feat : word-dictionary

Author: ekgns33

design-add-and-search-words-data-structure/ekgns33.java

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+/**
+ design ds that supports add, find
+ find if string matches any previously added string.
+ >> some kind of dictionary
+
+ example
+ add bad
+ add dad
+ add mad
+ search pad >> false
+ search bad >> true
+ search .ad >> true . can be 'b' or 'd' or 'm'
+ search b.. >> true .. can be "ad"
+
+ constraints:
+ 1) empty string can be valid input?
+ nope. lenght of string is in range of [1, 25]
+ 2) string only contiains alphanumeric? or alphabet
+ only lowercase English letters
+ 3) how many queries can be given as input?
+ at most 10^4
+
+ solution 1) brute force
+ save to data structure. for add
+ check every character for all the saved words
+ O(kn) when k is the number of saved words,
+ n is the length of input word token
+ 25 * 25 * 10^4
+ tc : O(kn) + O(kn)
+ add : O(1)
+ search : O(kn)
+ sc : O(kn)
+
+ solution 2) trie?
+
+ save words to trie.
+ when searching
+ do bfs or backtracking dfs
+ if current charcter is . add all childs
+ else add matching childs
+ overall tc : O(sum(m)) + O(n),
+ add : O(m), when m is the length of saved word
+ search : O(n), when n is the length of searh word
+ sc : O(sum(m))
+
+ if volume of search query is much bigger than add query
+ trie solution would be better
+ O(n) search time vs O(mn) search time
+ */
+class WordDictionary {
+  class TrieNode {
+    TrieNode[] childs;
+    boolean isEnd;
+    TrieNode() {
+      childs = new TrieNode[26];
+      isEnd = false;
+    }
+  }
+  TrieNode root;
+  public WordDictionary() {
+    root = new TrieNode();
+  }
+
+  public void addWord(String word) {
+    TrieNode curNode = root;
+    for(char c : word.toCharArray()) {
+      if(curNode.childs[c-'a'] == null) {
+        curNode.childs[c-'a'] = new TrieNode();
+      }
+      curNode = curNode.childs[c-'a'];
+    }
+    curNode.isEnd = true;
+  }
+
+  public boolean search(String word) {
+    return dfsHelper(root, word, 0);
+  }
+
+  public boolean dfsHelper(TrieNode node, String word, int p) {
+    //end clause
+    if(p == word.length()) {
+      return node.isEnd;
+    }
+
+    char curC = word.charAt(p);
+    if(curC == '.') {
+      for(TrieNode next : node.childs) {
+        if(next != null) {
+          if(dfsHelper(next, word, p + 1)) return true;
+        }
+      }
+      return false;
+    } else {
+      if(node.childs[curC-'a'] != null) {
+        if(dfsHelper(node.childs[curC - 'a'], word, p + 1)) return true;
+      }
+      return false;
+    }
+
+
+  }
+}
+
+/**
+ * Your WordDictionary object will be instantiated and called as such:
+ * WordDictionary obj = new WordDictionary();
+ * obj.addWord(word);
+ * boolean param_2 = obj.search(word);
+ */