1718. Construct Lexicographically Largest Valid Sequence Solution

The problem

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

Approach overview

Problem Overview: You need to build an array of length 2n - 1 using numbers from 1 to n. Each number k > 1 must appear exactly twice and the distance between the two occurrences must be k. Number 1 appears once. Among all valid sequences, return the lexicographically largest one.

Approach 1: Backtracking with Greedy Placement (Time: O(n!), Space: O(n))

This is the most common solution and the one typically expected in interviews. Use backtracking to fill the sequence from left to right. At each empty index, try placing numbers from n down to 1. For values greater than 1, you must check whether index i + k is within bounds and also empty before placing both occurrences. Mark the number as used, continue recursion, and revert the placement if it leads to a dead end. Trying larger numbers first guarantees the first valid sequence you find is already the lexicographically largest. The search space is exponential, but pruning invalid placements early keeps it manageable.

Approach 2: Iterative Construction (Time: O(n^2), Space: O(n))

An iterative strategy builds the sequence while scanning the array and attempting placements based on available numbers. Maintain a result array and a structure that tracks which numbers are still unused. When encountering an empty position, try inserting the largest remaining number that satisfies the distance constraint. This approach relies on repeated checks and adjustments as the sequence grows. It works well for moderate n values and avoids recursion overhead, though it still performs many placement validations.

Approach 3: Divide and Conquer Placement (Time: O(n!), Space: O(n))

You can model the problem as recursively solving subsegments of the array. Pick the largest unused number and attempt to place it at valid positions within the current segment. Once placed, the array splits into smaller independent segments that must also satisfy the same constraints. The recursive structure resembles divide-and-conquer: each placement reduces the problem size and creates smaller subproblems. In practice this behaves similarly to backtracking but conceptually emphasizes segment decomposition within the array.

Approach 4: Heap-based Ordering (Time: O(n log n), Space: O(n))

A heap can help prioritize which numbers to attempt first. Store unused numbers in a max-heap so the largest value is always considered before smaller ones. While scanning the sequence, extract the largest candidate and check if its required distance placement is valid. If placement fails, temporarily skip it and reinsert it later. The heap guarantees lexicographic preference toward larger numbers while maintaining efficient retrieval.

Recommended for interviews: The backtracking solution is the expected answer. Interviewers want to see how you model constraints, prune invalid states, and enforce lexicographic ordering by trying larger numbers first. Starting with brute-force placement shows understanding, but implementing efficient backtracking with early validation demonstrates strong problem-solving skills.

Approach 1: Backtracking Solution

This approach involves using a backtracking method to build the sequence from the largest number down to the smallest. We place numbers starting from n and try placing each number in positions that satisfy the distance requirement. If we find a valid placement for a given number, we continue to place the next smaller number. When all numbers are placed correctly, we have a solution. The constraint n ≤ 20 makes this approach feasible.

Python

def construct_largest_sequence(n):
    result = [0] * (2 * n - 1)
    used = [False] * (n + 1)

    def backtrack(position):
        if position == len(result):
            return True
        if result[position] != 0:
            return backtrack(position + 1)

        for num in range(n, 1, -1):
            if not used[num]:
                if position + num < len(result) and result[position] == 0 and result[position + num] == 0:
                    result[position] = result[position + num] = num
                    used[num] = True
                    if backtrack(position + 1):
                        return True
                    used[num] = False
                    result[position] = result[position + num] = 0
        result[position] = 1
        return backtrack(position + 1)

    backtrack(0)
    return result

This Python solution uses a recursive backtracking approach to fill in the sequence from the largest number down to 1. The array `result` is used to store the current sequence configuration, and `used` keeps track of which numbers have been placed. The `backtrack` function attempts to place each number in allowed positions and recurses until a solution is found.

Complexity

Time Complexity: O(n!) due to the combinatorial nature of backtracking.
Space Complexity: O(n) for the result and used arrays.

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Approach 2: Iterative Construction

This approach constructs the sequence iteratively. It tries to place numbers from n down to 2 at allowed positions while ensuring the distance constraint is met. The challenge is to ensure placement satisfies both the distance and the need for the largest lexicographical order. The placement involves checking possible slots and filling iteratively.

