diff --git a/AlienDictionary.java b/AlienDictionary.java
new file mode 100644
index 0000000..c6e59d4
--- /dev/null
+++ b/AlienDictionary.java
@@ -0,0 +1,117 @@
+
+// In this problem, we have to identify that if the claim that words are sorted in alien order is correct or not. If correct than return
+// order string else return "" empty string. So here, we are first building the graph, when comparing two consecutive words at a time
+// if two chars are different we get to know that first char comes before second char, so there is a relation, there is a dependency,
+// therefore it is a graph problem. Keeping one map to store the dependancy(char and the list of char depending on the key char), and
+// indegrees list, so we will decrement the indegree of the node by 1, if it has some dependancy. And all the independent nodes we put
+// in the queue, we can start from any char that is independent(indegree as 0) so we put all in queue. Then we start the bfs, poll the
+// current char, append to stringbuilder sb, and check for it's dependant nodes, reduce their indegree by 1, and if it becomes zero 
+// put them in the queue.
+
+// Time Complexity : O(nk) n length of words array and k avg length of each word
+// Space Complexity : O(nk)  dependancy Hashmap 
+// Did this code successfully run on Leetcode : yes
+// Any problem you faced while coding this : no
+import java.util.*;
+
+public class AlienDictionary {
+    int[] indegrees;
+    HashMap<Character, List<Character>> m;
+
+    public String alienOrder(String[] words) {
+        // Base case
+        if (words == null || words.length == 0) {
+            return "";
+        }
+        // indegrees array of length 26, each index represent a char (a-z)
+        indegrees = new int[26];
+        // Map for storing the char and list<char>
+        m = new HashMap<>();
+        // constructGraph method will fill this two data structures
+        constructGraph(words);
+        // Queue for bfs
+        Queue<Character> q = new java.util.LinkedList<>();
+        // Ans
+        StringBuilder sb = new StringBuilder();
+        // Iterate over the map and check if any key from map having indegree zero, put
+        // them in queue
+        for (Character key : m.keySet()) {
+            if (indegrees[key - 'a'] == 0) {
+                q.add(key);
+            }
+        }
+        // Start bfs
+        while (!q.isEmpty()) {
+            // poll the current
+            char curr = q.poll();
+            // append to sb
+            sb.append(curr);
+            // Get the list of all dependant nodes(chars)
+            List<Character> allChar = m.get(curr);
+            // if it is null, continue
+            if (allChar.size() == 0) {
+                continue;
+            }
+            // Else for each char, reduce their indegree by 1 and if it becomes 0, put the
+            // char in queue
+            for (char singleChar : allChar) {
+                indegrees[singleChar - 'a']--;
+                if (indegrees[singleChar - 'a'] == 0) {
+                    q.add(singleChar);
+                }
+            }
+        }
+        // At last if the string size is not equal to the map size, means we have not
+        // got the correct order
+        if (sb.length() != m.size()) {
+            return "";
+        }
+        // Else return string
+        return sb.toString();
+    }
+
+    private void constructGraph(String[] words) {
+        // First go over each word, each char, put all the unique chars in map with
+        // empty list as value
+        for (int i = 0; i < words.length; i++) {
+            for (int j = 0; j < words[i].length(); j++) {
+                if (!m.containsKey(words[i].charAt(j))) {
+                    m.put(words[i].charAt(j), new ArrayList<>());
+                }
+            }
+        }
+        // Go over each words
+        for (int i = 0; i < words.length - 1; i++) {
+            // Take two consecutive words at a time
+            String first = words[i];
+            String second = words[i + 1];
+            // Take their length
+            int m1 = first.length();
+            int n = second.length();
+            // If the first word startswith second, that means not sorted, so clear map and
+            // return
+            if (first.startsWith(second) && m1 > n) {
+                m.clear();
+                return;
+            }
+            // Else compare each char
+            for (int j = 0; j < n && j < m1; j++) {
+                char firstChar = first.charAt(j);
+                char secondChar = second.charAt(j);
+                // If different
+                if (firstChar != secondChar) {
+                    // Increase the indegree of second char by 1
+                    indegrees[secondChar - 'a']++;
+                    // And add to the list in map
+                    m.get(firstChar).add(secondChar);
+                }
+            }
+        }
+    }
+
+    public static void main(String[] args) {
+        String[] words = new String[] { "wrt", "wrf", "er", "ett", "rfttz" };
+        AlienDictionary a = new AlienDictionary();
+        System.out.println(a.alienOrder(words));
+    }
+}
diff --git a/FindACelebrity.java b/FindACelebrity.java
new file mode 100644
index 0000000..211adf5
--- /dev/null
+++ b/FindACelebrity.java
@@ -0,0 +1,82 @@
+// In this problem, simply traversing through the 2 by 2 matrix and maintainin a indegrees array. As we know that for a celeb since 
+// everyone knows celeb, the indegree value will be n-1, and since celeb doesnt know anyone the outdegree value will be 0. So in this
+// indegree array if we find that i knows j, we decrement the indegree[i] by 1 indicating an outgoing edge, and increment the indegree[j]
+// by 1, indicating an incoming edge. At end check if any index having n-1, that is our ans else -1.
