Leetcode 1718: Construct the Lexicographically Largest Valid Sequence
Given an integer n, find a sequence of length 2n - 1 that satisfies the following conditions:
- The integer
1occurs once in the sequence. - Each integer between
2andnoccurs twice in the sequence. - For every integer
ibetween2andn, the distance between the two occurrences ofiis exactlyi. - The sequence should be lexicographically the largest possible sequence.
The sequence is considered lexicographically larger if, at the first position where the two sequences differ, the number in the first sequence is greater.
Problem
Approach
Steps
Complexity
Input: You are given a single integer `n`.
Example: Input: n = 4
Constraints:
• 1 <= n <= 20
Output: Return the lexicographically largest sequence that satisfies the conditions described.
Example: Output: [4, 1, 3, 2, 3, 4, 2]
Constraints:
• The returned sequence will always be valid.
Goal: The goal is to construct a sequence that satisfies the given conditions while ensuring it is lexicographically the largest possible sequence.
Steps:
• 1. Use backtracking to attempt filling the sequence while adhering to the constraints.
• 2. For each integer from `n` to `1`, try placing the number in the sequence and ensuring the required distance between its occurrences.
• 3. If a number can be placed, move to the next index and repeat the process. If the sequence is complete, return it.
Goal: The sequence should be constructed under the following constraints:
Steps:
• The sequence must have a length of `2n - 1`.
• Each number from `2` to `n` must appear exactly twice with a distance of `i` between its occurrences.
• The sequence must be lexicographically largest.
Assumptions:
• It is guaranteed that a valid sequence exists for any input `n`.
• Input: Input: n = 3
• Explanation: The lexicographically largest sequence that satisfies the conditions is `[3, 1, 2, 3, 2]`. It places the largest number (3) first, then places the smaller numbers (1 and 2) in the remaining positions, ensuring the distances are maintained.
• Input: Input: n = 5
• Explanation: The lexicographically largest sequence that satisfies the conditions is `[5, 3, 1, 4, 3, 5, 2, 4, 2]`. This sequence follows the same logic, ensuring the largest numbers come first while adhering to the distances.
Approach: The problem can be solved using a backtracking approach to construct the sequence by filling the positions while ensuring the required distances between numbers and maximizing lexicographical order.
Observations:
• This problem involves filling the sequence while adhering to a strict set of rules, including maintaining the required distances for each number.
• A backtracking approach seems suitable here, as it allows us to try placing each number while considering all possible positions in a systematic way.
Steps:
• 1. Initialize an array of size `2n - 1` with zeros to represent the sequence.
• 2. Use backtracking to try placing each number from `n` down to `1` in the sequence. For each number, ensure it can appear at two positions with the required distance.
• 3. Once a valid sequence is found, return it.
Empty Inputs:
• The input will always be a positive integer `n`, so no handling for empty inputs is required.
Large Inputs:
• The solution should handle `n = 20` efficiently. Ensure that the backtracking process terminates quickly for large inputs.
Special Values:
• For `n = 1`, the sequence will just be `[1]`.
Constraints:
• The backtracking approach must handle the largest cases efficiently within the time constraints.
int n;
vector<int> constructDistancedSequence(int n) {
this-> n = n;
vector<int> ans(2 * n - 1, 0);
vector<int> vis(n + 1, 0);
bt(ans, vis, 0);
return ans;
}
bool bt(vector<int> &ans, vector<int> &vis, int idx) {
if(idx == ans.size()) return true;
if(ans[idx] != 0) return bt(ans, vis, idx + 1);
else {
for(int i = n; i >= 1; i--) {
if(vis[i]) continue;
vis[i] = 1;
ans[idx] = i;
if(i == 1) {
if(bt(ans, vis, idx + 1)) return true;
} else if (idx + i < ans.size() && ans[idx + i] == 0) {
ans[idx + i] = i;
if(bt(ans, vis, idx + 1)) return true;
ans[idx + i] = 0;
}
ans[idx] = 0;
vis[i] = 0;
}
}
return false;
}
1 : Variable Initialization
int n;
Declaring the integer variable `n`, which represents the size of the distanced sequence to be generated.
