Nth Staged License Plate

License plates are generated in stages for a fixed total length length. Stage k has exactly k leading uppercase letters followed by length - k digits. Stages run from k = 0 through k = length.
Within one stage, enumerate the letter prefix as a zero-padded base-26 number with A = 0, then enumerate the digit suffix as a zero-padded base-10 number. The index n is zero-based across the concatenation of all stages. Return the plate at index n.

Function
nthStagedLicensePlate(n: long, length: int) → String

Examples
Example 1
n = 0
length = 5
return = "00000"
The first stage starts with the all-zero five-digit plate.
Example 2
n = 99999
length = 5
return = "99999"
This is the last all-digit plate.
Example 3
n = 100000
length = 5
return = "A0000"
After 100000 digit-only plates, the one-letter stage begins.
Example 4
n = 360
length = 2
return = "AA"
Length two has 100 digit plates and 260 one-letter plates before the all-letter stage.

Constraints
1 <= length <= 10.
0 <= n < sum(26^k * 10^(length-k)) for 0 <= k <= length.
All intermediate counts fit in a signed 64-bit integer.

The trick: stage k holds 26^k * 10^(length-k) plates. Start at k = 0 and while n is at least that count, subtract it and move to the next stage. Once n is smaller than the stage size, you're inside the right stage. The offset n then splits mixed-radix style. The digit suffix is n % 10^(length-k), and the letter prefix is n / 10^(length-k). Convert the prefix to base 26 with A = 0, padded to k letters. Convert the suffix to base 10, padded to length-k digits. Pitfalls: forgetting zero padding, computing powers with floats, and getting the order wrong (letters first, then digits). Check Example 4: length 2, n = 360. Stage 0 has 100 and stage 1 has 260, leaving offset 0 in stage 2, which gives AA. Everything fits in 64-bit, so no big integers are needed. Time is O(length).

If you see this problem in your OA tomorrow, the play is to recognize the pattern in 30 seconds. StealthCoder buys you that recognition.