Codeforces Round 670 (Div. 2) |
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Finished |
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:
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 $$$mex(A)+mex(B)$$$.
4 6 0 2 1 5 0 1 3 0 1 2 4 0 2 0 1 6 1 2 3 4 5 6
5 3 4 0
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.
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