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Leetcode #1718: Construct the Lexicographically Largest Valid Sequence

In this guide, we solve Leetcode #1718 Construct the Lexicographically Largest Valid Sequence in Python and focus on the core idea that makes the solution efficient.

You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.

Leetcode

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.

Quick Facts

  • Difficulty: Medium
  • Premium: No
  • Tags: Array, Backtracking

Intuition

We must explore combinations of choices, but many branches can be pruned early.

Backtracking enumerates valid candidates while keeping the search space under control.

Approach

Use DFS to build candidates step by step, and backtrack when constraints are violated.

Pruning keeps the exploration practical for typical constraints.

Steps:

  • Define the decision tree.
  • DFS through choices and backtrack.
  • Prune invalid paths early.

Example

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.

Python Solution

class Solution: def constructDistancedSequence(self, n: int) -> List[int]: def dfs(u): if u == n * 2: return True if path[u]: return dfs(u + 1) for i in range(n, 1, -1): if cnt[i] and u + i < n * 2 and path[u + i] == 0: cnt[i] = 0 path[u] = path[u + i] = i if dfs(u + 1): return True path[u] = path[u + i] = 0 cnt[i] = 2 if cnt[1]: cnt[1], path[u] = 0, 1 if dfs(u + 1): return True path[u], cnt[1] = 0, 1 return False path = [0] * (n * 2) cnt = [2] * (n * 2) cnt[1] = 1 dfs(1) return path[1:]

Complexity

The time complexity is Exponential (worst case). The space complexity is O(depth).

Edge Cases and Pitfalls

Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.

Summary

This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.


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