We define a harmonious array is an array where the difference between its maximum value and its minimum value is exactly 1.

Now, given an integer array, you need to find the length of its longest harmonious subsequence among all its possible subsequences.

Example 1:

Input: [1,3,2,2,5,2,3,7]
Output: 5
Explanation: The longest harmonious subsequence is [3,2,2,2,3].

Note: The length of the input array will not exceed 20,000.

Brute Force Method:

class Solution(object):
    def findLHS(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """
        nums=sorted(nums)
        i=0
        pre_count=1
        ans=0
        i=0
        while i<len(nums):
            count=1
            if i>0 and nums[i]-nums[i-1]==1:
                while i<len(nums)-1 and nums[i]==nums[i+1]:
                    i+=1
                    count+=1
                ans=max(ans,pre_count+count)
                pre_count=count
            else:
                while i<len(nums)-1 and nums[i]==nums[i+1]:
                    i+=1
                    count+=1
                pre_count=count
            i+=1
        return ans

HashMap Method:

from collections import Counter
class Solution(object):
    def findLHS(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """
        nc=Counter(nums)
        ans=0
        for i in nc:
            if i==1:
                if nc[1]>0 and nc[2]>0:
                    ans=max(ans,nc[i]+nc[i+1])
            else:
                if nc[i]>0 and nc[i+1]>0:
                    ans=max(ans,nc[i]+nc[i+1])
                if nc[i]>0 and nc[i-1]<0:
                    ans=max(ans,nc[i-1]+nc[i])
        return ans