diff --git a/coding-3.java b/coding-3.java
new file mode 100644
index 00000000..367b1902
--- /dev/null
+++ b/coding-3.java
@@ -0,0 +1,35 @@
+// Approach:
+// 1. Store the frequency of each element in a HashMap.
+// 2. If k == 0, count the numbers whose frequency is at least 2,
+//    since a pair requires duplicate elements.
+// 3. If k > 0, iterate through all unique keys in the map.
+// 4. For each key, check whether (key - k) exists in the map.
+// 5. If it exists, increment the count as it forms a unique k-diff pair.
+// 6. Return the total count of unique pairs.
+
+// Time Complexity: O(n)
+// Space Complexity: O(n)
+class Solution {
+    public int findPairs(int[] nums, int k) {
+        int n=nums.length;
+        HashMap<Integer,Integer> map=new HashMap<>();
+        int cnt=0;
+        for(int i=0;i<n;i++){
+            map.put(nums[i],map.getOrDefault(nums[i],0)+1);
+        }
+        if(k==0){
+            for(int freq:map.values()){
+                if(freq>=2){
+                    cnt++;
+                }
+            }
+        }else{
+            for(int key:map.keySet()){
+                if(map.containsKey(key-k)){
+                    cnt++;
+                }
+            }
+        }
+        return cnt;
+    }
+}