A. Subset Mex
greedy; implementation; math
Given a set of integers (it can contain equal elements).You have to split it into two subsets \(A\) and \(B\) (both of them can contain equal elements or be empty). You have to maximize the value of \(mex(A)+mex(B)\).Here \(mex\) of a set denotes the smallest non-negative integer that doesn't exist in the set. For example: \(mex(\{1,4,0,2,2,1\})=3\) \(mex(\{3,3,2,1,3,0,0\})=4\) \(mex(\varnothing)=0\) (\(mex\) for empty set) The set is splitted into two subsets \(A\) and \(B\) if for any integer number \(x\) the number of occurrences of \(x\) into this set is equal to the sum of the number of occurrences of \(x\) into \(A\) and the number of occurrences of \(x\) into \(B\).
The input consists of multiple test cases. The first line contains an integer \(t\) (\(1\leq t\leq 100\)) — the number of test cases. The description of the test cases follows.The first line of each test case contains an integer \(n\) (\(1\leq n\leq 100\)) — the size of the set.The second line of each testcase contains \(n\) integers \(a_1,a_2,\dots a_n\) (\(0\leq a_i\leq 100\)) — the numbers in the set.
For each test case, print the maximum value of \(mex(A)+mex(B)\).
In the first test case, \(A=\left\{0,1,2\right\},B=\left\{0,1,5\right\}\) is a possible choice.In the second test case, \(A=\left\{0,1,2\right\},B=\varnothing\) is a possible choice.In the third test case, \(A=\left\{0,1,2\right\},B=\left\{0\right\}\) is a possible choice.In the fourth test case, \(A=\left\{1,3,5\right\},B=\left\{2,4,6\right\}\) is a possible choice.