Problem Description

You are given two strings low and high that represent two integers where low <= high. Your task is to count how many strobogrammatic numbers exist within the range [low, high] (inclusive).

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

For example:

• The digit 0 rotated 180 degrees is still 0
• The digit 1 rotated 180 degrees is still 1
• The digit 8 rotated 180 degrees is still 8
• The digit 6 rotated 180 degrees becomes 9
• The digit 9 rotated 180 degrees becomes 6
• Other digits (2, 3, 4, 5, 7) don't have valid rotations

So numbers like 69, 88, 181, and 9006 are strobogrammatic because when you rotate them 180 degrees, they look the same. For instance, 69 becomes 69 (the 6 becomes 9 and the 9 becomes 6 when rotated), and 181 remains 181 (1 stays 1, 8 stays 8, and 1 stays 1).

The function should return the total count of strobogrammatic numbers that fall within the given range.

Intuition

To build strobogrammatic numbers, we need to think about which digits can be used and how they must be arranged. Only the digits 0, 1, 6, 8, 9 can appear in a strobogrammatic number since these are the only digits that have valid 180-degree rotations.

The key insight is that strobogrammatic numbers have a symmetric structure. When we rotate the number 180 degrees, the first digit becomes the last digit (rotated), the second digit becomes the second-to-last digit (rotated), and so on. This means we can build these numbers from the outside in - by adding pairs of digits to both ends.

For a strobogrammatic number of length n:

• If n is odd, there must be a middle digit that looks the same when rotated (only 0, 1, or 8 work)
• If n is even, we don't need a middle digit

We can recursively construct all strobogrammatic numbers of length n by:

1. Starting with strobogrammatic numbers of length n-2
2. Wrapping them with valid digit pairs: (1,1), (8,8), (6,9), (9,6), and (0,0)

The pair (0,0) is special - we can't use it for the outermost digits because that would create a number with leading zeros (like 00110), which isn't valid. However, we can use (0,0) for inner positions.

By building numbers of each possible length between the lengths of low and high, and then filtering those that fall within the actual numeric range, we can count all valid strobogrammatic numbers in the given range.

The base cases are straightforward:

• Length 0: empty string ""
• Length 1: only "0", "1", "8" (single digits that look the same when rotated)

From these base cases, we can build up all longer strobogrammatic numbers by recursively adding valid pairs around shorter ones.

Solution Approach

The solution uses a recursive approach with a helper function dfs(u) that generates all strobogrammatic numbers of length u.

Base Cases:

• When u = 0: Return [''] (empty string list)
• When u = 1: Return ['0', '1', '8'] (single digits that remain the same when rotated)

Recursive Case: For lengths greater than 1, we:

1. Get all strobogrammatic numbers of length u - 2 by calling dfs(u - 2)
2. For each number v from that list, wrap it with valid digit pairs:
   • '1' + v + '1' (pair 11)
   • '8' + v + '8' (pair 88)
   • '6' + v + '9' (pair 69)
   • '9' + v + '6' (pair 96)
3. Special handling for '0' + v + '0': Only add this if u != n (not the outermost layer) to avoid leading zeros

Main Algorithm:

1. Extract the lengths a and b from the input strings low and high
2. Convert low and high to integers for numerical comparison
3. Iterate through all possible lengths from a to b + 1:
   • For each length n, generate all strobogrammatic numbers using dfs(n)
   • Convert each generated string to an integer
   • Check if it falls within the range [low, high]
   • If yes, increment the counter

Example Walkthrough: If we want strobogrammatic numbers of length 3:

• Start with dfs(1) which returns ['0', '1', '8']
• Wrap each with valid pairs:
  • '0' becomes '101', '808', '609', '906'
  • '1' becomes '111', '818', '619', '916'
  • '8' becomes '181', '888', '689', '986'

The time complexity is O(2^{n+2} × log n) where the exponential factor comes from the branching in the recursion (each level can add up to 5 pairs), and the logarithmic factor comes from string to integer conversions.

Example Walkthrough

Let's walk through finding all strobogrammatic numbers in the range [50, 100].

