Sort Integers by The Power Value — LeetCode 1387 Python Solution
- Problem
- #1387
- Pattern
- Sorting
- Reading time
- 3 min
- Source
- leetcode.com
The problem
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.
Example
- 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)
Python solution
@cache
def f(x: int) -> int:
ans = 0
while x != 1:
if x % 2 == 0:
x //= 2
else:
x = 3 * x + 1
ans += 1
return ans
class Solution:
def getKth(self, lo: int, hi: int, k: int) -> int:
return sorted(range(lo, hi + 1), key=f)[k - 1]Complexity
| Measure | Complexity |
|---|---|
| Time | O(n \times \log n \times M) |
| Space | O(n) auxiliary |
Pattern: Sorting
Spend O(n log n) once to buy an ordering that makes the rest of the problem trivial. LeetCode 1387. Sort Integers by The Power Value is filed here because LeetCode tags it Sorting, which is the vocabulary this hub collects.
The sorting guide has the Python template for the pattern and the 401 LeetCode problems that use it.
Related problems
Frequently asked questions
- How hard is LeetCode 1387. Sort Integers by The Power Value?
- LeetCode 1387. Sort Integers by The Power Value is rated Medium on LeetCode.
- What is the time complexity of LeetCode 1387. Sort Integers by The Power Value?
- The Python solution on this page runs in O(n \times \log n \times M).
- What is the space complexity of LeetCode 1387. Sort Integers by The Power Value?
- The Python solution on this page uses O(n) auxiliary space.
- What topics does LeetCode 1387. Sort Integers by The Power Value cover?
- LeetCode 1387. Sort Integers by The Power Value is tagged Memoization, Dynamic Programming and Sorting on LeetCode.