K-diff Pairs in an Array - LeetCode
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
|nums[i] - nums[j]| == kNotice 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
- code
class Solution:
def findPairs(self, nums: List[int], k: int) -> int:
if k == 0:
nums.sort()
res = 0
i = 1
while i < len(nums):
if nums[i] == nums[i-1]:
res += 1
while i < len(nums) and nums[i-1] == nums[i]:
i += 1
i += 1
return res
nums = list(set(nums))
nums.sort()
res = 0
match = set()
for v in nums:
if v in match: res += 1
match.add(v + k)
return res
- code
class Solution:
def findPairs(self, nums: List[int], k: int) -> int:
res = 0
if k == 0:
c = Counter(nums)
for k, v in c.items():
if v >= 2: res += 1
return res
nums = list(set(nums))
nums.sort()
res = 0
m = set()
for v in nums:
if v in m: res += 1
m.add(v + k)
return res
- code
class Solution:
def findPairs(self, nums: List[int], k: int) -> int:
res = 0
c = Counter(nums)
for num, count in c.items():
if k == 0:
if count >= 2: res += 1
else:
if num + k in c: res += 1
return res