diff --git a/algorithms/Java/README.md b/algorithms/Java/README.md
index 006285eb..0aeeea66 100644
--- a/algorithms/Java/README.md
+++ b/algorithms/Java/README.md
@@ -72,6 +72,7 @@
 4. [Sequence](strings/sequence.java)
 5. [Split String](strings/SplitString.java)
 6. [Tokenizer](strings/tokenizer.java)
+7. [Anagram](strings/anagram.java)
 
 ## Trees
 
diff --git a/algorithms/Java/strings/anagram.java b/algorithms/Java/strings/anagram.java
new file mode 100644
index 00000000..8f9554f9
--- /dev/null
+++ b/algorithms/Java/strings/anagram.java
@@ -0,0 +1,55 @@
+package Strings;
+
+import java.util.HashMap;
+
+
+//Given two equal-size strings s and t. In one step you can choose any character of t and replace it with another character.
+//Return the minimum number of steps to make t an anagram of s.
+//An Anagram of a string is a string that contains the same characters with a different (or the same) ordering.
+
+public class anagram {
+    public static void main(String[] args) {
+        String s = "leetcode";
+        String t = "practice";
+        System.out.println(minSteps(s, t));
+        System.out.println(minSteps2(s, t));
+
+    }
+
+
+    //1st solution
+    private static int minSteps(String s, String t) {
+        HashMap<Integer, Character> map = new HashMap<>();
+        char[] charS = s.toCharArray();
+        char[] charT = t.toCharArray();
+
+        for (int i = 0; i < charT.length; i++) {
+            map.put(i, charT[i]);
+        }
+
+        for (char aChar : charS) {
+            if (map.containsValue(aChar)) {
+                map.values().remove(aChar);
+            }
+        }
+
+        return map.size();
+    }
+
+    //2nd solution better and fast
+    private static int minSteps2(String s, String t) {
+        int n = s.length();
+        int ans = 0;
+        int[] count = new int[26];//always keep in mind (26) when doing a string question
+        for (int i = 0; i < n; i++) {
+            count[s.charAt(i) - 'a']++;
+            count[t.charAt(i) - 'a']--;
+        }
+        for (int i = 0; i < 26; i++) {
+            if (count[i] > 0)
+                ans += count[i];
+        }
+        return ans;
+    }
+}
+