Class Solution
- java.lang.Object
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- g1301_1400.s1387_sort_integers_by_the_power_value.Solution
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public class Solution extends Object
1387 - Sort Integers by The Power Value.Medium
The power of an integer
x
is defined as the number of steps needed to transformx
into1
using the following steps:- if
x
is even thenx = x / 2
- if
x
is odd thenx = 3 * x + 1
For example, the power of
x = 3
is7
because3
needs7
steps to become1
(3 --> 10 --> 5 --> 16 --> 8 --> 4 --> 2 --> 1
).Given three integers
lo
,hi
andk
. 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 thatx
will transform into1
using these steps and that the power ofx
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
- if
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Constructor Summary
Constructors Constructor Description Solution()
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