Problem of the day
You have been given an array/list ‘ARR’ of integers consisting of ‘N’ integers. The factor value is the number of following operations it takes for the number to become 1.
The two operations are as follows:-
If ‘x’ is even then ‘x’ will become ‘x / 2’.
If ‘x’ is odd then ‘x’ will become ‘x * 3 + 1’.
Your need to sort them in increasing order of factor value i.e. if two integers have the same factor value then sort in increasing order of their value. Your task is to return the ‘K-th’ value in the list after sorting.
Example: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.
Output Format:
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.
Note:
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
Use Priority Queue to store the result after each operation.
We will find the factor value of each element in the array. Finally, we will use a priority queue to maintain the sorted list.
We will apply the algorithm as follows:-
O(N * log K), where ‘N’ denotes the size of the array/list and ‘K’ denotes the position whose value in the sorted list you need to return.
We are inserting ‘N’ elements in the priority queue which takes ‘log K’ time for each insertion.
O(K), where ‘K’ denotes a position whose value in the sorted list you need to return.
We build a priority queue whose maximum size is ‘K’.