adding longest common prefix problem

lectures/27_hybrid_data_structures/README.md

@@ -100,9 +100,8 @@ for the key (or NULL or a special value, e.g., ’-1’, if the path to that poi
 
 - What is the worst case # of children for a single node? What are the member variables for the Node class?
 - Unlike a hash table, we can iterate over the keys in a trie / prefix tree in sorted order.
-Exercise: Implement the trie sorted-order iterator (in code or pseudocode) and print the table on the right.
 
 ## 27.8 Leetcode Exercises
 
 - [Leetcode problem 208: Implement Trie (Prefix Tree)](https://leetcode.com/problems/implement-trie-prefix-tree/). Solution: [p208_trie.cpp](../../leetcode/p208_trie.cpp).
-
+- [Leetcode problem 14: Longest Common Prefix](https://leetcode.com/problems/longest-common-prefix/). Solution: [p14_longest_common_prefix.cpp](../../leetcode/p14_longest_common_prefix.cpp).

leetcode/p14_longest_common_prefix.cpp

@@ -0,0 +1,91 @@
+class TrieNode {
+public:
+    // initializing one node
+    TrieNode(){
+        count = 0;
+        // each node can have at most 26 children: a to z.
+        // and when accessing child, always remember do that -'a'
+        for(int i=0;i<26;i++){
+            children.push_back(nullptr);
+        }
+    }
+    // so that we can access private member variables.
+    friend class Trie;
+// private:
+    // count how many words use this node
+    int count;
+    std::vector<TrieNode*> children;
+};
+
+class Trie {
+public:
+    // initializing the prefix tree
+    Trie() {
+        // which basically is initializing the root
+        // and we guarantee that the root always exists
+        root = new TrieNode;
+    }
+    
+    void insert(string word) {
+        int len = word.length();
+        TrieNode* current = root;
+        for(int i=0; i<len; i++){
+            // inserting a node at the branch of [word[i]-'a'], as the children of current
+            // if no one has inserted here yet
+            if(current->children[word[i]-'a'] == nullptr){
+                // create a node
+                current->children[word[i]-'a'] = new TrieNode;
+                current->children[word[i]-'a']->count++;
+                // after the insertion, update current
+                current = current->children[word[i]-'a'];
+            }else{
+                // else, it means this node is already there, just update current
+                // we make the code duplicated here, but this way it's easier to understand.
+                current->children[word[i]-'a']->count++;
+                current = current->children[word[i]-'a'];   
+            }
+        }
+    }
+    
+// private:
+    // just like every tree, we should have a root.
+    TrieNode* root;
+};
+
+/**
+ * Your Trie object will be instantiated and called as such:
+ * Trie* obj = new Trie();
+ * obj->insert(word);
+ * bool param_2 = obj->search(word);
+ * bool param_3 = obj->startsWith(prefix);
+ */
+
+class Solution {
+public:
+    string longestCommonPrefix(vector<string>& strs) {
+        int size = strs.size();
+        std::string result;
+        Trie trie;
+        TrieNode* current = trie.root;
+        // step 1, insert every word in trie
+        for(int i=0;i<size;i++){
+            trie.insert(strs[i]);
+        }
+        // step 2, search prefix
+        int len = strs[0].length();
+        // the common prefix can never be longer than the length of strs[0].
+        for(int i=0;i<len;i++){
+            if(current && (current->children[strs[0][i]-'a']!=nullptr)){
+                // std::cout << current->children[strs[0][i]-'a']->count << ":" << size << std::endl;
+            }
+            if(current && (current->children[strs[0][i]-'a']!=nullptr) && current->children[strs[0][i]-'a']->count == size){
+                result.push_back(strs[0][i]);
+                // once it's pushed back, update current so as to move on to the next level.
+                current = current->children[strs[0][i]-'a'];
+            }else{
+                break;
+            }
+        }
+        return result;
+    }
+};