Start by clarifying the problem: whether pairs are index-based or value-based, and if duplicates count as unique pairs. Then propose an efficient solution using a hash set to track seen elements, achieving O(n) time and O(n) space, and discuss edge cases like k=0 and negative k.
Pro tip: Mention that for k=0, you need to count pairs of identical elements, which requires a frequency map instead of a set. Also, note that if k<0, the answer is 0 since absolute difference is non-negative.
Ask whether pairs are based on indices or values, and whether duplicate values count as unique pairs. Confirm that k is non-negative.
Use a hash set for k>0 to track seen elements, or a frequency map for k=0 to count duplicates. Explain why a set is efficient for O(1) lookups.
For k>0: iterate through the array, for each element check if element+k or element-k is in the set, and if so, increment count; then add the element to the set. For k=0: count frequencies and add freq*(freq-1)/2 for each element with freq>1.
State that the time complexity is O(n) and space complexity is O(n) in the worst case. Compare with the brute-force O(n^2) approach to highlight efficiency.
Discuss cases like empty array, k<0 (return 0), k=0 (use frequency map), and arrays with many duplicates. Also mention that if k is very large, the set may not help but still O(n).
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