Create coding3.java

coding3.java

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+// Approach:
+// 1. Store the frequency of each element using a HashMap.
+// 2. If k == 0, count the elements whose frequency is at least 2.
+// 3. Otherwise, iterate through the unique keys and check if (key + k) exists in the map.
+// 4. Return the total count of valid k-diff 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;
+    }
+}