Easy solution. WITHOUT  SORTING. Beats 96 %
\n# Code\npython []\nclass Solution:\n    def maxProductDifference(self, nums: List[int]) -> int:\n        max_1 = max(nums)\n        nums.remove(max_1)\n      
\n# Code\n```python []\nclass Solution:\n    def maxProductDifference(self, nums: List[int]) -> int:\n        max_1 = max(nums)\n        nums.remove(max_1)\n        max_2 = max(nums)\n\n        min_1 = min(nums)\n        nums.remove(min_1)\n        min_2 = min(nums)\n\n\n        return (max_1 * max_2) - (min_1 * min_2)\n```\n```javascript []\nvar maxProductDifference = function(nums) {\n    const max_1 = Math.max(...nums);\n    nums = removeElement(nums, max_1);\n    const max_2 = Math.max(...nums);\n\n    const min_1 = Math.min(...nums);\n    nums = removeElement(nums, min_1);\n    const min_2 = Math.min(...nums);\n\n    return ( max_1 * max_2 ) - ( min_1 * min_2 )\n};\n\nvar removeElement = (nums, target) => {\n    for (let i=0; i<nums.length; i++) {\n        if (nums[i] === target) {\n            nums = [...nums.slice(0, i), ...nums.slice(i+1, nums.length)];\n            return nums;\n        };\n    };\n    return nums;\n};\n```\n\n# Complexity\n- Time complexity:$$O(n)$$\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n- Space complexity:$$O(1)$$\n\n# ***Please Upvote !!!***
['Array', 'Python3', 'JavaScript']