How do you solve Numbers With Same Consecutive Differences on LeetCode?

Numbers With Same Consecutive Differences (LeetCode #967) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: The first digit cannot be zero to avoid leading zeros.

What is the time complexity of Numbers With Same Consecutive Differences?

The optimal solution for Numbers With Same Consecutive Differences runs in O(n) time and uses O(n) space. The time complexity is O(n) since we only build valid numbers directly without generating invalid combinations.

What is the brute force solution for Numbers With Same Consecutive Differences?

The brute force approach for Numbers With Same Consecutive Differences is Brute Force, with O(n²) time complexity and O(1) space complexity. We can generate all possible numbers of length n and filter out those that meet the conditions. This is straightforward but inefficient as it checks many invalid numbers. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Numbers With Same Consecutive Differences?

Numbers With Same Consecutive Differences uses the following concepts: Backtracking, Breadth-First Search. The recommended patterns to study are: Backtracking, Depth-First Search.

What are the interview tips for Numbers With Same Consecutive Differences?

Explain your thought process clearly. Consider edge cases, such as k=0 or k=9. Use backtracking for problems involving combinations or permutations.

What are common mistakes when solving Numbers With Same Consecutive Differences?

Not handling leading zeros correctly. Failing to check both possible digits when generating numbers.

#967

Numbers With Same Consecutive Differences

Medium

BacktrackingBreadth-First SearchBacktrackingDepth-First Search

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

💡

Intuition

Time O(n)Space O(n)

We can use a backtracking approach to build valid numbers directly, avoiding the generation of invalid combinations. This is more efficient as it only constructs valid numbers.

⚙️

Algorithm

3 steps

1Step 1: Start with each digit from 1 to 9 as the first digit.
2Step 2: Recursively add digits that differ from the last digit by k.
3Step 3: Stop when the number reaches the desired length n.

solution.py16 lines

1# Full working Python code
2
3def numsSameConsecDiff(n, k):
4    results = []
5    def backtrack(num, length):
6        if length == n:
7            results.append(num)
8            return
9        last_digit = num % 10
10        if last_digit + k < 10:
11            backtrack(num * 10 + (last_digit + k), length + 1)
12        if last_digit - k >= 0:
13            backtrack(num * 10 + (last_digit - k), length + 1)
14    for i in range(1, 10):
15        backtrack(i, 1)
16    return results

ℹ

Complexity note: The time complexity is O(n) since we only build valid numbers directly without generating invalid combinations.

1The first digit cannot be zero to avoid leading zeros.
2The difference between consecutive digits must equal k.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.