1718. Construct the Lexicographically Largest Valid Sequence - Detailed Explanation

Problem Statement

Given an integer n, we need to construct an array (sequence) of length 2*n - 1 that satisfies all of the following conditions:

• The integer 1 occurs once in the sequence.
• Every integer from 2 to n occurs exactly twice in the sequence.
• For each integer i from 2 to n, the distance between its two occurrences is exactly i. Here, the distance between two elements a[i] and a[j] is defined as the absolute difference of their indices, |j - i|.

Among all sequences that meet these rules, we must return the lexicographically largest sequence. Lexicographically larger means that at the first index where two sequences differ, the sequence with the bigger number at that index is considered larger.

Explanation of the Rules:

• The length of the sequence is fixed at 2*n - 1 because we place one 1 and two of each number 2..n (total elements = 1 + 2*(n-1) = 2n - 1).
• If a number i (≥2) is placed at positions p and q in the 0-indexed array, then |q - p| should equal i. For example, if i = 4 and one occurrence is at index 2, the other must be at index 2 + 4 = 6 (distance 4).
• The sequence should be as large as possible in lexicographic order. In other words, we want the leftmost elements to be as large as constraints allow.

Example 1:
Input: n = 3
Output: [3, 1, 2, 3, 2]
Explanation: The sequence length is 5 (which is 2*3-1). Here, 3 appears twice and the distance between the two 3s is 3 (indices 0 and 3 differ by 3). 2 appears twice with a distance of 2 (indices 2 and 4 differ by 2). 1 appears once. This sequence is valid and is lexicographically larger than other valid sequences. For instance, [2, 3, 2, 1, 3] is another valid arrangement, but [3, 1, 2, 3, 2] is lexicographically greater because at the first index they differ, it has 3 vs 2.

Example 2:
Input: n = 5
Output: [5, 3, 1, 4, 3, 5, 2, 4, 2]
Explanation: The output array has length 9 (which is 2*5-1). Check the distance rule for each number:

• 5 appears at indices 0 and 5 (0-indexed), and |5 - 0| = 5
• 4 appears at indices 3 and 7, and |7 - 3| = 4
• 3 appears at indices 1 and 4, and |4 - 1| = 3
• 2 appears at indices 6 and 8, and |8 - 6| = 2
• 1 appears once at index 2.
  This sequence meets all the conditions. It starts with 5, which makes it lexicographically maximum. (Any sequence starting with a number less than 5 would be lexicographically smaller.)

Constraints:

• 1 <= n <= 20
• It is guaranteed that under these constraints, there is always at least one valid sequence. (We do not have to handle the case of no solution.)

Hints

• Think in terms of positions: Instead of trying to place numbers arbitrarily, recognize that each number (except 1) fixes two positions in the sequence. For a number k, if you decide one occurrence at index i, the other must be at index i + k (since the distance must be k). This greatly reduces the possible placements for each number.

• Backtracking (DFS) is a natural approach: You can build the sequence one position at a time, placing numbers and undoing (backtracking) if a placement leads to a dead end. This is essentially a constraint satisfaction problem.

• Greedy choice for lexicographic maximum: To get the lexicographically largest sequence, try to place larger numbers first whenever possible. By starting with the biggest number at the earliest open position, you ensure the leftmost values are as high as they can be. This hint suggests that during backtracking, you should consider numbers in descending order (n down to 1) rather than ascending.

• Use a used marker: Keep track of which numbers have been placed (both occurrences for numbers ≥2, or the single occurrence for 1). You might use a boolean array or bitmask to mark numbers that are already fully placed. This prevents using a number more than allowed.

• Skip filled spots: While constructing the sequence, you can skip indices that are already occupied to focus only on empty positions. This way, you only make decisions at positions that need filling.

Approaches to Solve the Problem

Brute Force Solution (Unfeasible)

A straightforward brute force approach would be: generate all possible sequences of length 2n-1 containing the numbers 1..n with the required frequencies, then check which ones satisfy the distance rule, and finally pick the largest lexicographically. This approach is conceptually simple but computationally impossible for larger n. The number of permutations of the multiset of required numbers is huge (roughly (2n-1)!/(2^{(n-1)}) because numbers 2..n each repeat twice). For even moderate n (e.g. n = 7 gives length 13), this is astronomically large. So, a direct brute force permutation generation is not viable.

