# 967. Numbers With Same Consecutive Differences

## Description

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

## Solutions

DFS.

• class Solution {
public int[] numsSameConsecDiff(int n, int k) {
List<Integer> res = new ArrayList<>();
for (int i = 1; i < 10; ++i) {
dfs(n - 1, k, i, res);
}
int[] ans = new int[res.size()];
for (int i = 0; i < res.size(); ++i) {
ans[i] = res.get(i);
}
return ans;
}

private void dfs(int n, int k, int t, List<Integer> res) {
if (n == 0) {
return;
}
int last = t % 10;
if (last + k <= 9) {
dfs(n - 1, k, t * 10 + last + k, res);
}
if (last - k >= 0 && k != 0) {
dfs(n - 1, k, t * 10 + last - k, res);
}
}
}

• class Solution {
public:
vector<int> ans;

vector<int> numsSameConsecDiff(int n, int k) {
for (int i = 1; i < 10; ++i)
dfs(n - 1, k, i);
return ans;
}

void dfs(int n, int k, int t) {
if (n == 0) {
ans.push_back(t);
return;
}
int last = t % 10;
if (last + k <= 9) dfs(n - 1, k, t * 10 + last + k);
if (last - k >= 0 && k != 0) dfs(n - 1, k, t * 10 + last - k);
}
};

• class Solution:
def numsSameConsecDiff(self, n: int, k: int) -> List[int]:
ans = []

def dfs(n, k, t):
if n == 0:
ans.append(t)
return
last = t % 10
if last + k <= 9:
dfs(n - 1, k, t * 10 + last + k)
if last - k >= 0 and k != 0:
dfs(n - 1, k, t * 10 + last - k)

for i in range(1, 10):
dfs(n - 1, k, i)
return ans


• func numsSameConsecDiff(n int, k int) []int {
var ans []int
var dfs func(n, k, t int)
dfs = func(n, k, t int) {
if n == 0 {
ans = append(ans, t)
return
}
last := t % 10
if last+k <= 9 {
dfs(n-1, k, t*10+last+k)
}
if last-k >= 0 && k != 0 {
dfs(n-1, k, t*10+last-k)
}
}

for i := 1; i < 10; i++ {
dfs(n-1, k, i)
}
return ans
}