Solution Explanation for Sequential Digits

The problem asks to find all integers within a given range [low, high] that have sequential digits (each digit is one more than the previous digit). The solution uses a straightforward enumeration approach.

Approach

The core idea is to generate all possible sequential digit numbers and then filter those that fall within the specified range.

1. Generate Sequential Numbers: We iterate through possible starting digits (1 to 8). For each starting digit, we build sequential numbers by appending the next digit until we reach 9. This efficiently generates all possible sequential digit numbers.

2. Filtering: We check if each generated number is within the [low, high] range. If it is, we add it to the result list.

3. Sorting: Finally, we sort the result list to satisfy the requirement of a sorted output.

Time and Space Complexity Analysis

• Time Complexity: The time complexity is dominated by the nested loops that generate sequential numbers. The outer loop runs 8 times (starting digits 1 to 8), and the inner loop runs a maximum of 9 times (appending digits until 9). Thus, the generation step takes O(1) time, as it's bounded by a constant (a maximum of 72 iterations). Sorting the result list adds O(N log N) time, where N is the number of sequential numbers within the range. In the worst case, N could be large, but it's still polynomial. Overall, the time complexity is effectively O(N log N), dominated by sorting. If we consider the constant upper bound for generation, it can also be viewed as just O(N log N).

• Space Complexity: The space complexity is determined by the size of the ans list (the list of sequential numbers found within the range). In the worst-case scenario, the list could contain a large number of elements (though still polynomial in the size of the input range). So, the space complexity is O(N), where N is the number of sequential numbers within the range.

Code Explanation (Python Example)

class Solution:
    def sequentialDigits(self, low: int, high: int) -> List[int]:
        ans = []
        for i in range(1, 9):  # Iterate through possible starting digits
            x = i
            for j in range(i + 1, 10):  # Build sequential number
                x = x * 10 + j
                if low <= x <= high:  # Check if within range
                    ans.append(x)
        return sorted(ans)  # Sort the result

The other languages (Java, C++, Go, TypeScript) follow the same algorithmic approach; only the syntax differs. They all have the same time and space complexity.