+// Time Complexity : O(n^2)
+// Space Complexity : O(n)
+// Did this code successfully run on Leetcode : yes
+// Any problem you faced while coding this : no
+public class FindACelebrity extends Relation {
+    public int findCeleb(int n) {
+        // Base case
+        if (n == 0) {
+            return -1;
+        }
+        // Indegree array
+        int[] indegree = new int[n];
+        // Loop through n*n
+        for (int i = 0; i < n; i++) {
+            for (int j = 0; j < n; j++) {
+                // Check for all the combinations except i==j
+                if (i != j) {
+                    // Check if the i knows j
+                    if (knows(i, j)) {
+                        // Decrement
+                        indegrees[i]--;
+                        // Increment
+                        indegrees[j]++;
+                    }
+                }
+            }
+        }
+        // Check what contains the n-1 value, return that i
+        for (int i = 0; i < n; i++) {
+            if (indegrees[i] == n - 1) {
+                return i;
+            }
+        }
+        // Else -1
+        return -1;
+    }
+}
+
+// In this approach, we are making assumption that 0 is our celeb, than running
+// a loop and checking if there is anyone to whom our
+// current celeb (0) knows, if yes than making that person as the celeb, becoz 0
+// is knowing someone so 0 cant be celeb. Now with this
+// second celeb, we check if everyone knows him and he doenst know anyone, if
+// any breach found, return -1, else return this celeb.
+
+// Time Complexity : O(n)
+// Space Complexity : O(1)
+// Did this code successfully run on Leetcode : yes
+// Any problem you faced while coding this : no
+public class FindACelebrity extends Relation {
+    public int findCeleb(int n) {
+        // Base case
+        if (n == 0) {
+            return -1;
+        }
+        int celeb = 0;
+        // Loop over and check if the first assumed celeb is knowing anyone
+        for (int i = 0; i < n; i++) {
+            if (i != celeb) {
+                if (knows(celeb, i)) {
+                    // Then 0 cannot be celeb, so change the celeb to the person whom 0 knows that
+                    // is i
+                    celeb = i;
+                }
+            }
+        }
+        // Now again loop over and check if this celeb knows anyone or anyone doesnt
+        // know this celeb, then return -1, else return this celeb
+        for (int i = 0; i < n; i++) {
+            if (i != celeb) {
+                if (knows(i, celeb) == 0 || knows(celeb, i) == 1) {
+                    return -1;
+                }
+            }
+        }
+        return celeb;
+    }
+}
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diff --git a/VerifyingAlienDictionary.java b/VerifyingAlienDictionary.java
new file mode 100644
index 0000000..9643e0a
--- /dev/null
+++ b/VerifyingAlienDictionary.java
@@ -0,0 +1,64 @@
+// In this problem, using hashmap to store the order string with it's index. Then going over the array and comparing each consecutive
+// words, if there is any char that is not matching, checking its index value in map, if the first word char has lower index value
+// than the second word char, than returning true, else false. Also if the chars match and first word has length greater than the second
+// word, returning false.
+
+// Time Complexity : O(nk) n length of words array and k avg length of each word
+// Space Complexity : O(1) 
+// Did this code successfully run on Leetcode : yes
+// Any problem you faced while coding this : no
+class Solution {
+    HashMap<Character, Integer> map;
+
+    public boolean isAlienSorted(String[] words, String order) {
+        // Base case
+        if (words == null || words.length == 0) {
+            return true;
+        }
+        map = new HashMap<>();
+        // Loop and put the order string in the map
+        for (int i = 0; i < order.length(); i++) {
+            char c = order.charAt(i);
+            map.put(c, i);
+        }
+        // Now loop over all words
+        for (int i = 0; i < words.length - 1; i++) {
+            // Take two consecutive words at a time
+            String first = words[i];
+            String second = words[i + 1];
+            // Check if not ordered
+            if (isNotOrdered(first, second)) {
+                // Return false
+                return false;
+            }
+        }
+        // return true
+        return true;
+    }
+
+    private boolean isNotOrdered(String first, String second) {
+        // Take the length of two words
+        int m = first.length();
+        int n = second.length();
+        // Loop over each char
+        for (int i = 0; i < m && i < n; i++) {
+            // Take two chars
+            char firstChar = first.charAt(i);
+            char secondChar = second.charAt(i);
+            // Check if not equal
+            if (firstChar != secondChar) {
+                // Check their index position in map
+                if (map.get(firstChar) > map.get(secondChar)) {
+                    // If first greater, means not sorted, so return true
+                    return true;
+                } else {
+                    // Else false
+                    return false;
+                }
+            }
+        }
+        // If no chars mismatch and the length of the first is greater than second, not
+        // sorted, so return true else false
+        return (m > n);
+    }
+}
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