Runtime: 290 ms (Top 55.58%) | Memory: 13.9 MB (Top 85.26%)

Author: AnasImloul

scripts/algorithms/W/Word Ladder/Word Ladder.cpp

@@ -1,12 +1,13 @@
+// Runtime: 290 ms (Top 55.58%) | Memory: 13.9 MB (Top 85.26%)
 // For approach: ref: https://www.youtube.com/watch?v=ZVJ3asMoZ18&t=0s
 
-class Solution 
+class Solution
 {
 public:
-    int ladderLength(string beginWord, string endWord, vector<string>& wordList) 
+    int ladderLength(string beginWord, string endWord, vector<string>& wordList)
     {
         // 1. Create a set and insert all wordList to the set to have
-        //    easy search of O(1)
+        // easy search of O(1)
         unordered_set<string> wordSet;
         bool isEndWordPresent = false;
         for (string s:wordList)
@@ -20,34 +21,34 @@ class Solution
             }
             wordSet.insert(s);
         }
-        
+
         // 2. If endWord is not present in the wordList return
         if (!isEndWordPresent)
         {
             return 0;
         }
-        
+
         // 3. Create a queue for BST insert the elements of currentlevel to the queue
-        //    currentLevel is defined as all the strings that can be reached from 
-        //    all the strings of the previous string by just replacing one character
-        //    and the one-char-replaced-string is present in the wordSet / wordList
+        // currentLevel is defined as all the strings that can be reached from
+        // all the strings of the previous string by just replacing one character
+        // and the one-char-replaced-string is present in the wordSet / wordList
         queue<string> q;
         q.push(beginWord);
         int level = 0; // every time all the strings at this level are processed increment it
-        
+
         // 4. Loop through all the elements in the queue
         while (!q.empty())
         {
             level++;
             int numStringsInCurLevel = q.size();
-            // loop through all the strings in this current level 
+            // loop through all the strings in this current level
             // and find out if destination can be reached before
             // jumping to the next level of strings added to the queue
             while (numStringsInCurLevel--)
             {
                 string curStr = q.front();
                 q.pop();
-                
+
                 for (int i=0; i<curStr.length(); i++)
                 {
                     string tempStr = curStr;
@@ -76,8 +77,8 @@ class Solution
                     }
                 }
             }
-        }       
-        
+        }
+
         // If we are here then we couldn't reach destination
         // even though the destination string is present in the wordList
         // there is no path from source to destination for the given rules