LeetCode #967 — MEDIUM

Numbers With Same Consecutive Differences

Move from brute-force thinking to an efficient approach using backtracking strategy.

The Problem

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.

Note that the integers should not have leading zeros. Integers as 02 and 043 are not allowed.

Example 1:

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.

Example 2:

Input: n = 2, k = 1
Output: [10,12,21,23,32,34,43,45,54,56,65,67,76,78,87,89,98]

Constraints:

2 <= n <= 9
0 <= k <= 9

Step 01

Brute Force Baseline

Problem summary: 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. Note that the integers should not have leading zeros. Integers as 02 and 043 are not allowed.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Backtracking

Example 1

3
7

Example 2

2
1

Step 02

Core Insight

What unlocks the optimal approach

No official hints in dataset. Start from constraints and look for a monotonic or reusable state.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #967: Numbers With Same Consecutive Differences
class Solution {
    private List<Integer> ans = new ArrayList<>();
    private int boundary;
    private int k;

    public int[] numsSameConsecDiff(int n, int k) {
        this.k = k;
        boundary = (int) Math.pow(10, n - 1);
        for (int i = 1; i < 10; ++i) {
            dfs(i);
        }
        return ans.stream().mapToInt(i -> i).toArray();
    }

    private void dfs(int x) {
        if (x >= boundary) {
            ans.add(x);
            return;
        }
        int last = x % 10;
        if (last + k < 10) {
            dfs(x * 10 + last + k);
        }
        if (k != 0 && last - k >= 0) {
            dfs(x * 10 + last - k);
        }
    }
}

// Accepted solution for LeetCode #967: Numbers With Same Consecutive Differences
func numsSameConsecDiff(n int, k int) (ans []int) {
	bounary := int(math.Pow10(n - 1))
	var dfs func(int)
	dfs = func(x int) {
		if x >= bounary {
			ans = append(ans, x)
			return
		}
		last := x % 10
		if last+k < 10 {
			dfs(x*10 + last + k)
		}
		if k > 0 && last-k >= 0 {
			dfs(x*10 + last - k)
		}
	}
	for i := 1; i < 10; i++ {
		dfs(i)
	}
	return
}

# Accepted solution for LeetCode #967: Numbers With Same Consecutive Differences
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

// Accepted solution for LeetCode #967: Numbers With Same Consecutive Differences
struct Solution;

impl Solution {
    fn nums_same_consec_diff(n: i32, k: i32) -> Vec<i32> {
        let n = n as usize;
        let mut cur: Vec<i32> = vec![];
        let mut res = vec![];
        for i in 0..10 {
            if i == 0 && n != 1 {
                continue;
            }
            cur.push(i);
            Self::dfs(1, &mut cur, &mut res, k, n);
            cur.pop();
        }
        res
    }
    fn dfs(start: usize, cur: &mut Vec<i32>, all: &mut Vec<i32>, k: i32, n: usize) {
        if start == n {
            let mut x = 0;
            for i in 0..n {
                x *= 10;
                x += cur[i];
            }
            all.push(x);
        } else {
            for i in 0..10 {
                if (cur[start - 1] - i).abs() != k {
                    continue;
                }
                cur.push(i);
                Self::dfs(start + 1, cur, all, k, n);
                cur.pop();
            }
        }
    }
}

#[test]
fn test() {
    let n = 3;
    let k = 7;
    let res = vec![181, 292, 707, 818, 929];
    assert_eq!(Solution::nums_same_consec_diff(n, k), res);
    let n = 2;
    let k = 1;
    let res = vec![
        10, 12, 21, 23, 32, 34, 43, 45, 54, 56, 65, 67, 76, 78, 87, 89, 98,
    ];
    assert_eq!(Solution::nums_same_consec_diff(n, k), res);
}

// Accepted solution for LeetCode #967: Numbers With Same Consecutive Differences
function numsSameConsecDiff(n: number, k: number): number[] {
    const ans: number[] = [];
    const boundary = 10 ** (n - 1);
    const dfs = (x: number) => {
        if (x >= boundary) {
            ans.push(x);
            return;
        }
        const last = x % 10;
        if (last + k < 10) {
            dfs(x * 10 + last + k);
        }
        if (k > 0 && last - k >= 0) {
            dfs(x * 10 + last - k);
        }
    };
    for (let i = 1; i < 10; i++) {
        dfs(i);
    }
    return ans;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n!)

Space

O(n)

Approach Breakdown

EXHAUSTIVE

O(nⁿ) time

O(n) space

Generate every possible combination without any filtering. At each of n positions we choose from up to n options, giving nⁿ total candidates. Each candidate takes O(n) to validate. No pruning means we waste time on clearly invalid partial solutions.

BACKTRACKING + PRUNING

O(n!) time

O(n) space

Backtracking explores a decision tree, but prunes branches that violate constraints early. Worst case is still factorial or exponential, but pruning dramatically reduces the constant factor in practice. Space is the recursion depth (usually O(n) for n-level decisions).

Shortcut: Backtracking time = size of the pruned search tree. Focus on proving your pruning eliminates most branches.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Missing undo step on backtrack

Wrong move: Mutable state leaks between branches.

Usually fails on: Later branches inherit selections from earlier branches.

Fix: Always revert state changes immediately after recursive call.