532. K-diff Pairs in an Array

https://leetcode.com/problems/k-diff-pairs-in-an-array/#/description

 

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:

1. The pairs (i, j) and (j, i) count as the same pair.
2. The length of the array won't exceed 10,000.
3. All the integers in the given input belong to the range: [-1e7, 1e7].

 

 

 

Sol 1: 

 

In short, check if the smaller element plus difference is in the input list.

 

Sort the list first.

K is the absolute difference here, we put the smaller element in the element in the candidate list and then check if the bigger element, whose value is the sum of value of smaller element plus difference.  

 

Use a count list to store all qualified pairs.

 

Then create a dictionary. The key of dictionary is the sum of smaller value plus difference, and we will check if this bigger value is in the input list later. The value of the dictionary is the smaller value.

 

(P.S. Always put what we wanna find in the key of the dictionary.)  

 

 

class Solution(object):
    def findPairs(self, nums, k):
        """
        :type nums: List[int]
        :type k: int
        :rtype: int
        """
        
        nums.sort()
        count = []
        dict = {}
        for i in range(len(nums)):
            if nums[i] in dict:
                count.append((dict[nums[i]],nums[i]))
            dict[nums[i]+k] = nums[i]
        return len(set(count))

 

 

Note:

 

1 When elements are small, add keys using values(element). When elements are big, find values(element) using keys. Elegent!  

 

2 Length of set() returns unique pairs.

 

 

ex. 

 

count = [(22,33),(66,66),(66,66)]
print len(count)                                     ==> 3
print len(set(count))                              ==> 2

 

 

 

Sol 2:

 

Use count method to track the appearance of elements. 

 

 

class Solution(object):
    def findPairs(self, nums, k):
        """
        :type nums: List[int]
        :type k: int
        :rtype: int
        """
        
        res = 0
        c = collections.Counter(nums)
        for i in c:
            if (k > 0 and i + k in c) or (k == 0 and c[i] > 1):
                res += 1
        return res

 

 

Note:

 

1 collestions.counter is a python built-in method to count the appearance of elements.  

 

import collections
nums = 'aaaaaaaaabbcccccddddeeef'
c = collections.Counter(nums)
print c

 

Counter({'a': 9, 'c': 5, 'd': 4, 'e': 3, 'b': 2, 'f': 1})

for i in c:
    print i

 

a
c
b
e
d
f