Step 1: Determine the length range

• low = "50" has length 2
• high = "100" has length 3
• We need to generate strobogrammatic numbers of lengths 2 and 3

Step 2: Generate length 2 strobogrammatic numbers

Starting with dfs(2):

• First, we call dfs(0) which returns ['']
• For each empty string, we wrap with valid pairs:
  • '1' + '' + '1' = '11'
  • '8' + '' + '8' = '88'
  • '6' + '' + '9' = '69'
  • '9' + '' + '6' = '96'
• Since u = n = 2 (outermost layer), we skip '0' + '' + '0' to avoid leading zeros
• Result: ['11', '88', '69', '96']

Step 3: Generate length 3 strobogrammatic numbers

Starting with dfs(3):

• First, we call dfs(1) which returns ['0', '1', '8']

• For each single digit, we wrap with valid pairs:

  For '0':

  • '1' + '0' + '1' = '101'
  • '8' + '0' + '8' = '808'
  • '6' + '0' + '9' = '609'
  • '9' + '0' + '6' = '906'

  For '1':

  • '1' + '1' + '1' = '111'
  • '8' + '1' + '8' = '818'
  • '6' + '1' + '9' = '619'
  • '9' + '1' + '6' = '916'

  For '8':

  • '1' + '8' + '1' = '181'
  • '8' + '8' + '8' = '888'
  • '6' + '8' + '9' = '689'
  • '9' + '8' + '6' = '986'

• Since u = n = 3 (outermost layer), we don't add '0' + v + '0' pairs

• Result: ['101', '808', '609', '906', '111', '818', '619', '916', '181', '888', '689', '986']

Step 4: Filter by range [50, 100]

From length 2 numbers:

• 11 → Not in range (< 50)
• 88 → In range ✓
• 69 → In range ✓
• 96 → In range ✓

From length 3 numbers:

• All are > 100, so none are in range

Final Answer: 3 (the numbers 88, 69, and 96)

This example demonstrates how the recursive approach builds strobogrammatic numbers layer by layer, avoiding leading zeros, and then filters the results to count only those within the specified range.

Solution Implementation

Time and Space Complexity

Time Complexity: O((b - a + 1) * 5^(b/2)) where a is the length of the low string and b is the length of the high string.

• The outer loop iterates through each possible length from a to b, which is O(b - a + 1) iterations.
• For each length n, the dfs(n) function is called which generates all strobogrammatic numbers of length n.
• The recurrence relation for counting strobogrammatic numbers is approximately T(n) = 5 * T(n-2) for n > 2 (since we add 5 pairs for each recursive call - '00', '11', '88', '69', '96', though '00' is conditional).
• This gives us approximately 5^(n/2) strobogrammatic numbers for length n.
• In the worst case, when n = b, we generate O(5^(b/2)) numbers.
• For each generated number, we perform an integer conversion and comparison which takes O(b) time.
• Overall time complexity: O((b - a + 1) * 5^(b/2) * b), which simplifies to O((b - a + 1) * 5^(b/2)) since b is typically much smaller than the exponential term.

Space Complexity: O(5^(b/2) * b)

• The dfs function uses recursion with maximum depth O(b/2), contributing O(b) to the call stack.
• At each recursion level, we store intermediate results. The largest storage occurs at the top level when n = b.
• For length b, we store approximately 5^(b/2) strings, each of length b.
• Therefore, the space needed to store all strobogrammatic numbers of length b is O(5^(b/2) * b).
• The space for storing the answer list ans in each recursive call is dominated by the final result at the top level.
• Overall space complexity: O(5^(b/2) * b)

Common Pitfalls

1. Leading Zeros Issue

The most critical pitfall in this problem is incorrectly handling leading zeros. The recursive function might generate numbers like "00", "000", or "0110" which are not valid integers.

Why it happens: When building strobogrammatic numbers recursively, we can add '0' + middle + '0' as a valid pair. However, if this happens at the outermost layer (when length == target_length), it creates invalid numbers with leading zeros.

The Fix: Only add the '0' + middle + '0' pair when we're NOT at the outermost layer:

# Correct approach
if length != target_length:
    result.append('0' + middle + '0')

2. String vs Integer Comparison Confusion

Another common mistake is comparing string representations directly instead of integer values:

Wrong approach:

# This would fail because string comparison is lexicographic
if low <= num_str <= high:  # WRONG!
    count += 1

Correct approach:

# Convert to integers for numerical comparison
low_value = int(low)
high_value = int(high)
num_value = int(num_str)
if low_value <= num_value <= high_value:
    count += 1

3. Off-by-One Error in Length Iteration

Forgetting to include the maximum length in the range:

Wrong:

for current_length in range(min_length, max_length):  # Excludes max_length

Correct:

for current_length in range(min_length, max_length + 1):  # Includes max_length

4. Not Handling Edge Cases for Single-Digit Numbers

When the range includes single-digit numbers, ensure the base case correctly identifies which single digits are strobogrammatic (0, 1, 8) and excludes invalid ones (2, 3, 4, 5, 7).

5. Memory Inefficiency from Redundant Calculations

The current solution recalculates strobogrammatic numbers for each length independently. If you're calling this function multiple times or with large ranges, consider memoization to cache previously computed results for specific lengths.