1387. Sort Integers by The Power Value Medium

1/**
2 * [1387] Sort Integers by The Power Value
3 *
4 * The power of an integer x is defined as the number of steps needed to transform x into 1 using the following steps:
5 * 
6 * 	if x is even then x = x / 2
7 * 	if x is odd then x = 3 * x + 1
8 * 
9 * For example, the power of x = 3 is 7 because 3 needs 7 steps to become 1 (3 --> 10 --> 5 --> 16 --> 8 --> 4 --> 2 --> 1).
10 * Given three integers lo, hi and k. The task is to sort all integers in the interval [lo, hi] by the power value in ascending order, if two or more integers have the same power value sort them by ascending order.
11 * Return the k^th integer in the range [lo, hi] sorted by the power value.
12 * Notice that for any integer x (lo <= x <= hi) it is guaranteed that x will transform into 1 using these steps and that the power of x is will fit in a 32-bit signed integer.
13 *  
14 * Example 1:
15 * 
16 * Input: lo = 12, hi = 15, k = 2
17 * Output: 13
18 * Explanation: The power of 12 is 9 (12 --> 6 --> 3 --> 10 --> 5 --> 16 --> 8 --> 4 --> 2 --> 1)
19 * The power of 13 is 9
20 * The power of 14 is 17
21 * The power of 15 is 17
22 * The interval sorted by the power value [12,13,14,15]. For k = 2 answer is the second element which is 13.
23 * Notice that 12 and 13 have the same power value and we sorted them in ascending order. Same for 14 and 15.
24 * 
25 * Example 2:
26 * 
27 * Input: lo = 7, hi = 11, k = 4
28 * Output: 7
29 * Explanation: The power array corresponding to the interval [7, 8, 9, 10, 11] is [16, 3, 19, 6, 14].
30 * The interval sorted by power is [8, 10, 11, 7, 9].
31 * The fourth number in the sorted array is 7.
32 * 
33 *  
34 * Constraints:
35 * 
36 * 	1 <= lo <= hi <= 1000
37 * 	1 <= k <= hi - lo + 1
38 * 
39 */
40pub struct Solution {}
41
42// problem: https://leetcode.com/problems/sort-integers-by-the-power-value/
43// discuss: https://leetcode.com/problems/sort-integers-by-the-power-value/discuss/?currentPage=1&orderBy=most_votes&query=
44
45// submission codes start here
46
47impl Solution {
48    pub fn get_kth(lo: i32, hi: i32, k: i32) -> i32 {
49        0
50    }
51}
52
53// submission codes end
54
55#[cfg(test)]
56mod tests {
57    use super::*;
58
59    #[test]
60    fn test_1387() {
61    }
62}
63