KdiffPairsInAnArray.js

// Source : https://leetcode.com/problems/k-diff-pairs-in-an-array/
// Author : Dean Shi
// Date   : 2018-08-29

/***************************************************************************************
 *
 * Given an array of integers and an integer k, you need to find the number of unique k-
 * diff pairs in the array. Here a k-diff pair is defined as an integer pair (i, j),
 * where i and j are both numbers in the array and their absolute difference is k.
 *
 * Example 1:
 * Input: [3, 1, 4, 1, 5], k = 2
 * Output: 2
 * Explanation: There are two 2-diff pairs in the array, (1, 3) and (3, 5).Although we
 * have two 1s in the input, we should only return the number of unique pairs.
 *
 * Example 2:
 * Input:[1, 2, 3, 4, 5], k = 1
 * Output: 4
 * Explanation: There are four 1-diff pairs in the array, (1, 2), (2, 3), (3, 4) and (4,
 * 5).
 *
 * Example 3:
 * Input: [1, 3, 1, 5, 4], k = 0
 * Output: 1
 * Explanation: There is one 0-diff pair in the array, (1, 1).
 *
 * Note:
 *
 * The pairs (i, j) and (j, i) count as the same pair.
 * The length of the array won't exceed 10,000.
 * All the integers in the given input belong to the range: [-1e7, 1e7].
 *
 ***************************************************************************************/

/**
 * @param {number[]} nums
 * @param {number} k
 * @return {number}
 */
var findPairs = function(nums, k) {
    if (k < 0) return 0

    const numsSet = new Set()
    const data = new Set(nums)
    let count = 0

    nums.forEach((n) => {
        if (numsSet.has(n)) return

        if (k !== 0) {
            if (data.has(k + n) && !numsSet.has(k + n)) {
                count++
            }
            if (data.has(n - k) && !numsSet.has(n - k)) {
                count++
            }
        } else {
            if (nums.indexOf(n) !== nums.lastIndexOf(n)) {
                count++;
            }
        }
        numsSet.add(n)
    })

    return count
};