Word Ladder in C++

C++Server Side ProgrammingProgramming

Suppose we have two words (beginWord and endWord), and we have dictionary's word list, find the length of shortest transformation sequence from beginWord to endWord, such that −

  • Only one letter can be converted at a time.

  • In each transformed word must exist in the word list. The beginWord is not a transformed word.

We have to keep in mind that −

  • Return 0 when there is no such change sequence.

  • All words have the same length.

  • All words contain only lowercase characters.

  • We can assume no duplicates in the word list.

So if the input is like: beginWord = "hit", endWord = "cog", and wordlist = ["hot", "dot", "dog", "lot", "log", "cog"]

Then the output will be 5, as one shortest transformation is hit → hot → dot → dog → cog

To solve this, we will follow these steps −

  • Define a method called putStar, this will take j and string s. This will work as follows −

  • temp := empty string

  • for i in range 0 to size of s – 1

    • if i = j, then update temp by concatenating “*” with it, otherwise update temp by concatenating s[i] with temp.

  • in the main method it will take string b, string e and list of words w, this will work like −

  • if e is not in w, or b is empty, or e is empty, or w is empty, then return 0

  • define a map m for string type key and array type value.

  • for i in range 0 to size of w

    • x := w[i]

    • for j := 0 to size of x

      • inter := putStar(j, x)

      • insert x into m[inter]

  • Define a queue q, insert a pair (b, 1) into q

  • make a map called visited

  • while q is not empty

    • s := front pair from q, delete front element from q

    • x := first element of the pair s, l := second element of the pair s

    • for i in range 0 to size of x

      • temp := putStar(i, x)

      • for j in range 0 to size of m[temp]

        • aa := m[temp, j]

        • if aa is same as e, then return l + 1

        • if visited[aa] is not set, then insert pair (aa, l + 1), and set visited[aa] = 1

  • level := 0

  • return 0

Let us see the following implementation to get better understanding −

Example (C++)

class Solution {
public:
   string putStar(int j, string s){
      string temp = "";
      for(int i = 0; i < s.size(); i++){
         if(i == j)temp += "*";
            else temp += s[i];
         }
      return temp;
   }
   int ladderLength(string b, string e, vector<string>& w) {
      if(find(w.begin(), w.end(), e) == w.end() || !b.size() || !e.size() || !w.size())return 0;
         map < string , vector <string> > m;
      for(int i = 0; i < w.size(); i++){
         string x = w[i];
         for(int j = 0; j < x.size(); j++){
            string inter = putStar(j,x);
            m[inter].push_back(x);
         }
      }
      queue < pair <string, int> > q;
      q.push({b, 1});
      map <string, int> visited;
      while(!q.empty()){
         pair < string, int > s = q.front();
         q.pop();
         string x = s.first;
         int l = s.second;
         for(int i = 0; i < x.size(); i++){
            string temp = putStar(i ,x);
            for(int j = 0; j < m[temp].size(); j++){
               string aa = m[temp][j];
               if(aa == e)return l+1;
                  if(!visited[aa]){
                     q.push({aa, l+1});
                     visited[aa] = 1;
                  }
               }
            }
         }
         int level = 0;
      return 0;
   }
};

Input

"hit"
"cog"
["hot","dot","dog","lot","log","cog"]

Output

5
raja
Published on 17-Mar-2020 05:40:32
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