Numbers With Same Consecutive Differences
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
Solution
We want to build up the numbers from the leftmost digit to the rightmost.
Since n
is >= 2 and there should not be leading zeros, we will start from one digit (1-9) and call our backtracking helper to construct the next digits.
Here, our path
(num
in the implementation) will store an integer instead of a list to save space and time.
We want to apply backtracking1 template. We fill in the logic:
is_leaf
:start_index == n
, when the constucted number containsn
digits.get_edges
: the next digit can becur_digit +- k
is_valid
:cur_digit + k <= 9
andcur_digit - k >= 0
to avoid the next digit being out of bound. Note that in the implementation, ifk=0
thencur_digit+k == cur_digit - k
, we must avoid the duplicate by checking whetherk=0
.
Implementation
1def numsSameConsecDiff(self, n: int, k: int) -> List[int]:
2 def dfs(start_index, num):
3 if start_index == n: # num is n digits, add to ans
4 ans.append(num)
5 return
6 cur_digit = num % 10 # current digit
7 if (cur_digit - k >= 0):
8 dfs(start_index + 1, num * 10 + (cur_digit - k))
9 if (cur_digit + k <= 9 and k != 0): # avoid duplicate when k = 0
10 dfs(start_index + 1, num * 10 + (cur_digit + k))
11 ans = []
12 for i in range(1, 10): # first digit can be 1-9
13 dfs(1, i)
14 return ans
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