Introduction

In this blog post, we will explore how to solve the LeetCode problem "1718" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

Given an integer n, find a sequence with elements in the range [1, n] 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

Explanation

Here's the breakdown of the solution:

• Backtracking with Lexicographical Prioritization: The solution uses a backtracking algorithm to explore possible sequences. It tries to place numbers from n down to 1, prioritizing larger numbers to achieve the lexicographically largest sequence.

• Constraint Checking: Before placing a number, the algorithm checks if the placement satisfies the problem constraints (number range, occurrence count, and distance requirement).

• Optimization via Early Termination: If a valid sequence is found, the backtracking process terminates early to avoid unnecessary exploration.

• Runtime Complexity: O(n!), Storage Complexity: O(n^2)

Code

    def construct_distanced_sequence(n: int) -> list[int]:
    """
    Finds the lexicographically largest sequence with elements in the range [1, n] that satisfies the problem constraints.
    """

    length = 2 * n - 1
    sequence = [0] * length
    used = [False] * (n + 1)

    def backtrack(index: int) -> bool:
        """
        Recursive backtracking function to construct the sequence.
        """
        if index == length:
            return True

        if sequence[index] != 0:
            return backtrack(index + 1)

        for num in range(n, 0, -1):
            if not used[num]:
                if num == 1:
                    sequence[index] = num
                    used[num] = True
                    if backtrack(index + 1):
                        return True
                    used[num] = False
                    sequence[index] = 0  # Backtrack
                else:
                    if index + num < length and sequence[index + num] == 0:
                        sequence[index] = num
                        sequence[index + num] = num
                        used[num] = True
                        if backtrack(index + 1):
                            return True
                        used[num] = False
                        sequence[index] = 0  # Backtrack
                        sequence[index + num] = 0  # Backtrack

        return False

    backtrack(0)
    return sequence