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1718. Construct the Lexicographically Largest Valid Sequence

Difficulty Topics

Description

Given an integer n, find a sequence that satisfies all of the following:

  • The integer 1 occurs once in the sequence.
  • Each integer between 2 and n occurs twice in the sequence.
  • For every integer i between 2 and n, the distance between the two occurrences of i is exactly i.

The distance between two numbers on the sequence, a[i] and a[j], is the absolute difference of their indices, |j - i|.

Return the lexicographically largest sequence. It is guaranteed that under the given constraints, there is always a solution.

A sequence a is lexicographically larger than a sequence b (of the same length) if in the first position where a and b differ, sequence a has a number greater than the corresponding number in b. For example, [0,1,9,0] is lexicographically larger than [0,1,5,6] because the first position they differ is at the third number, and 9 is greater than 5.

 

Example 1:

Input: n = 3
Output: [3,1,2,3,2]
Explanation: [2,3,2,1,3] is also a valid sequence, but [3,1,2,3,2] is the lexicographically largest valid sequence.

Example 2:

Input: n = 5
Output: [5,3,1,4,3,5,2,4,2]

 

Constraints:

  • 1 <= n <= 20

Solution

construct-the-lexicographically-largest-valid-sequence.py
class Solution:
    def constructDistancedSequence(self, n: int) -> List[int]:
        res = [0] * (n * 2 - 1)
        used = [0] * (n + 1)

        def backtrack(i):
            if i == len(res): return True

            if res[i]: return backtrack(i + 1)

            for k in range(n, 1, -1):
                if used[k]: continue

                if i + k < len(res) and res[i] == res[i + k] == 0:
                    res[i] = res[i + k] = k
                    used[k] = 1

                    if backtrack(i + 1): return True

                    res[i] = res[i + k] = 0
                    used[k] = 0

            if not used[1]:
                res[i] = 1
                used[1] = 1

                if backtrack(i + 1): return True

                res[i] = 0
                used[1] = 0

            return False

        backtrack(0)

        return res