Is Strobogrammatic Number III easy or hard?

Strobogrammatic Number III is classified as Hard because it combines recursion, string-based numeric comparison, and combinatorial generation. Handling variable digit lengths and avoiding invalid leading zeros adds complexity. Efficient pruning and careful boundary checks are key to passing large test cases.

Strobogrammatic Number III Python/Java solution

Python and Java implementations typically use recursion to generate numbers from the center outward. The algorithm fills positions with valid strobogrammatic pairs, skips leading zeros, and checks whether the final string lies within the given bounds. The same logic easily translates to C++ and Go.

How to solve Strobogrammatic Number III in O(n)?

A true O(n) solution is not possible because you must enumerate possible strobogrammatic combinations. The closest optimal strategy is recursive generation, which builds only valid numbers and avoids scanning the entire numeric range. Its complexity is roughly O(5^(n/2)), far better than brute force enumeration.

What is the best approach for Strobogrammatic Number III?

Recursive generation of strobogrammatic numbers by length is the most effective approach. Instead of checking every value in the range, construct numbers using valid digit pairs such as (0,0), (1,1), (6,9), (8,8), and (9,6). Generate candidates for lengths between len(low) and len(high) and filter them against the range. This reduces the search space to O(5^(n/2)).

Is Strobogrammatic Number III asked at Google/Amazon/Meta?

Strobogrammatic problems frequently appear in interviews at companies like Google, Amazon, and Meta because they test recursion, string manipulation, and combinatorial generation. Variants such as Strobogrammatic Number I, II, and III are common in interview prep platforms and coding rounds.

What data structure is used in Strobogrammatic Number III?

The solution primarily uses strings or character arrays to construct numbers symmetrically. A mapping of valid digit pairs like {0↔0, 1↔1, 6↔9, 8↔8, 9↔6} guides recursive placement. Recursion manages the generation process while comparing strings against the range boundaries.

What is the time complexity of Strobogrammatic Number III?

The optimal recursive approach runs in O(5^(n/2)) time, where n is the number of digits in the longest bound. Each recursive level adds symmetric digit pairs, producing at most five choices per outer position. Space complexity is O(n) due to recursion depth and temporary string construction.

Description Solution Video 19:26

248. Strobogrammatic Number III

Hard42.3% acceptancePremium elsewhere, free here

Given two strings low and high that represent two integers low and high where low <= high, return the number of strobogrammatic numbers in the range [low, high].

A strobogrammatic number is a number that looks the same when rotated 180 degrees (looked at upside down).

Example 1:

Input: low = "50", high = "100"
Output: 3

Example 2:

Input: low = "0", high = "0"
Output: 1

Constraints:

1 <= low.length, high.length <= 15
low and high consist of only digits.
low <= high
low and high do not contain any leading zeros except for zero itself.

Topics

19:26

Still stuck?Read the written solution, or watch a walkthrough.

Open solution

248. Strobogrammatic Number III

Topics

Similar problems