A slightly better brute force is to construct the sequence by placing numbers one by one with backtracking without any specific heuristic. For example, you could try to place numbers 1 through n in increasing order (or any order) using recursion: place two 2's with one space between, two 3's with two spaces between, etc., and try all possibilities. This will eventually find all valid sequences. However, without a strategy to prune the search or choose good candidates first, you'll end up exploring a massive search tree. In the worst case, the state space grows super-exponentially (considered around O(n!) in complexity).

Why brute force fails: Even though the constraints (n ≤ 20) might seem small, exploring all arrangements is infeasible. For example, for n = 20, the sequence length is 39, and the number of potential arrangements is astronomical (~10^40). We need to leverage the problem's constraints to cut down the search drastically.

Optimized Backtracking Approach (Greedy DFS)

The feasible solution is to use backtracking with a greedy twist: always place the largest possible number at each step (to ensure the result is lexicographically largest). The idea is to construct the array step by step, filling empty positions, and backtrack whenever a constraint is violated or no further placement is possible.

Key idea: We fill the sequence from left to right. At each empty position, we try to place a number (with its pair, if applicable) that hasn't been used yet. We attempt numbers in descending order (n down to 1) at each step, so that the first valid arrangement we complete will be the lexicographically largest.

Backtracking algorithm outline:

1. Setup: Create an array result of length 2*n - 1 filled with 0 or some placeholder to indicate empties. Also prepare a boolean array (or bit mask) used of size n+1 (indexed by number) to mark which numbers are already placed. Initially, all entries in used are false (nothing placed yet).

2. Recursive function (DFS): Define a recursive function dfs(index) that will try to fill the sequence starting from a given index.

   • If index has reached the end of the array (past the last position), it means the sequence is complete; return true to signal success.
   • If the position at index is already filled (we might have filled it as the second part of a pair when placing a larger number earlier), then call dfs(index+1) to skip to the next position.
   • If the position is empty, then iterate over candidate numbers k from n down to 1:

     • If k is already used (meaning we've placed both occurrences of k already, or the single occurrence for 1), skip it.

     • If k == 1: This number only needs one slot. Place 1 at result[index] (if it's free, which it is in this branch), mark 1 as used, and recursively call dfs(index+1). If the recursion succeeds (returns true), bubble up true (solution found). If it fails, backtrack: un-place the 1 (set that slot back to empty and mark 1 as unused) and try the next number.

     • If k > 1: We need to place two occurrences of k. Check if the slot at index and the slot at index + k are both within bounds and currently empty. The second slot must be index + k because we need a distance of k between them (i.e., indices differ by k). If either position is occupied or out of range, k cannot be placed starting at index. If both positions are free and k is not used yet, place k at result[index] and result[index+k], mark k as used, and recursively call dfs(index+1). If recursion succeeds, return true. If it fails, backtrack: remove k from both positions, mark it as unused, and try the next candidate.

   • If no number can be placed at this index (all attempts failed), return false to signal dead-end, which triggers backtracking to previous indices.

3. Start recursion: Call dfs(0) to begin filling from the start of the array. Because we always try larger numbers first, as soon as we find a valid complete sequence, it is guaranteed to be the lexicographically largest solution (we don't need to search further once found).

4. Result: The array result will be filled with the valid sequence when the algorithm finishes. Return this array.

This approach ensures correctness by exploring all valid configurations via backtracking, but prunes large parts of the search space by enforcing placement rules and marking used numbers. The greedy ordering (descending) does not affect correctness, but it guides the search to the lexicographically largest solution first. Essentially, it finds the solution without needing to try every possibility, by intelligently skipping invalid placements and ordering tries in a way that yields the best sequence early.

