Solution Explanation: Numbers With Same Consecutive Differences

This problem asks to find all integers of length n where the absolute difference between consecutive digits is k. The solution uses Depth-First Search (DFS) or Backtracking to explore all possible number combinations.

Approach:

The core idea is to build numbers digit by digit. We start with a single digit (1-9, excluding 0 to avoid leading zeros). For each digit, we explore two possibilities: adding k and subtracting k. We recursively build the number until we reach the desired length n.

Algorithm (DFS):

1. Base Case: If the current number has n digits, add it to the result list and return.
2. Recursive Step:
   • Get the last digit of the current number.
   • If lastDigit + k is a valid digit (0-9), recursively call the function with the new number formed by appending lastDigit + k.
   • If lastDigit - k is a valid digit (0-9) and k is not 0 (to handle the case where k is 0 and lastDigit - k would be the same as lastDigit), recursively call the function with the new number formed by appending lastDigit - k.

Time Complexity Analysis:

In the worst-case scenario, for each digit, we explore two branches (adding and subtracting k). Since we have n digits, the total number of branches explored is approximately 2n. The initial loop iterates through 9 possible starting digits. Therefore, the time complexity is O(9 * 2n), which simplifies to O(2n). This is because the constant factor 9 is insignificant compared to the exponential growth.

Space Complexity Analysis:

The space complexity is determined by the recursion depth and the size of the result list. The recursion depth is n, and the size of the result list can be at most 9 * 2(n-1) in the worst case. Therefore, the space complexity is O(2n) to account for the recursion stack and the output list.

Code Examples (Python):

class Solution:
    def numsSameConsecDiff(self, n: int, k: int) -> List[int]:
        ans = []
        
        def backtrack(current_num, length):
            if length == n:
                ans.append(current_num)
                return
            
            last_digit = current_num % 10
            next_digits = [last_digit + k, last_digit - k]
            
            for next_digit in next_digits:
                if 0 <= next_digit <= 9:
                    backtrack(current_num * 10 + next_digit, length + 1)
        
        for i in range(1, 10):  # Iterate through possible starting digits
            backtrack(i, 1)
            
        return ans

This Python code directly implements the DFS algorithm described above. The backtrack function performs the recursive exploration. Note that error handling (checking for valid digits) is included within the backtrack function. Other language examples provided previously demonstrate the same approach with minor syntax differences.