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Asked: 2024-09-24T08:23:45+05:30 In:

You are tasked with finding all the stepping numbers within a given range. A stepping number is defined as a number where every two adjacent digits have an absolute difference of exactly one. Your goal is to identify all stepping numbers that fall between two specified integers, inclusive. For instance, if you are given a lower bound X and an upper bound Y, you need to generate a list of all numbers in that range that qualify as stepping numbers. To illustrate, given the range from 0 to 21, the stepping numbers would be: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12, 21. You can implement a function that efficiently computes the stepping numbers and returns them in ascending order. Consider the constraints for X and Y, which can be as large as 10^9. Make sure your solution can handle edge cases and large values gracefully.

Let’s dive into a fun and challenging problem! Imagine you’re given a task that’s all about finding stepping numbers in a specific range. Now, stepping numbers are pretty cool – they’re numbers where the absolute difference between every two adjacent digits is exactly one. For example, 10 and 12 are valid stepping numbers since, in each case, the differences between the digits are 1 (1 and 0 in 10, and 1 and 2 in 12).

Now, let’s spice things up a bit! Suppose you need to find all the stepping numbers that fall between two given integers, X and Y, where X is the lower bound and Y is the upper bound. Your goal is to generate a complete list of these stepping numbers, including both X and Y if they are stepping numbers themselves.

Here’s a quick example to get you started: if your range is from 0 to 21, the stepping numbers you’d find are 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12, and 21. Pretty neat, right? But things could get tricky. What if X and Y are close to the upper limit, say 10^9? You’d need a solution that runs efficiently even for those big numbers.

How would you go about implementing this? Think about which methods you could use to explore possible stepping numbers efficiently without just brute-forcing through every single number. You need to consider edge cases, too! For example, what if X is greater than Y? Or what if the range contains no stepping numbers at all?

So, can you come up with a function that does this for you? Visualize how you’d tackle generating this list and ensure it’s sorted in ascending order. It’s a fun challenge, and I bet you’ll get a sense of accomplishment when you figure it out! What’s your approach going to be? Would love to hear your thoughts on this!

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    2 Answers

    1. To tackle the problem of finding stepping numbers between a given range X and Y, we can implement a breadth-first search (BFS) algorithm. The BFS approach is particularly beneficial because it allows us to generate potential stepping numbers layer by layer, starting from each digit (0 to 9) and exploring their valid successors based on the stepping number properties. We will enqueue initial stepping numbers (0 through 9) and continuously dequeue numbers while checking if they lie within the range [X, Y]. For each number, we inspect its last digit; valid next digits can be either one less or one more than the last digit. If these new digits fall within the bounds of 0-9, we append them to the current number and enqueue the new number for further exploration until all possibilities are exhausted.

      Additionally, we must account for edge cases such as when X exceeds Y or when there are no valid stepping numbers in the provided range. To handle cases where X > Y, we can simply return an empty list. After gathering potential stepping numbers, we can sort this list to ensure the output is in ascending order. Given the constraints and the possibility of high upper limits (up to 10^9), the BFS will efficiently generate valid numbers without directly iterating through each possible number within the range, thus optimizing performance while ensuring correctness.






    2. Stepping Numbers Challenge

      Finding Stepping Numbers

      So, stepping numbers are these cool numbers where the difference between adjacent digits is always one. Like, 10 and 12 are stepping numbers, but 23 is not (because 2 and 3 are fine, but if you had a 4, 3 and 4 aren’t stepping). And now, I’ve got this task where I need to find all those stepping numbers between X and Y.

      First off, I’d probably want to check if X is greater than Y. If they are flipped, I could just say there are no stepping numbers in that case and stop right there. But if not, here’s what I think I should do!

      I’ll consider creating a function that starts from each digit from 0 to 9 and then builds numbers by adding or subtracting 1 to the last digit. For example:

      • Start from 0, then you can only build up to 1.
      • Start from 1, and you can create 10 or 12 (because 0 and 2 are the only choices).
      • Do this recursively until the number reaches Y.

      Once I’ve got those numbers, I’d then check if they fall between X and Y and add them to my final list. Oh, and I need to keep them sorted. I guess that can be done naturally with the way I’m building them?

      And if I hit some giant numbers like 109, I’d have to be careful not to waste time checking every single number. Going digit by digit is probably way more efficient, right?

      Here’s a quick pseudo-code I was thinking about:

      function findSteppingNumbers(X, Y) {
    if (X > Y) return [];
    
    steppingNums = [];
    
    for (digit in 0 to 9) {
        findSteppingNumbersFrom(digit, X, Y, steppingNums);
    }

    return sort(steppingNums);
}

function findSteppingNumbersFrom(number, X, Y, steppingNums) {
    if (number != 0 && number >= X && number <= Y) {
        steppingNums.push(number);
    }

    lastDigit = number % 10;

    if (lastDigit > 0) {
        findSteppingNumbersFrom(number * 10 + (lastDigit - 1), X, Y, steppingNums);
    }
    if (lastDigit < 9) {
        findSteppingNumbersFrom(number * 10 + (lastDigit + 1), X, Y, steppingNums);
    }
}

      I think this will help me get all the stepping numbers without going through every number in between! And if there are no stepping numbers found, that’s okay too. It’s just part of the challenge! Let’s go for it!


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