Leetcode #967: Numbers With Same Consecutive Differences
In this guide, we solve Leetcode #967 Numbers With Same Consecutive Differences in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
Given two integers n and k, return an array of all the integers of length n where the difference between every two consecutive digits is k. You may return the answer in any order.
Quick Facts
- Difficulty: Medium
- Premium: No
- Tags: Breadth-First Search, Backtracking
Intuition
We must explore combinations of choices, but many branches can be pruned early.
Backtracking enumerates valid candidates while keeping the search space under control.
Approach
Use DFS to build candidates step by step, and backtrack when constraints are violated.
Pruning keeps the exploration practical for typical constraints.
Steps:
- Define the decision tree.
- DFS through choices and backtrack.
- Prune invalid paths early.
Example
Input: n = 3, k = 7
Output: [181,292,707,818,929]
Explanation: Note that 070 is not a valid number, because it has leading zeroes.
Python Solution
class Solution:
def numsSameConsecDiff(self, n: int, k: int) -> List[int]:
def dfs(x: int):
if x >= boundary:
ans.append(x)
return
last = x % 10
if last + k <= 9:
dfs(x * 10 + last + k)
if last - k >= 0 and k != 0:
dfs(x * 10 + last - k)
ans = []
boundary = 10 ** (n - 1)
for i in range(1, 10):
dfs(i)
return ans
Complexity
The time complexity is , where represents the set of digits, and in this problem . The space complexity is .
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.