Why this approach is efficient enough: Each number is placed at most once (for 1) or once as a pair (for 2..n) in the recursion. The search tree is pruned by two major factors: (a) constraint pruning – we only place numbers where they fit (e.g., we can't place a k if its required paired position is occupied or out of bounds), and (b) used array – we never reuse a number once placed, limiting the branching. The worst-case time complexity is exponential in n, but with n ≤ 20 the backtracking is feasible. In practice, the complexity is around O(2^n) or similar, which for n=20 is on the order of 1 million possibilities. This is a drastic improvement over raw brute force (which would be on the order of n! possibilities in concept).

Step-by-Step Walkthrough with Visual Examples

Let's walk through the algorithm with a concrete example to see how it builds the sequence and uses backtracking. We’ll use n = 5 (from Example 2) because it’s large enough to illustrate backtracking, yet small enough to follow.

• Initial state: result = [_, _, _, _, _, _, _, _, _] (9 positions, indexed 0 to 8, all empty initially). used = [False, False, False, False, False, False] for numbers 0..5 (we won’t use index 0). We start at index = 0.

• Step 1 (index 0): Position 0 is empty. We try the largest unused number:

  • Try 5: not used yet. We need to place two 5s with a distance of 5. That means one at index 0 and another at index 0 + 5 = 5. Index 5 is within bounds (0–8) and currently empty, so this placement is possible. Place 5 at index 0 and index 5. Mark 5 as used.
  • Now result = [5, _, _, _, _, 5, _, _, _], used[5] = True.
  • Recurse to index = 1.

• Step 2 (index 1): Position 1 is empty. Again, try the largest unused number at this point:

  • 5 is already used, so skip 5.
  • Try 4: not used. To place 4, we need index 1 and index 1 + 4 = 5. But index 5 is not free (there's a 5 at index 5). So we cannot place 4 at index 1. Skip 4 for this index.
  • Try 3: not used. For 3, we need index 1 and index 1 + 3 = 4. Index 4 is free, so we can place 3. Put 3 at index 1 and index 4, and mark 3 as used.
  • Now result = [5, 3, _, _, 3, 5, _, _, _], used[3] = True.
  • Recurse to index = 2.

• Step 3 (index 2): Position 2 is empty. Try largest unused:

  • 5 used, skip; 4 is still unused. Try 4: it requires index 2 and index 2 + 4 = 6. Index 6 is free and in range. Good – place 4 at positions 2 and 6. Mark 4 as used.
  • Now result = [5, 3, 4, _, 3, 5, 4, _, _], used[4] = True.
  • Recurse to index = 3.

• Step 4 (index 3): Position 3 is empty. Try largest unused:

  • 5, 4, 3 are used; next is 2 (unused). Try 2: it needs index 3 and index 3 + 2 = 5. Index 5 is occupied (by 5), so we cannot place 2 at index 3. Skip 2.
  • Try 1: not used. 1 needs only one slot, so we can place 1 at index 3. Mark 1 as used.
  • Now result = [5, 3, 4, 1, 3, 5, 4, _, _], used[1] = True.
  • Recurse to index = 4.

• Step 5 (index 4): Index 4 is not empty (we placed a 3 there in Step 2), so skip to index = 5.

• Step 6 (index 5): Index 5 is also filled (with 5 from Step 1), skip to index = 6.

• Step 7 (index 6): Index 6 is filled (with 4 from Step 3), skip to index = 7.

• Step 8 (index 7): Position 7 is empty. Try largest unused:

  • Used so far: 5,4,3,1 are used; the only remaining unused number is 2. Try 2: it requires index 7 and 7 + 2 = 9. Index 9 is out of bounds (our array indices go 0–8). This means we cannot place a 2 starting at index 7 (there isn't a valid position for the second 2).
  • No other unused numbers to try (1 is used, and 2 was the last candidate). This is a dead end – we can't fill position 7 under the current configuration.

At this point, the algorithm will backtrack because we hit a dead end. Backtracking means we undo some of the recent placements and try a different route:

• Backtrack Step (from index 7 failure): The failure at index 7 was caused by having only 2 left but being unable to place it. We go back to the previous index where we had choices – that was at index 3 (Step 4), where we chose to place 1. We'll undo that choice and try something else at index 3.