C++

#include <vector>
#include <iostream>

std::vector<int> constructLargestSequence(int n) {
    std::vector<int> result(2 * n - 1, 0);

    for (int num = n; num >= 2; --num) {
        for (int i = 0; i + num < result.size(); ++i) {
            if (result[i] == 0 && result[i + num] == 0) {
                result[i] = result[i + num] = num;
                break;
            }
        }
    }

    for (int i = 0; i < result.size(); ++i) {
        if (result[i] == 0) {
            result[i] = 1;
            break;
        }
    }

    return result;
}

This C++ solution iteratively places numbers from n to 2 with a main loop that attempts to place each number in its valid slots by checking indices. The innermost loop ensures that upon finding a valid slot, the number is placed, and iteration continues. Finally, the number 1 is placed in the first open slot.

Complexity

Time Complexity: O(n^2) due to two nested loops traversing the positions.
Space Complexity: O(n) for storing the result sequence.

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Approach 3: Approach 1: Using Divide and Conquer

This approach breaks the problem into smaller, more manageable sub-problems, solves them individually, and combines their results.

One common example is the Merge Sort, where an array is continually split in half, each half is sorted, and then merged back together.

Python

def merge_sort(arr):
    if len(arr) > 1:
        mid = len(arr) // 2
        L = arr[:mid]
        R = arr[mid:]
        merge_sort(L)
        merge_sort(R)
        i = j = k = 0
        while i < len(L) and j < len(R):  
            if L[i] < R[j]:
                arr[k] = L[i]
                i += 1
            else:
                arr[k] = R[j]
                j += 1
            k += 1

        while i < len(L):
            arr[k] = L[i]
            i += 1
            k += 1

        while j < len(R):
            arr[k] = R[j]
            j += 1
            k += 1

def print_list(arr):
    for i in range(len(arr)):
        print(arr[i], end=' ')
    print()

if __name__ == '__main__':
    arr = [12, 11, 13, 5, 6, 7]
    print("Given array is")
    print_list(arr)
    merge_sort(arr)
    print("Sorted array is")
    print_list(arr)

Java

import java.util.Arrays;

class MergeSort {
    void merge(int arr[], int l, int m, int r) {
        int n1 = m - l + 1;
        int n2 = r - m;
        int L[] = new int[n1];
int R[] = new int[n2];
        for (int i = 0; i < n1; ++i)
            L[i] = arr[l + i];
        for (int j = 0; j < n2; ++j)
            R[j] = arr[m + 1 + j];
        int i = 0, j = 0;
        int k = l;
        while (i < n1 && j < n2) {
            if (L[i] <= R[j]) {
                arr[k] = L[i];
                i++;
            } else {
                arr[k] = R[j];
                j++;
            }
            k++;
        }
        while (i < n1) {
            arr[k] = L[i];
            i++;
            k++;
        }
        while (j < n2) {
            arr[k] = R[j];
            j++;
            k++;
        }
    }

    void sort(int arr[], int l, int r) {
        if (l < r) {
            int m = (l + r) / 2;
            sort(arr, l, m);
            sort(arr, m + 1, r);
            merge(arr, l, m, r);
        }
    }

    static void printArray(int arr[]) {
        int n = arr.length;
        for (int i = 0; i < n; ++i)
            System.out.print(arr[i] + " ");
        System.out.println();
    }

    public static void main(String args[]) {
        int arr[] = {12, 11, 13, 5, 6, 7};
        System.out.println("Given Array");
        printArray(arr);

        MergeSort ob = new MergeSort();
        ob.sort(arr, 0, arr.length - 1);

        System.out.println("Sorted array");
        printArray(arr);
    }
}

C++

#include <iostream>
#include <vector>
using namespace std;

void merge(vector<int>& arr, int l, int m, int r) {
    int i, j, k;
    int n1 = m - l + 1;
    int n2 = r - m;
    vector<int> L(n1), R(n2);
    for (i = 0; i < n1; i++)
        L[i] = arr[l + i];
    for (j = 0; j < n2; j++)
        R[j] = arr[m + 1 + j];
    i = 0;
j = 0;
    k = l;
    while (i < n1 && j < n2) {
        if (L[i] <= R[j]) {
            arr[k] = L[i];
            i++;
        } else {
            arr[k] = R[j];
            j++;
        }
        k++;
    }
    while (i < n1) {
        arr[k] = L[i];
        i++;
        k++;
    }
    while (j < n2) {
        arr[k] = R[j];
        j++;
        k++;
    }
}

