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Community-reported · source · updated 2026-10-08. ● asked in the last 6 months.
Go step by step: try it yourself before the video and the visual. Progress saves on this device.
0 / 130 pointsSolved (Medium) 0/40Reviews 0/50Mastered 0/25Practice steps 0/15
Step 1
Know these before you start
Practise each one here, or tick what you already know
Concept videoStep 2
Try it on paper
5–10 minutes, no code yet
In the wild · Folders and JSON: Finding every file in a folder means listing the folder and doing the same for each folder inside it. Code that reads nested JSON walks it the same way.
About this problem
Solve a problem with a smaller copy of itself: a base case, one step, and trust the call.
Read the full statement and examples on LeetCode, try it for 20–30 minutes, then come back for the walkthrough.
Examples
Full statement on LeetCode ↗Example 1
- Input
- n = 1
- Output
- 5
Example 2
- Input
- n = 2
- Output
- 20
Example 3
- Input
- n = 4
- Output
- 400
Code it here
Tab indents · ⌘/Ctrl+Enter runs · Esc leaves the editor
Step 6
Get it accepted
All test cases pass
Step 7
Review it so it sticks
Re-solve from memory after
The step that matters most: without reviews most of a solved problem fades within weeks. Three short re-solves, each just as you start to forget, are what make it yours.
Your notes
Three lines in your own words. They come back with this problem's reviews, so you can check yourself after recalling the solution.
Solve it yourself before you look
Reading the solution first feels like progress, but it makes the next similar problem — and the interview version — much harder, because you skipped the part where you figure it out. Give it an honest 20–30 minutes. Stuck? Re-read the pattern, watch the concept video, or try the brute force first.
Hidden: optimal solution
Solution code
Optimal solutionEven positions take one of 5 even digits and odd positions one of 4 , so the answer is 5^(n/2) · 4^floor(n/2) mod 10⁹ + 7. Fast power halves the exponent each call, so even n = 10¹⁵ is about 50 steps.
class Solution:
def countGoodNumbers(self, n: int) -> int:
MOD = 10**9 + 7
def power(x: int, e: int) -> int:
if e == 0:
return 1
half = power(x, e // 2)
result = half * half % MOD
return result * x % MOD if e % 2 else result
return power(5, (n + 1) // 2) * power(4, n // 2) % MODReference solution, tested in Python, Java and C++.
Other ways to solve it
- Thought of another approach (slower, or cleverer)? Code it here: the same tests check any approach.
- Community solutions on LeetCode ↗ for many more versions, in many languages. Read them after you've solved it.