diff --git a/README.md b/README.md
index 23aaf3c..991e485 100644
--- a/README.md
+++ b/README.md
@@ -6,7 +6,7 @@
 
 | Current Status|     Stats     |
 | :------------: | :----------: |
-| Total Problems | 186 |
+| Total Problems | 187 |
 
 </center>
 
@@ -219,6 +219,7 @@ Include contains single header implementation of data structures and some algori
 | :------------ | :----------: |
 | You are given a digit string (e.g "1234", "567" etc), provide all possible letter combinations we could generate from this digit string, based on the mapping we see on the telphone/mobile dialpad. If you have typed SMS in old style phones, you would know. For e.g. "1" is mapped to "abc", 2 is mapped to "def". You can refer to the <a href="http://techotv.com/wp-content/uploads/2013/03/Lumia-620-dial-pad-screen-1.jpg" target="_blank">image.</a>. <ul><li> Example: "34" will give output: {"dg","dh","di","eg","eh","ei","fg","fh","fi"} </li></ul> Note that order does not matter in result set.|[dialpad_combinations.cpp](backtracking_problems/dialpad_combinations.cpp)|
 | Implement wildcard pattern maching with support for '?' & '*'. <ul><li>'?' Matches any single character.</li></ul><ul><li>'*' Matches any sequence of character.</li></ul>. Checkout examples in file for more details.|[wild_card_matching.cpp](backtracking_problems/wild_card_matching.cpp)|
+| Given a 2D board and list of words from a dictionary, find all the possible words on board fromt the list. (Check example in the solution)| [word_search.cpp](backtracking_problems/word_search.cpp)|
 
 
 
diff --git a/backtracking_problems/word_search.cpp b/backtracking_problems/word_search.cpp
new file mode 100644
index 0000000..9b51627
--- /dev/null
+++ b/backtracking_problems/word_search.cpp
@@ -0,0 +1,142 @@
+/*
+ * Given a 2D board and list of words from a dictionary,
+ * find all the possible words on board fromt the list.
+ * Example:
+ * words = ["oath","pea","eat","rain"]
+ * board = 
+ * [
+ *  ['o','a','a','n'],
+ *  ['e','t','a','e'],
+ *  ['i','h','k','r'],
+ *  ['i','f','l','v']
+ * ]
+ * 
+ * Output: ["eat", "oath"]
+ * 
+ * Source : LeetCode Problem 212
+ * 
+ * Approach:
+ * 
+ * We can use backtracking(dfs) to traverse from each character on board
+ * to all four direction. In order to do a fast searching, we can use
+ * a trie from words in dictionary, as soon as we find a word from dictionary
+ * in trie, we remove it from trie, so we don't have duplicates.
+ * Also, we can use the board to mark '#' as visited character to 
+ * achieve backtracking.
+ */
+
+#include <vector>
+#include <iostream>
+
+struct TrieNode {
+    std::string word;
+    std::vector<TrieNode*> children;
+    TrieNode():
+        word{""}, children{std::vector<TrieNode*>(26, nullptr)}{}
+};
+
+TrieNode* build_trie(const std::vector<std::string>& dictionary)
+{
+    TrieNode* root = new TrieNode();
+    for (auto word: dictionary) {
+        TrieNode* curr = root;
+        for (auto c: word) {
+            int index = c - 'a';
+            if (curr->children[index] == nullptr) {
+                curr->children[index] = new TrieNode();
+            }
+            curr = curr->children[index];
+        }
+        curr->word = word;
+    }
+    return root;
+}
+
+void backtrack(std::vector<std::vector<char>>& board, 
+    TrieNode* root, int i, int j, std::vector<std::string>& result)
+{
+    char c = board[i][j];
+    if (c == '#' || root == nullptr) {
+        return;
+    }
+    int index = c - 'a';
+    TrieNode* curr = root->children[index];
+    if (curr == nullptr) {
+        return;
+    }
+
+    if (curr->word != "") {
+        result.push_back(curr->word);
+        // Reset to avoid duplicates.
+        curr->word = "";
+    }
+
+    board[i][j] = '#';
+    if (i > 0) {
+        backtrack(board, curr, i - 1, j, result);
+    }
+    if (j > 0) {
+        backtrack(board, curr, i, j - 1, result);
+    }
+    if (i < board.size() - 1) {
+        backtrack(board, curr, i + 1, j, result);
+    }
+    if (j < board[0].size() - 1) {
+        backtrack(board, curr, i, j + 1, result);
+    }
+    board[i][j] = c;
+}
+
+std::vector<std::string> find_words(std::vector<std::vector<char>>& board,
+    std::vector<std::string>& dictionary)
+{
+    std::vector<std::string> result;
+    for (int i = 0; i < board.size(); ++i) {
+        for (int j = 0; j < board[0].size(); ++j) {
+            TrieNode* root = build_trie(dictionary);
+            backtrack(board, root, i, j, result);
+        }
+    }
+    return result;
+}
+
+template <typename T>
+void print_vector(const std::vector<T>& vec)
+{
+    std::cout << "{ ";
+    for (auto t : vec) {
+        std::cout << t << " ";
+    }
+    std::cout << " }" << std::endl;
+}
+
+void print_board(const std::vector<std::vector<char>>& board)
+{
+    for (auto v: board) {
+        print_vector(v);
+    }
+}
+
+
+int main()
+{
+    std::vector<std::vector<char>> board = {
+        {'o','a','a','n'},
+        {'e','t','a','e'},
+        {'i','h','k','r'},
+        {'i','f','l','v'}
+    };
+
+    std::vector<std::string> dictionary = {
+        "oath","pea","eat","rain"
+    };
+
+    std::vector<std::string> result = find_words(board, dictionary);
+    std::cout << "Board:" << std::endl;
+    print_board(board);
+    std::cout << "Dictionary:" << std::endl;
+    print_vector(dictionary);
+    std::cout << "Result:" << std::endl;
+    print_vector(result);
+    return 0;
+}