1718. Construct the Lexicographically Largest Valid Sequence

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

Solutions

DFS.

• 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;
}
}

• class Solution {
public:
int n;
vector<int> cnt, path;

vector<int> constructDistancedSequence(int _n) {
n = _n;
cnt.resize(n * 2, 2);
path.resize(n * 2);
cnt[1] = 1;
dfs(1);
vector<int> ans;
for (int i = 1; i < n * 2; ++i) ans.push_back(path[i]);
return ans;
}

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

• 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:]


• 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:]
}