[Python] straightforward solution with 2 heaps
\nfrom heapq import *\nclass Solution:\n    def minDifference(self, nums: List[int]) -> int:\n        if len(nums) <= 4:\n            return 0\n        \n      
```\nfrom heapq import *\nclass Solution:\n    def minDifference(self, nums: List[int]) -> int:\n        if len(nums) <= 4:\n            return 0\n        \n        maxHeap, minHeap = [], []\n        \n        for n in nums:\n            heappush(maxHeap, -n)\n            heappush(minHeap, n)\n            if len(maxHeap) > 4:\n                heappop(maxHeap)\n                heappop(minHeap)\n                \n        # e.g. [1,5,0,10,14]\n\t\t#print(maxHeap)        #[-10, -5, 0, -1]\n        #print(minHeap)         #[1, 5, 14, 10]\n        \n        minNums = sorted([-1*n for n in maxHeap])            # [0, 1, 5, 10]\n        maxNums = sorted(minHeap, reverse = True)            # [10, 14, 5, 1]\n        \n        res = float("inf")      \n        for i in range(4):\n            res = min(res, maxNums[3-i]-minNums[i])\n        return res        \n\t\t\n```
['Heap (Priority Queue)', 'Python', 'Python3']