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LC532 - K-Diff Pairs in an Array

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Leetcode: 130/2658 (53+63+14)

Description

Given an array of integers nums and an integer k, return the number of unique k-diff pairs in the array.

A k-diff pair is an integer pair (nums[i], nums[j]), where the following are true:

0 <= i, j < nums.length
i != j
nums[i] - nums[j] == k
Notice that |val| denotes the absolute value of val.

Example 1:

Input: nums = [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: nums = [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: nums = [1,3,1,5,4], k = 0
Output: 1
Explanation: There is one 0-diff pair in the array, (1, 1).

Constraints:

1 <= nums.length <= 104
-107 <= nums[i] <= 107
0 <= k <= 107

Solutions

  • Sort the nums list
  • Then, maintain two pointers i, j, and guarantee i < j < n, if nums[j] = nums[i] + k is counted in the answer
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# O(nlogn) time | O(nlogn) space
class Solution:
    def findPairs(self, nums: List[int], k: int) -> int:
        ans = set()
        l, r = 0, 0
        nums.sort()

        while r < len(nums):
            left, right = nums[l], nums[r]
            if right - left == k:
                if l < r and (left, right) not in ans:
                    ans.add((left, right))
                r+=1
            elif right-left < k:
                r +=1
            elif right - left > k:
                l +=1
        return len(ans)