Problem of the day
If ‘x’ is even then ‘x’ will become ‘x / 2’.
If ‘x’ is odd then ‘x’ will become ‘x * 3 + 1’.
Let’s say you have an array/list [1, 3, 4, 5] and ‘K’=2. The factor values are [0, 7, 2, 5] respectively. Finally, our array will look like [1, 4, 5, 3]. Since ‘K’ is 2 return 4.
The first line contains a single integer ‘T’ representing the number of test cases.
The first line of each test case contains two space-separated integers ‘N’ and ‘K’ representing the size of the array/list ‘ARR’ and position whose value in the sorted list you need to return.
The second line and the last line of input contain ‘N’ single space-separated integers representing the array/list elements.
For each test case, print a single line containing a single integer denoting the ‘K-th’ element in sorted list.
The output of each test case will be printed in a separate line.
You do not need to print anything; it has already been taken care of. Just implement the function.
1 <= T <= 10
1 <= N <= 1000
1 <= K <= N
1 <= ‘ARR[i]’ <= 10 ^ 4
Where ‘ARR[i]’ is an element of array/list ARR.
Time Limit: 1sec
2
4 4
1 2 3 4
2 1
13 12
3
12
Test case 1:
The factor values of [1, 2, 3, 4] are [0, 1, 7, 2]. The array/list will become [1, 2, 4, 3] on sorting according to factor value.
Therefore the answer is 3.
Test case 2:
Both 12 and 13 have a factor value of 9. Since they have the same factor value we will sort them according to their value. Therefore on sorting ‘arr’ will be [12,13].
Therefore the answer is 12.
2
4 3
1 2 6 4
4 2
7 8 9 10
4
10