Yet another MEX problem solution codechef

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The MEX (minimum excluded) of an array is the smallest non-negative integer that does not belong to the array. For instance:

The MEX of $[2, 2, 1]$ is $0$ , because $0$ does not belong to the array.
The MEX of $[3, 1, 0, 1]$ is $2$ , because $0$ and $1$ belong to the array, but $2$ does not.
The MEX of $[0, 3, 1, 2]$ is $4$ , because $0, 1, 2$ and $3$ belong to the array, but $4$ does not.

You are given an array $A$ of length $N$ . You create a list $B$ consisting of the MEX-es of all subarrays of the array $A$ . Formally, for all pairs $(l, r)$ such that $1 \leq l \leq r \leq N$ , you calculate $M E X (A_{l}, A_{l + 1}, \dots, A_{r})$ and append the value in the list $B$ . Find the $K$ -th smallest value in the list $B$ . Yet another MEX problem solution codechef

Note: Since the size of the input and output is large, please use fast input-output methods.

Yet another MEX problem solution codechef

The first line contains $T$ denoting the number of test cases. Then the test cases follow.
The first line of each test case contains two space-separated integers $N$ and $K$ .
The second line contains $N$ space-separated integers $A_{1}, A_{2}, \dots, A_{N}$ denoting the given array.

Yet another MEX problem solution codechef

For each test case, output on a single line the $K$ -th smallest value in the list $B$ .

Constraints

$1 \leq T \leq 3 \cdot 10^{4}$
$1 \leq N \leq 10^{5}$
$1 \leq K \leq \frac{N \cdot (N + 1)}{2}$
$0 \leq A_{i} \leq N$
Sum of $N$ over all test cases does not exceed $2 \cdot 10^{6}$ .

Yet another MEX problem solution codechef

Subtask 1 (10 points):

$1 \leq N \leq 5 \cdot 10^{3}$
Sum of $N$ over all test cases does not exceed $5 \cdot 10^{4}$ .

Subtask 2 (90 points): Original constraints

Yet another MEX problem solution codechef

Sample Output 1

1
0
3

Yet another MEX problem solution codechef

Test case $1$ : $M E X (A_{1}) = 0$ , $M E X (A_{1}, A_{2}) = 2$ , $M E X (A_{1}, A_{2}, A_{3}) = 3$ , $M E X (A_{2}) = 1$ , $M E X (A_{2}, A_{3}) = 1$ , $M E X (A_{3}) = 0$ . Hence the list $B = [0, 2, 3, 1, 1, 0]$ and the $4$ -th smallest value in $B$ is $1$ .

Test case $2$ : The MEX of all subarrays of the array $A$ is $0$ . Hence the $2$ -nd smallest element in the list $B$ is $0$ .

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Yet another MEX problem solution codechef

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Yet another MEX problem solution codechef

Yet another MEX problem solution codechef

Constraints

Yet another MEX problem solution codechef

Sample Output 1

Yet another MEX problem solution codechef

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