Python Hash Solution
Intuition\n Describe your first thoughts on how to solve this problem. \nTo qualify, each member + difference for nums1 must exist in nums2.  \nSo, for the set 
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nTo qualify, each member + difference for nums1 must exist in nums2.  \nSo, for the set with added difference, check all the values exist by counting the frequency. The first difference will be the answer.\n# Approach\n<!-- Describe your approach to solving the problem. -->\n1. Sort the numbers to know the minimum and maximum value of the solution. Every number in nums2 must be used, so the lowest difference is the difference between the lowest value in nums2 - lowest value in nums2. On the other hand, since 2 members may be deserted, the maximum value is the lowest value in nums2 - 3rd lowest value in nums1. \n2. Count the frequency of each number in nums1. \n3. As loop through the difference, compare the frequency of each member in nums2 + difference and the frequency in nums1 calculated in step2. If all the nums2 + difference exist in nums1, it is the answer to return \n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\nSorting takes O(nlogn), which may be reduced to n because the only information is the 1st and 3rd smalled in nums1 & the smallest in nums2.\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```python3 []\nclass Solution:\n    def check(self,nums2,d,add):\n        for n in nums2:\n            if d[n-add]==0: return False\n            d[n-add]-=1\n        return True\n\n\n    def minimumAddedInteger(self, nums1: List[int], nums2: List[int]) -> int:\n        nums1.sort()\n        nums2.sort()\n        maxval = nums2[0]-nums1[0]\n        minval = nums2[0]-nums1[2]\n\n        d = defaultdict(int)\n        for n in nums1: d[n]+=1\n\n        for i in range(minval,maxval+1):\n            if self.check(nums2,d.copy(),i):\n                return i\n        return 100000\n\n```
['Python3']