Completed Interview - 3

KDiffPairsInArray.java

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+// Time Complexity : O(n)
+// Space Complexity : O(n)
+// Did this code successfully run on Leetcode : Yes
+// Any problem you faced while coding this : No
+
+
+// Your code here along with comments explaining your approach
+// 1: We first create a frequency map for each element from the input array
+// 2: For each key in the set, we check to see whether the complement exists
+// 3: For k=0, we check to see if duplicates exist in the array
+class Solution {
+    public int findPairs(int[] nums, int k) {
+        Map<Integer, Integer> map = new HashMap<>();
+
+        for(int i = 0; i<nums.length;i++){
+            if(map.containsKey(nums[i])){
+                int count = map.get(nums[i]);
+                map.put(nums[i], count+1 );
+            }
+            else{
+                map.put(nums[i], 1);
+            }
+        }
+
+        int result = 0;
+        // iterate through the map
+        for(int key : map.keySet()){
+            int complement = key + k;
+            if(k > 0 && map.containsKey(complement)){
+                result++;
+            }
+            else if(k==0 && map.get(key) >= 2 ){
+                result++;
+            }
+        }
+        return result;
+
+    }
+}