Construct the Lexicographically Largest Valid Sequence

This approach involves finding every possible valid sequence that satisfies the given conditions. We use a standard backtracking algorithm to explore all placement possibilities. After generating all valid sequences, we compare them to find the one that is lexicographically the largest.

• Create a recursive backtracking function, say generateAll(index), to build the sequence.
• Use a boolean array used to keep track of numbers from 1 to n that have been placed in the sequence.
• The recursive function will try to fill a sequence array of size 2n - 1.
• Base Case: If the index reaches the end of the sequence (2n - 1), a valid sequence has been found. Add a copy of it to a list of solutions.
• Recursive Step:
  • If the current position sequence[index] is already filled, recurse on the next index: generateAll(index + 1).
  • Otherwise, iterate through numbers i from 1 to n.
  • For each i that has not been used:
    • Try to place i at sequence[index]. If i > 1, its second occurrence must be placed at index + i.
    • If the placement is valid (i.e., the second position is within bounds and is empty), mark i as used, update the sequence, and make a recursive call: generateAll(index + 1).
    • After the recursive call returns, backtrack by undoing the placement and unmarking i as used.
• After the initial call generateAll(0) completes, iterate through the list of all found solutions to find and return the lexicographically largest one.

The fundamental idea is to treat this as a permutation problem with constraints. We can build a sequence recursively. A helper function, say generateAll(index), attempts to fill the sequence from the given index. It tries placing each unused number i (from 1 to n) at the current position and, if i > 1, its corresponding pair at index + i. If a placement is valid, it recurses to the next position. When a full valid sequence is formed, it's stored in a list. This process continues until all possibilities are exhausted. Finally, the list of solutions is scanned to find the lexicographically greatest sequence.

import java.util.ArrayList;
import java.util.Arrays;
import java.util.List;

class Solution {
    List<int[]> solutions;
    boolean[] used;
    int n;
    int[] currentSequence;

    public int[] constructDistancedSequence(int n) {
        this.n = n;
        this.solutions = new ArrayList<>();
        this.used = new boolean[n + 1];
        this.currentSequence = new int[2 * n - 1];
        
        generateAll(0);
        
        // This part is inefficient as it requires finding all solutions first.
        if (solutions.isEmpty()) return new int[0];

        int[] largest = solutions.get(0);
        for (int i = 1; i < solutions.size(); i++) {
            if (isLexicographicallyLarger(solutions.get(i), largest)) {
                largest = solutions.get(i);
            }
        }
        return largest;
    }

    private void generateAll(int index) {
        if (index == currentSequence.length) {
            solutions.add(Arrays.copyOf(currentSequence, currentSequence.length));
            return;
        }

        if (currentSequence[index] != 0) {
            generateAll(index + 1);
            return;
        }

        // Iterate through numbers in an arbitrary order (e.g., 1 to n)
        for (int i = 1; i <= n; i++) {
            if (used[i]) continue;

            currentSequence[index] = i;
            used[i] = true;

            if (i == 1) {
                generateAll(index + 1);
            } else {
                int secondPos = index + i;
                if (secondPos < currentSequence.length && currentSequence[secondPos] == 0) {
                    currentSequence[secondPos] = i;
                    generateAll(index + 1);
                    currentSequence[secondPos] = 0; // Backtrack
                }
            }
            
            currentSequence[index] = 0; // Backtrack
            used[i] = false;
        }
    }
    
    private boolean isLexicographicallyLarger(int[] a, int[] b) {
        for (int i = 0; i < a.length; i++) {
            if (a[i] > b[i]) return true;
            if (a[i] < b[i]) return false;
        }
        return false;
    }
}