  • Remove 1 from index 3. Mark 1 as unused again. Now result becomes [5, 3, 4, _, 3, 5, 4, _, _] and used[1] = False.
  • We had already tried 2 at index 3 (it failed) and then 1. After undoing 1, the next candidate (descending) after 1 would be none (since 1 was the smallest). That means no other options at index 3. So we backtrack further.

• Backtrack Step (to index 2): Undo the placement at index 2. We had placed 4 at indices 2 and 6. Remove those: now indices 2 and 6 become empty again, and mark 4 as unused. result reverts to [5, 3, _, _, 3, 5, _, _, _], used[4] = False.

  • Now we are back at index 2 with 3 used, 5 used, 4 now free (since we removed it), 1 free, 2 free. We will try the next option for index 2 (since 4 eventually led to a dead end).
  • At index 2, after skipping used numbers 5 and 3, the next largest to try was 4 (which we did). The next after 4 is 2 (since 3 is used). Try 2 at index 2: it needs index 2 and index 2 + 2 = 4. But index 4 currently has a 3, so we cannot place 2 here (that spot is taken). Skip 2.
  • Next try 1 at index 2: 1 can occupy a single slot. Place 1 at index 2, mark 1 used. Now result = [5, 3, 1, _, 3, 5, _, _, _].
  • Recurse to index = 3 again (with this new configuration).

• Resume (index 3 after backtracking): Now, result is [5, 3, 1, _, 3, 5, _, _, _], and used numbers are 5 and 3 and 1. At index 3 (empty):

  • Try largest unused: 5 used, 4 is free (we backtracked it, so it's available again). Try 4 at index 3: it requires index 3 + 4 = 7. Index 7 is free (note: our current placements are 5 at 0 & 5; 3 at 1 & 4; 1 at 2; so empty spots are 3, 6, 7, 8). Index 7 is within bounds and empty, so we can place 4 now. Put 4 at index 3 and index 7. Mark 4 as used.
  • Now result = [5, 3, 1, 4, 3, 5, _, 4, _], used[4] = True.
  • Recurse to index = 4 (but 4 is filled with 3, skip to 5, skip 5, skip 6).

• Next (index 6): Index 6 is empty (after placing 4, index 6 remained empty). At index 6:

  • Try largest unused: 5,4,3,1 are used; 2 is unused. Try 2 at index 6: it needs index 6 + 2 = 8. Index 8 is free and in range. Good – place 2 at index 6 and index 8. Mark 2 as used.
  • Now result = [5, 3, 1, 4, 3, 5, 2, 4, 2], used[2] = True.
  • Recurse to index = 7. Index 7 is filled (with 4), skip to 8, filled with 2, skip to 9. Index 9 is beyond the array, which means we've filled all positions successfully!

The recursion ends successfully, and we have constructed a valid sequence: [5,3,1,4,3,5,2,4,2]. This matches the expected output. During this process, we placed numbers in descending order whenever possible, and used backtracking to adjust earlier placements when we encountered a dead end. The final sequence is indeed the lexicographically largest valid sequence for n=5.

Visualization of the final sequence construction:

[5, _, _, _, _, 5, _, _, _]         Place 5 at 0 and 5  
[5, 3, _, _, 3, 5, _, _, _]         Place 3 at 1 and 4  
[5, 3, 4, _, 3, 5, 4, _, _]         Place 4 at 2 and 6  
[5, 3, 4, 1, 3, 5, 4, _, _]         Place 1 at 3 (dead end later, backtrack)  
[5, 3, _, _, 3, 5, _, _, _]         Backtrack 1 and 4 placements (index 3 and 2)  
[5, 3, 1, _, 3, 5, _, _, _]         Place 1 at 2 (after backtracking)  
[5, 3, 1, 4, 3, 5, _, 4, _]         Place 4 at 3 and 7  
[5, 3, 1, 4, 3, 5, 2, 4, 2]         Place 2 at 6 and 8 – done!

Each step ensures the distance rule and adjusts to maintain lexicographic order as much as possible. Whenever a placement choice led to an impossible situation (no valid continuation), we backtracked and tried a different option for the previous number.

Code Implementations

Python Solution