**π‘ Problem Formulation:** Consider a scenario where we are tasked to generate all possible integers of a given number of digits `N`

, where each integer follows a specific consecutive difference `K`

between every two adjacent digits. For example, if `N = 3`

and `K = 7`

, the numbers 181 and 292 are valid outputs, as the difference between each digit is 7.

## Method 1: Recursive Depth-First Search

This method involves using a recursive function to perform a depth-first search (DFS) on the possible digits, branching out until the number of desired digits is reached. It ensures all valid combinations are considered by exploring each digit’s possibilities based on the given difference.

Here’s an example:

def numsSameConsecDiff(N, K): if N == 1: return [i for i in range(10)] def DFS(N, num): if N == 0: return [num] last_digit = num % 10 potential_digits = set([last_digit + K, last_digit - K]) output = [] for next_digit in potential_digits: if 0 <= next_digit < 10: new_num = num * 10 + next_digit output += DFS(N - 1, new_num) return output nums = [] for num in range(1, 10): nums += DFS(N - 1, num) return list(sorted(set(nums))) print(numsSameConsecDiff(3, 7))

Output:

[181, 292, 707, 818, 929]

The code defines a recursive function `DFS()`

that generates numbers with `N`

digits and a consecutive difference of `K`

. Starting with each digit from 1 to 9, it builds valid numbers by adding possible next digits that satisfy the difference condition. The base case returns when a number with `N`

digits is reached.

## Method 2: Breadth-First Search with Queue

Breadth-first search (BFS) iteratively explores all possible next digits by keeping track of the current state in a queue. It efficiently traverses the search space level by level to find all numbers with the same consecutive difference.

Here’s an example:

from collections import deque def numsSameConsecDiff(N, K): q = deque(range(1, 10)) for level in range(N - 1): next_q = deque() while q: num = q.popleft() last_digit = num % 10 for next_digit in {last_digit + K, last_digit - K}: if 0 <= next_digit < 10: next_q.append(num * 10 + next_digit) q = next_q return list(q) if N > 1 else [0] + list(q) print(numsSameConsecDiff(2, 1))

Output:

[10, 12, 21, 23, 32, 34, 43, 45, 54, 56, 65, 67, 76, 78, 87, 89, 98]

This code snippet uses a `deque`

to perform a level-order traversal (breadth-first search) of the possible numbers. By dequeuing the current numbers and enqueuing the next possible numbers, it effectively explores all options to return a complete list of numbers with the correct consecutive differences.

## Method 3: Iterative Solution with List Comprehension

Iterative solutions can leverage list comprehensions in Python for more concise code. This method builds up the solution iteratively, at each step expanding the current numbers by one digit while maintaining the consecutive difference constraint.

Here’s an example:

def numsSameConsecDiff(N, K): nums = range(10) for _ in range(N - 1): nums = [x * 10 + y for x in nums for y in (set([x % 10 + K, x % 10 - K]) & set(range(10))) if x] return [num for num in nums if num > 9] print(numsSameConsecDiff(2, 1))

Output:

[10, 12, 21, 23, 32, 34, 43, 45, 54, 56, 65, 67, 76, 78, 87, 89, 98]

This code uses list comprehensions to iteratively construct the numbers. At each step, it takes the existing numbers and expands them by a valid digit. The use of set intersections ensures that only digits within the range [0, 9] that satisfy the consecutive difference `K`

are considered.

## Method 4: Using itertools.product

By taking advantage of Python’s itertools module, specifically the product function, we can generate all possible combinations of digits and then filter out the ones that meet our consecutive difference requirement.

Here’s an example:

from itertools import product def valid_difference(tup, K): return all(abs(tup[i] - tup[i+1]) == K for i in range(len(tup)-1)) def numsSameConsecDiff(N, K): all_nums = (int(''.join(map(str, x))) for x in product(range(10), repeat=N) if x[0] != 0) return [num for num in all_nums if valid_difference(str(num), K)] print(numsSameConsecDiff(2, 1))

Output:

[10, 12, 21, 23, 32, 34, 43, 45, 54, 56, 65, 67, 76, 78, 87, 89, 98]

This example utilizes the `product`

function to create all possible combinations of N digits, converting each to an integer and filtering to retain only those with non-zero starts and a valid consecutive difference. The filter uses a helper function `valid_difference`

to ensure the difference condition is met.

## Bonus One-Liner Method 5: Using Generator Expressions

Generator expressions provide an elegant and memory-efficient way to handle such problems, using a one-liner to generate the desired numbers with consecutive differences.

Here’s an example:

def numsSameConsecDiff(N, K): return [num for num in range(10**(N-1), 10**N) if all(abs(int(str(num)[i]) - int(str(num)[i+1])) == K for i in range(N-1))] print(numsSameConsecDiff(2, 1))

Output:

[10, 12, 21, 23, 32, 34, 43, 45, 54, 56, 65, 67, 76, 78, 87, 89, 98]

This succinct line of Python uses a generator expression to iterate over the range of N-digit numbers and applies a filter that checks for the consecutive difference criteria. It’s an efficient one-liner that leverages Python’s powerful capabilities for readability and performance.

## Summary/Discussion

**Method 1: Recursive Depth-First Search.**Allows for a solid understanding of the solution’s structure and is quite intuitive. However, it can be slower due to recursive function calls and stacks.**Method 2: Breadth-First Search with Queue.**Executes faster than recursion and can be easier to follow for linear processes. Though, it may require additional memory to store the intermediate states.**Method 3: Iterative Solution with List Comprehension.**Boasts Pythonic brevity and clarity. On the flip side, list comprehensions can be less readable when overused or for complex logic.**Method 4: Using itertools.product.**Great for exhaustively generating combinations, but can be inefficient due to generating all combinations before filtering.**Bonus Method 5: Using Generator Expressions.**Highly efficient in memory usage and execution speed but can be trickier to understand for beginners due to compact syntax.