Pairs with difference k
Given an integer array nums and an integer k (k >= 0), return the number of UNIQUE pairs of distinct array positions whose values differ by exactly k. A pair is identified by its value pair, so (1,3) counts once even if 1 or 3 repeats. For k = 0, count values that appear at least twice. Array length up to 1e4.
Implement
count_pairs_diff(nums: list[int], k: int) → intExamples
in
[[3,1,4,1,5],2]out2What a strong answer looks like
State your approach and its time/space complexity out loud before you optimize. Handle the edge cases (empty input, duplicates, overflow), and say why you chose this over the brute force. Green tests are the floor, not the grade.
0:00 of about 18 min
solution.py
InputExpectedGot
[[3,1,4,1,5],2]2not run yetsample