leetcode

Author: Ashish-kumar7

leetcode-com-explore-challenge-card-august-leetcoding-challenge-549-week-1-august-1st-august-7th-3413.cpp

@@ -0,0 +1,71 @@
+class TrieNode{
+  public:
+    bool isEnd;
+	vector<TrieNode*> dict;
+    TrieNode(){
+        isEnd = false;
+        dict.resize(26,nullptr);
+    }
+};
+class WordDictionary {
+public:
+    /** Initialize your data structure here. */
+    TrieNode* root;
+    WordDictionary() {
+        root = new TrieNode();
+    }
+    
+    /** Adds a word into the data structure. */
+	// Addition of word in the trie is fairly simple.
+	// If you are not able to understand it, then solve the below question first :
+	// https://leetcode.com/problems/implement-trie-prefix-tree/
+    void addWord(string word) {
+        TrieNode* itr = root;
+        for(int i=0;i<word.length();i++){
+            if(itr->dict[word[i]-'a']==nullptr){
+                itr->dict[word[i]-'a'] = new TrieNode();
+            }
+            itr = itr->dict[word[i]-'a'];
+        }
+        itr->isEnd = true;
+    }
+
+
+	// Recursive function for searching the string in the trie.
+    bool func(TrieNode* root, string word, int pos){
+		// When pos becomes equal to length of the word i.e. we are at the end of the string.
+        if(word.length()==pos){
+			// If the string is present in the trie then "root" will be at the end of the trie.
+            if(root->isEnd)
+                return true;
+            return false;
+        }
+		// If the character is not '.' then it is fairly simple.
+		// We just need to check whether the word[pos] is present in the trie or not.
+        if(word[pos]!='.'){
+            if(root->dict[word[pos]-'a']!=nullptr)
+                return func(root->dict[word[pos]-'a'], word,pos+1);
+            else
+                return false;
+        }
+        else{
+			// Now, if we encounter '.' then we need to check every possible alphabet i.e. from a-z
+			// If we find any alphabet which is present in the trie,
+			// then we have to make the recursive call to the function with now --> pos + 1.
+			// If we encounter null node then we will continue and check the rest of the alphabets.
+            for(int i=0;i<26;i++){
+                if(root->dict[i]==nullptr)
+                    continue;
+                else{
+                    if(func(root->dict[i],word,pos+1))
+                        return true;
+                }
+            }
+        }
+		// If after checking everything, we don't find the alphabet in the trie ->
+        return false;
+    }
+    bool search(string word) {
+        return func(root, word, 0);
+    }
+};