nothing
  34.             }
  35.             else mismatch++;
  36.  
  37.             if (mismatch > 1)
  38.                 return false;
  39.         }
  40.  
  41.         return true;
  42.     }
  43.  
  44.     int dfs(string &begin, string &end, vector<string>& wordlife, bool visited[],int counting) {
  45.        //cout << "begin" << begin <<endl;
  46.         if(onedist(begin,end)) {
  47.  
  48.             return counting+1; 
  49.             // Should exit, but doesn't. Causing run time error
  50.         }
  51.  
  52.         int n = wordlife.size();
  53.         int c=0,min=n+1;//min is used to find minimum possible transformations required
  54.         for(int i = 0;i<n;i++)
  55.         {
  56.             if(!visited[i] && onedist(begin, wordlife[i])) {
  57.             	visited[i] = true;
  58.                  c=dfs(wordlife[i],end,wordlife,visited,counting+1);//we store the value returned by dfs in c so that we can find minimum required transformation
  59.                  //cout << "Test flow" << c << endl;
  60.                  if(c<min)
  61.                  min=c;//to find minimum transformations required
  62.  
  63.             }
  64.  
  65.         }
  66.         if(c!=0)
  67.         return min; //return min possible transformations
  68.          else
  69.          return 0;//if no possible transformations are possible
  70.     }
  71.  
  72.     int ladderLength(string begin, string end, vector<string>& wordlife)
  73.     {
  74.  
  75.         int n = wordlife.size();
  76.         bool visited[n];
  77.         for(int i = 0;i<n;i++)
  78.             visited[i] = false;
  79.         int count = 1;
  80.  
  81.         int ans = dfs(begin,end,wordlife,visited,count);
  82.  
  83.         return ans;
  84.  
  85.  
  86.  
  87.     }
  88.  
  89.     int main()
  90.     {
  91.     	string begin = "hit";
  92.     	string end = "cog";
  93.     	vector<string> wordlife;
  94.     	wordlife.push_back("hot");
  95.     	wordlife.push_back("dot");
  96.     	wordlife.push_back("dog");
  97.     	wordlife.push_back("lot");
  98.     	wordlife.push_back("log");
  99.     	int a = ladderLength( begin,  end,  wordlife);
  100.     	cout <<a<<endl;
  101.     	return 0;//the run time error was because you were returning 'a' instead of 0
  102.     }
Success #stdin #stdout 0s 3416KB
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https://ideone.com/JBcECN
language:
C++14 (gcc 8.3)
created:
8 years ago
visibility:
public

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