A good way of finding disappeared numbers, no times of appearance limitation
tianqi zhu
275
exploreFun With Arrays!
class Solution:
    def findDisappearedNumbers(self, nums: List[int]) -> List[int]:
        output=[]
        for each in nums:
            while each != nums[each-1]:
                nums[each-1],temp=each,nums[each-1]
                each=temp
        for i in range(1,len(nums)+1):
            if i != nums[i-1]:
                output.append(i)
        return output

This way will work for array like [1,1,1,2,4,5,8]
The original problem only asked for array where some number appear once, some twice. This solution has no such limitation.
In short, you put 4 to nums[3] and if nums[3] was 7, you put that in nums[6], until finally every element is in place by this rule.
Whatever isn't in the final list is the missing ones.

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