LeetCode 1718. Construct the Lexicographically Largest Valid Sequence

Q: What is the time complexity of the LeetCode 1718 solution?

The time complexity is O(N!) in the worst case due to the backtracking permutations. However, because we greedily place the largest numbers first, the valid lexicographically largest sequence is typically found much faster in practice.

Q: What does lexicographically largest mean?

Lexicographically largest means that when comparing two sequences from left to right, the sequence with the larger number at the first differing position is considered larger. Thus, prioritizing larger numbers at earlier indices ensures the result is optimal.

LeetCode 1718. Construct the Lexicographically Largest Valid Sequence

Constructing the lexicographically largest valid sequence for LeetCode 1718 requires a greedy backtracking algorithm. By iterating from the largest integer $n$ down to $1$, we attempt to place each number at the earliest available index in an array of size $2n-1$. For numbers greater than 1, we must place them at index i and i + num. This systematic depth-first search ensures that the first valid configuration discovered is inherently the lexicographically largest possible sequence.

Understanding the Problem: LeetCode 1718

The problem "Construct the Lexicographically Largest Valid Sequence" presents a fascinating challenge in algorithmic design. Given an integer n, the goal is to construct a sequence of size 2n - 1 containing integers from 1 to n. The core constraints dictate that the number 1 occurs exactly once, while all other numbers from 2 to n occur exactly twice. Furthermore, the distance between the two occurrences of any integer x (where x > 1) must be exactly x.

To maximize the lexicographical value, our objective is to place the largest possible numbers at the very beginning of the sequence. For instance, if n = 3, our available numbers are 3, 2, and 1. We want to place the 3 as early as possible. If placed at index 0, the second 3 must be placed at index 3. Navigating these constraints optimally requires a robust search strategy that can gracefully retreat when a path leads to an invalid state. This is where backtracking proves indispensable.

Greedy Backtracking Strategy and State Space Pruning

To find the lexicographically largest sequence, our algorithm must be inherently greedy. We initialize an array of size 2n - 1 with zeros, representing empty slots. We also maintain a boolean array to track which numbers have already been utilized in our current recursive path.

At each step of our backtracking function, we identify the first empty slot in our sequence array. Instead of iterating from 1 to n, we iterate backwards from n down to 1. This downward loop is the linchpin of our greedy approach. By always trying the largest available number first, we guarantee that the first complete, valid sequence we construct is the lexicographically maximum sequence.

For $num > 1$: We verify if the current slot is empty AND if the slot at current_index + num is also within bounds and empty. If both conditions are met, we place the number in both slots, mark it as used, and recursively call our function for the next index.
For $num = 1$: Since 1 only appears once, we simply place it in the current slot, mark it as used, and proceed recursively.

If a recursive call returns false, it means the current placement led to a dead end. We then execute the "backtrack" step: we remove the numbers from the slots, mark the number as unused, and the loop continues to try the next largest available number. This systematic exploration prunes invalid state spaces efficiently.

TypeScript Implementation and Code Walkthrough

Let's delve into the TypeScript code that brings this greedy backtracking logic to life. The implementation utilizes nested functions to encapsulate the recursive state, keeping the main namespace clean.


function constructDistancedSequence(n: number): number[] {
    const size = 2 * n - 1;
    const result: number[] = new Array(size).fill(0);
    const used: boolean[] = new Array(n + 1).fill(false);
    
    function backtrack(index: number): boolean {
        // Base Case: If we have reached the end of the array, a valid sequence is found.
        if (index === size) return true; 
        
        // Skip indices that have already been populated by previous number placements.
        if (result[index] !== 0) return backtrack(index + 1); 
        
        // Greedily iterate from largest number down to 1.
        for (let num = n; num >= 1; num--) { 
            if (used[num]) continue; // Skip if number is already in the sequence.
            
            if (num === 1) {
                // Place 1 (only requires a single slot).
                result[index] = 1;
                used[1] = true;
                if (backtrack(index + 1)) return true;
                
                // Backtrack: Undo placement if it led to failure.
                used[1] = false;
                result[index] = 0;
            } else {
                // Check if the second required slot is within bounds and available.
                if (index + num < size && result[index + num] === 0) {
                    result[index] = result[index + num] = num;
                    used[num] = true;
                    if (backtrack(index + 1)) return true;
                    
                    // Backtrack: Undo placement of both instances.
                    used[num] = false;
                    result[index] = result[index + num] = 0;
                }
            }
        }
        // If no number can be placed at the current index, this path is invalid.
        return false;
    }
    
    backtrack(0);
    return result;
}

Algorithmic Complexity Analysis

Understanding the performance characteristics of our backtracking algorithm is crucial for evaluating its efficiency on varying input sizes.

Time Complexity: The theoretical worst-case time complexity is $\mathcal{O}(N!)$ due to the nature of backtracking permutations. We are attempting to place $N$ numbers in various combinations. However, the theoretical worst-case is highly misleading here. Because we implement a greedy heuristic—searching from the largest number to the smallest—the algorithm actively prioritizes the optimal path. In practice, the search tree is heavily pruned, and the valid sequence is found exceptionally early. The time complexity is well within acceptable limits for the typical problem constraints (e.g., $N \le 20$).
Space Complexity: The space complexity is $\mathcal{O}(N)$. This encompasses the memory required for the result array of size $2N - 1$, the used boolean array of size $N + 1$, and the maximum depth of the call stack during the recursive backtrack function, which will not exceed $2N - 1$. This linear space requirement is highly efficient and easily scales.

Frequently Asked Questions (FAQ)

What is the time complexity of the LeetCode 1718 solution?

The time complexity is $\mathcal{O}(N!)$ in the worst case due to the backtracking permutations. However, because we greedily place the largest numbers first, the valid lexicographically largest sequence is typically found much faster in practice, often making the performance feel closer to linear or polynomial for small constraints.

Why do we use backtracking to solve this LeetCode problem?

Backtracking allows us to systematically explore all possible sequence combinations while respecting distance constraints. By placing the largest available number at the earliest possible index and backtracking if a conflict occurs down the line, we can guarantee that the first valid sequence completely filled is definitively the lexicographically largest.

What does lexicographically largest mean in this context?

Lexicographically largest means that when comparing two valid sequences from left to right, the sequence with the larger number at the first differing position is considered larger. For example, [5, ...] is lexicographically larger than [4, ...]. Thus, prioritizing larger numbers at earlier indices ensures the result is mathematically optimal.