LeetCode #1718 — MEDIUM

Construct the Lexicographically Largest Valid Sequence

Move from brute-force thinking to an efficient approach using array strategy.

The Problem

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

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Backtracking

Example 1

Example 2

Core Insight

What unlocks the optimal approach

Heuristic algorithm may work.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #1718: Construct the Lexicographically Largest Valid Sequence
class Solution {
    private int[] path;
    private int[] cnt;
    private int n;

    public int[] constructDistancedSequence(int n) {
        this.n = n;
        path = new int[n * 2];
        cnt = new int[n * 2];
        Arrays.fill(cnt, 2);
        cnt[1] = 1;
        dfs(1);
        int[] ans = new int[n * 2 - 1];
        for (int i = 0; i < ans.length; ++i) {
            ans[i] = path[i + 1];
        }
        return ans;
    }

    private boolean dfs(int u) {
        if (u == n * 2) {
            return true;
        }
        if (path[u] > 0) {
            return dfs(u + 1);
        }
        for (int i = n; i > 1; --i) {
            if (cnt[i] > 0 && u + i < n * 2 && path[u + i] == 0) {
                cnt[i] = 0;
                path[u] = i;
                path[u + i] = i;
                if (dfs(u + 1)) {
                    return true;
                }
                cnt[i] = 2;
                path[u] = 0;
                path[u + i] = 0;
            }
        }
        if (cnt[1] > 0) {
            path[u] = 1;
            cnt[1] = 0;
            if (dfs(u + 1)) {
                return true;
            }
            cnt[1] = 1;
            path[u] = 0;
        }
        return false;
    }
}

// Accepted solution for LeetCode #1718: Construct the Lexicographically Largest Valid Sequence
func constructDistancedSequence(n int) []int {
	path := make([]int, n*2)
	cnt := make([]int, n*2)
	for i := range cnt {
		cnt[i] = 2
	}
	cnt[1] = 1
	var dfs func(u int) bool
	dfs = func(u int) bool {
		if u == n*2 {
			return true
		}
		if path[u] > 0 {
			return dfs(u + 1)
		}
		for i := n; i > 1; i-- {
			if cnt[i] > 0 && u+i < n*2 && path[u+i] == 0 {
				cnt[i] = 0
				path[u], path[u+i] = i, i
				if dfs(u + 1) {
					return true
				}
				cnt[i] = 2
				path[u], path[u+i] = 0, 0
			}
		}
		if cnt[1] > 0 {
			cnt[1] = 0
			path[u] = 1
			if dfs(u + 1) {
				return true
			}
			cnt[1] = 1
			path[u] = 0
		}
		return false
	}
	dfs(1)
	return path[1:]
}

# Accepted solution for LeetCode #1718: Construct the Lexicographically Largest Valid Sequence
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:]

// Accepted solution for LeetCode #1718: Construct the Lexicographically Largest Valid Sequence
struct Solution;

impl Solution {
    fn construct_distanced_sequence(n: i32) -> Vec<i32> {
        let mut nums: Vec<i32> = vec![];
        for i in (2..=n).rev() {
            nums.push(i);
        }
        nums.push(1);
        let n = n as usize;
        let m = 2 * n - 1;
        let mut cur = vec![0; m];
        let mut visited: u32 = (1 << n) - 1;
        Self::dfs(0, &mut cur, &mut visited, &nums, m, n);
        cur
    }

    fn dfs(
        start: usize,
        cur: &mut Vec<i32>,
        visited: &mut u32,
        nums: &[i32],
        m: usize,
        n: usize,
    ) -> bool {
        if start == m {
            true
        } else {
            if cur[start] == 0 {
                for i in 0..n {
                    if 1 << i & *visited != 0 {
                        let distance = nums[i] as usize;
                        if distance == 1 {
                            *visited &= !(1 << i);
                            cur[start] = nums[i];
                            let found = Self::dfs(start + 1, cur, visited, nums, m, n);
                            if found {
                                return true;
                            }
                            cur[start] = 0;
                            *visited |= 1 << i;
                        } else {
                            if start + distance < m && cur[start + distance] == 0 {
                                cur[start] = nums[i];
                                cur[start + distance] = nums[i];
                                *visited &= !(1 << i);
                                let found = Self::dfs(start + 1, cur, visited, nums, m, n);
                                if found {
                                    return true;
                                }
                                cur[start + distance] = 0;
                                cur[start] = 0;
                                *visited |= 1 << i;
                            }
                        }
                    }
                }
                false
            } else {
                Self::dfs(start + 1, cur, visited, nums, m, n)
            }
        }
    }
}

#[test]
fn test() {
    let n = 3;
    let res = vec![3, 1, 2, 3, 2];
    assert_eq!(Solution::construct_distanced_sequence(n), res);
    let n = 5;
    let res = vec![5, 3, 1, 4, 3, 5, 2, 4, 2];
    assert_eq!(Solution::construct_distanced_sequence(n), res);
}

// Accepted solution for LeetCode #1718: Construct the Lexicographically Largest Valid Sequence
// Auto-generated TypeScript example from java.
function exampleSolution(): void {
}
// Reference (java):
// // Accepted solution for LeetCode #1718: Construct the Lexicographically Largest Valid Sequence
// class Solution {
//     private int[] path;
//     private int[] cnt;
//     private int n;
// 
//     public int[] constructDistancedSequence(int n) {
//         this.n = n;
//         path = new int[n * 2];
//         cnt = new int[n * 2];
//         Arrays.fill(cnt, 2);
//         cnt[1] = 1;
//         dfs(1);
//         int[] ans = new int[n * 2 - 1];
//         for (int i = 0; i < ans.length; ++i) {
//             ans[i] = path[i + 1];
//         }
//         return ans;
//     }
// 
//     private boolean dfs(int u) {
//         if (u == n * 2) {
//             return true;
//         }
//         if (path[u] > 0) {
//             return dfs(u + 1);
//         }
//         for (int i = n; i > 1; --i) {
//             if (cnt[i] > 0 && u + i < n * 2 && path[u + i] == 0) {
//                 cnt[i] = 0;
//                 path[u] = i;
//                 path[u + i] = i;
//                 if (dfs(u + 1)) {
//                     return true;
//                 }
//                 cnt[i] = 2;
//                 path[u] = 0;
//                 path[u + i] = 0;
//             }
//         }
//         if (cnt[1] > 0) {
//             path[u] = 1;
//             cnt[1] = 0;
//             if (dfs(u + 1)) {
//                 return true;
//             }
//             cnt[1] = 1;
//             path[u] = 0;
//         }
//         return false;
//     }
// }

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n!)

Space

O(n)

Approach Breakdown

EXHAUSTIVE

O(nⁿ) time

O(n) space

Generate every possible combination without any filtering. At each of n positions we choose from up to n options, giving nⁿ total candidates. Each candidate takes O(n) to validate. No pruning means we waste time on clearly invalid partial solutions.

BACKTRACKING + PRUNING

O(n!) time

O(n) space

Backtracking explores a decision tree, but prunes branches that violate constraints early. Worst case is still factorial or exponential, but pruning dramatically reduces the constant factor in practice. Space is the recursion depth (usually O(n) for n-level decisions).

Shortcut: Backtracking time = size of the pruned search tree. Focus on proving your pruning eliminates most branches.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Missing undo step on backtrack

Wrong move: Mutable state leaks between branches.

Usually fails on: Later branches inherit selections from earlier branches.

Fix: Always revert state changes immediately after recursive call.