659. Split Array into Consecutive Subsequences-优快云博客

探讨如何将升序排列且可能包含重复元素的整数数组拆分成多个子序列，每个子序列至少包含三个连续整数。通过使用两个哈希映射实现解决方案，并确保所有元素都能正确分配到相应的子序列中。

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You are given an integer array sorted in ascending order (may contain duplicates), you need to split them into several subsequences, where each subsequences consist of at least 3 consecutive integers. Return whether you can make such a split.

Example 1:

Input: [1,2,3,3,4,5]
Output: True
Explanation:
You can split them into two consecutive subsequences : 
1, 2, 3
3, 4, 5

Example 2:

Input: [1,2,3,3,4,4,5,5]
Output: True
Explanation:
You can split them into two consecutive subsequences : 
1, 2, 3, 4, 5
3, 4, 5

Example 3:

Input: [1,2,3,4,4,5]
Output: False

I don't know how to solve this problem from the beginning, then I search for the answer in the discussion. Basically, for every element, we have two ways to handle it. We can either start a new consequence using this element or we can add this element to an existed consequence. If we can't, return False, because there is no way to deal with this element.

We have to maintain two hashmaps. One is Remaining, which indicates those elements we haven't placed yet. Another one is end[i], which indicates the counts end with i.

class Solution(object):
    def isPossible(self, nums):
        """
        :type nums: List[int]
        :rtype: bool
        """
        remain=collections.Counter(nums)
        end=collections.Counter()
        for item in nums:
            if not remain[item]:
                continue
            remain[item]-=1
            if remain[item+1] and remain[item+2]:
                remain[item+1]-=1
                remain[item+2]-=1
                end[item+2]+=1
            elif end[item-1]>0:
                end[item-1]-=1
                end[item]+=1
            else:
                return False
        return True