Class Solution
java.lang.Object
g1301_1400.s1387_sort_integers_by_the_power_value.Solution
1387 - Sort Integers by The Power Value\.
Medium
The power of an integer `x` is defined as the number of steps needed to transform `x` into `1` using the following steps:
* if `x` is even then `x = x / 2`
* if `x` is odd then `x = 3 * x + 1`
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`).
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**.
Return the
kth integer in the range `[lo, hi]` sorted by the power value.
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.
**Example 1:**
**Input:** lo = 12, hi = 15, k = 2
**Output:** 13
**Explanation:** The power of 12 is 9 (12 --> 6 --> 3 --> 10 --> 5 --> 16 --> 8 --> 4 --> 2 --> 1)
The power of 13 is 9
The power of 14 is 17
The power of 15 is 17
The interval sorted by the power value [12,13,14,15]. For k = 2 answer is the second element which is 13.
Notice that 12 and 13 have the same power value and we sorted them in ascending order. Same for 14 and 15.
**Example 2:**
**Input:** lo = 7, hi = 11, k = 4
**Output:** 7
**Explanation:** The power array corresponding to the interval [7, 8, 9, 10, 11] is [16, 3, 19, 6, 14].
The interval sorted by power is [8, 10, 11, 7, 9].
The fourth number in the sorted array is 7.
**Constraints:**
* `1 <= lo <= hi <= 1000`
* `1 <= k <= hi - lo + 1`-
Constructor Summary
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Solution
public Solution()
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Method Details
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getKth
public int getKth(int lo, int hi, int k)
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