Count Unique Pairs With Difference K

Count the number of unique pairs `(a, b)` in an integer array such that `b - a = k`. The variant where `k` may be zero is a common trip-up. Requirements Given numbers[] and a non-negative integer k, count the number of unique pairs (a, b) from numbers with b - a == k. "Unique pair" means each value pair (a, b) is counted at most once, regardless of how many times a or b appears in the input. k may be 0 — in that case the answer counts each value that appears at least twice once. public static int countPair(int[] numbers, int k) Notes Standard hashset approach: put every value into a set; for each unique value v, check whether v + k is in the set. Count one pair per matching v. k == 0 case is the trap: with the hashset-only approach you must instead count values that appear ≥ 2 times in the original input (use a count map for that branch). O(n) time, O(n) space. A naïve implementation has a subtle bug — storing num + "" + (num + k) as the pair key but comparing against num + k (an int) in the set; rewrite cleanly before relying on it. Preparation Implement the general version that handles k == 0 correctly with a Map<Integer, Integer> count, not just a Set. Interviewers love this edge case as a follow-up. LC 532 "K-diff Pairs in an Array" is the canonical equivalent; LC 1 "Two Sum" is the same fingerprint pattern.