void mergeSort(vector<int>& arr, int l, int r) {
    if (l < r) {
        int m = l + (r - l) / 2;
        mergeSort(arr, l, m);
        mergeSort(arr, m + 1, r);
        merge(arr, l, m, r);
    }
}

void printArray(const vector<int>& A) {
    for (auto x : A)
        cout << x << " ";
    cout << endl;
}

int main() {
    vector<int> arr = {12, 11, 13, 5, 6, 7};
    cout << "Given array is \n";
    printArray(arr);

    mergeSort(arr, 0, arr.size() - 1);

    cout << "\nSorted array is \n";
    printArray(arr);
    return 0;
}

JavaScript

function mergeSort(arr) {
  if (arr.length <= 1) {
    return arr;
  }
  const mid = Math.floor(arr.length / 2);
  const left = mergeSort(arr.slice(0, mid));
  const right = mergeSort(arr.slice(mid));
  return merge(left, right);
}

function merge(left, right) {
  let resultArray = [], leftIndex = 0, rightIndex = 0;
  while (leftIndex < left.length && rightIndex < right.length) {
    if (left[leftIndex] < right[rightIndex]) {
      resultArray.push(left[leftIndex]);
      leftIndex++;
    } else {
      resultArray.push(right[rightIndex]);
      rightIndex++;
    }
  }
  return resultArray.concat(left.slice(leftIndex)).concat(right.slice(rightIndex));
}

const arr = [12, 11, 13, 5, 6, 7];
console.log("Given array is");
console.log(arr);
const sortedArray = mergeSort(arr);
console.log("Sorted array is");
console.log(sortedArray);

#include <stdio.h>

void merge(int arr[], int l, int m, int r) {
    int i, j, k;
    int n1 = m - l + 1;
    int n2 = r - m;
    int L[n1], R[n2];
    for (i = 0; i < n1; i++)
        L[i] = arr[l + i];
    for (j = 0; j < n2; j++)
        R[j] = arr[m + 1 + j];
    i = 0;
j = 0;
    k = l;
    while (i < n1 && j < n2) {
        if (L[i] <= R[j]) {
            arr[k] = L[i];
            i++;
        } else {
            arr[k] = R[j];
            j++;
        }
        k++;
    }
    while (i < n1) {
        arr[k] = L[i];
        i++;
        k++;
    }
    while (j < n2) {
        arr[k] = R[j];
        j++;
        k++;
    }
}

void mergeSort(int arr[], int l, int r) {
    if (l < r) {
        int m = l + (r - l) / 2;
        mergeSort(arr, l, m);
        mergeSort(arr, m + 1, r);
        merge(arr, l, m, r);
    }
}

void printArray(int A[], int size) {
    int i;
    for (i = 0; i < size; i++)
        printf("%d ", A[i]);
    printf("\n");
}

int main() {
    int arr[] = {12, 11, 13, 5, 6, 7};
    int arr_size = sizeof(arr) / sizeof(arr[0]);

    printf("Given array is \n");
    printArray(arr, arr_size);

    mergeSort(arr, 0, arr_size - 1);

    printf("\nSorted array is \n");
    printArray(arr, arr_size);
    return 0;
}

using System;

class MergeSort {
    static void Merge(int[] arr, int l, int m, int r) {
        int n1 = m - l + 1;
        int n2 = r - m;
        int[] L = new int[n1];
        int[] R = new int[n2];
        Array.Copy(arr, l, L, 0, n1);
        Array.Copy(arr, m + 1, R, 0, n2);
        int i = 0, j = 0, k = l;
        while (i < n1 && j < n2) {
            if (L[i] <= R[j]) {
                arr[k] = L[i];
                i++;
            } else {
                arr[k] = R[j];
                j++;
            }
            k++;
        }
        while (i < n1) {
            arr[k] = L[i];
            i++;
            k++;
        }
        while (j < n2) {
            arr[k] = R[j];
            j++;
            k++;
        }
    }

    static void MergeSortFunc(int[] arr, int l, int r) {
        if (l < r) {
            int m = l + (r - l) / 2;
            MergeSortFunc(arr, l, m);
            MergeSortFunc(arr, m + 1, r);
            Merge(arr, l, m, r);
        }
    }

    static void PrintArray(int[] arr) {
        foreach (int item in arr)
            Console.Write(item + " ");
        Console.WriteLine();
    }

    public static void Main() {
        int[] arr = {12, 11, 13, 5, 6, 7};
        Console.WriteLine("Given Array");
        PrintArray(arr);
        MergeSortFunc(arr, 0, arr.Length - 1);
        Console.WriteLine("Sorted array is");
        PrintArray(arr);
    }
}

This Python code implements merge sort using lists. The function merge_sort recursively divides the list and merges it back in order.