2 : Function Declaration
vector<int> constructDistancedSequence(int n) {
The function `constructDistancedSequence` is declared, taking an integer `n` as input, and returning a vector of integers representing the constructed sequence.
3 : Variable Assignment
this-> n = n;
Assigning the input value of `n` to the member variable `n` of the current object.
4 : Vector Initialization
vector<int> ans(2 * n - 1, 0);
Initializing a vector `ans` of size `2 * n - 1`, filled with zeros. This will hold the distanced sequence.
5 : Vector Initialization
vector<int> vis(n + 1, 0);
Creating a vector `vis` to track which numbers have been used in the sequence, initialized with zeros.
6 : Backtracking Call
bt(ans, vis, 0);
Calling the backtracking function `bt` to fill the sequence `ans` using the visitation array `vis`, starting from index `0`.
7 : Return Statement
return ans;
Returning the constructed distanced sequence after it has been completed by the backtracking function.
8 : Backtracking Function Declaration
bool bt(vector<int> &ans, vector<int> &vis, int idx) {
Declaring the recursive backtracking function `bt`, which attempts to construct the sequence by filling the vector `ans` while ensuring each number `i` appears at positions `idx` and `idx + i`.
9 : Base Case Check
if(idx == ans.size()) return true;
The base case of the recursion, returning `true` if the entire sequence has been filled correctly.
10 : Recursive Call
if(ans[idx] != 0) return bt(ans, vis, idx + 1);
If the current index `idx` is already filled, the function moves on to the next index by recursively calling `bt`.
11 : Loop Through Numbers
for(int i = n; i >= 1; i--) {
Iterating over the numbers from `n` down to `1`, trying to place each number `i` at index `idx` and `idx + i`.
12 : Skip Used Numbers
if(vis[i]) continue;
Skipping the number `i` if it has already been used in the sequence (tracked by the `vis` array).
13 : Set Number Visited
vis[i] = 1;
Marking the number `i` as used by setting the corresponding index in the `vis` array to `1`.
14 : Assign Value to Sequence
ans[idx] = i;
Placing the number `i` at the current index `idx` in the sequence `ans`.
15 : Recursive Check for 1
if(i == 1) {
If the number `i` is `1`, attempt to move to the next index and fill the sequence.
16 : Backtracking Call
if(bt(ans, vis, idx + 1)) return true;
Recursively calling `bt` to check if the current assignment leads to a valid sequence.
17 : Check for Valid Pairing
} else if (idx + i < ans.size() && ans[idx + i] == 0) {
If `i` is greater than `1`, checking if the position `idx + i` is available for placing `i`.
18 : Assign Value to Pairing Position
ans[idx + i] = i;
Placing the number `i` at the position `idx + i` in the sequence `ans`.
19 : Backtracking Call for Pairing
if(bt(ans, vis, idx + 1)) return true;
Recursively calling `bt` to check if the current assignment leads to a valid sequence.
20 : Undo Assignment
ans[idx + i] = 0;
Undoing the assignment if the current choice does not lead to a valid sequence.
21 : Undo Assignment
ans[idx] = 0;
Undoing the assignment of `i` at index `idx` if it does not lead to a valid sequence.
22 : Reset Visitation
vis[i] = 0;
Resetting the visitation state for number `i` in the `vis` array.
23 : Final Return
return false;
Returning `false` if no solution has been found after trying all possibilities.
Best Case: O(n), when the solution is found early in the backtracking process.
Average Case: O(n!) in the worst case for larger values of `n` due to the backtracking approach.
Worst Case: O(n!) due to the nature of the backtracking algorithm exploring all possible placements for each number.
Description: The time complexity is dominated by the backtracking search for placing numbers in the sequence.
Best Case: O(n), if the solution is found early and requires minimal backtracking.
Worst Case: O(n), as we only need to store the sequence of length `2n - 1` and a few additional variables for backtracking.
Description: The space complexity is linear in terms of the size of the sequence.
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