README_EN.md

1718. Construct the Lexicographically Largest Valid Sequence

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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.

Python3

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:]

Java

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;
    }
}

C++

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;
    }
};

Go

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:]
}

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