K-diff Pairs in an Array
Given an array of integers and an integerk, you need to find the number ofuniquek-diff pairs in the array. Here ak-diffpair is defined as an integer pair (i, j), whereiandjare both numbers in the array and theirabsolute differenceisk.
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).
分析
就是2sum的思想,排序后,a和a-k pair存入map
class Solution:
def findPairs(self, nums: List[int], k: int) -> int:
nums.sort()
ll = len(nums)
mm = dict()
res = set()
for i in range(ll):
if nums[i]-k in mm:
res.add((nums[i]-k,nums[i]))
mm[nums[i]] = i
return len(res)
Last updated
Was this helpful?