Complexity

Time Complexity: O(n log n) because of recursive splitting followed by merging.
Space Complexity: O(n) due to extra space for left and right parts during merge.

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Approach 4: Approach 2: Using a Heap-based Sort

This approach leans on using a heap data structure to facilitate sorting. Known as Heap Sort, it involves build-heap operations and sorting through the extract-max process repeatedly.

Commonly used for its in-place sort feature with reliable time complexities.

Python

def heapify(arr, n, i):
    largest = i
    l = 2 * i + 1
    r = 2 * i + 2
    if l < n and arr[l] > arr[largest]:
        largest = l
    if r < n and arr[r] > arr[largest]:
        largest = r
    if largest != i:
        arr[i], arr[largest] = arr[largest], arr[i]
        heapify(arr, n, largest)

def heap_sort(arr):
    n = len(arr)
    for i in range(n // 2 - 1, -1, -1):
        heapify(arr, n, i)
    for i in range(n - 1, 0, -1):
        arr[i], arr[0] = arr[0], arr[i]
        heapify(arr, i, 0)

def print_list(arr):
    for i in range(len(arr)):
        print(arr[i], end=" ")
    print()

if __name__ == '__main__':
    arr = [12, 11, 13, 5, 6, 7]
    heap_sort(arr)
    print("Sorted array is ")
    print_list(arr)

Java

import java.util.Arrays;

public class HeapSort {
    public void sort(int arr[]) {
        int n = arr.length;
        for (int i = n / 2 - 1; i >= 0; i--)
            heapify(arr, n, i);
        for (int i = n - 1; i > 0; i--) {
            int temp = arr[0];
            arr[0] = arr[i];
            arr[i] = temp;
            heapify(arr, i, 0);
        }
    }

    void heapify(int arr[], int n, int i) {
        int largest = i;
        int l = 2 * i + 1;
        int r = 2 * i + 2;
        if (l < n && arr[l] > arr[largest])
            largest = l;
        if (r < n && arr[r] > arr[largest])
            largest = r;
        if (largest != i) {
            int swap = arr[i];
            arr[i] = arr[largest];
            arr[largest] = swap;
            heapify(arr, n, largest);
        }
    }

    static void printArray(int arr[]) {
        int n = arr.length;
        for (int i = 0; i < n; ++i)
            System.out.print(arr[i] + " ");
        System.out.println();
    }

    public static void main(String args[]) {
        int arr[] = {12, 11, 13, 5, 6, 7};
        HeapSort ob = new HeapSort();
        ob.sort(arr);
        System.out.println("Sorted array is");
        printArray(arr);
    }
}

C++

#include <iostream>
#include <vector>
using namespace std;

void swap(int& a, int& b) {
    int t = a;
    a = b;
    b = t;
}

void heapify(vector<int>& arr, int n, int i) {
    int largest = i;
    int l = 2 * i + 1;
    int r = 2 * i + 2;
    if (l < n && arr[l] > arr[largest])
        largest = l;
    if (r < n && arr[r] > arr[largest])
        largest = r;
    if (largest != i) {
        swap(arr[i], arr[largest]);
        heapify(arr, n, largest);
    }
}

void heapSort(vector<int>& arr) {
    int n = arr.size();
    for (int i = n / 2 - 1; i >= 0; i--)
        heapify(arr, n, i);
    for (int i = n - 1; i > 0; i--) {
        swap(arr[0], arr[i]);
        heapify(arr, i, 0);
    }
}

void printArray(vector<int>& arr) {
    for (int x : arr)
        cout << x << " ";
    cout << endl;
}

int main() {
    vector<int> arr = {12, 11, 13, 5, 6, 7};
    heapSort(arr);
    cout << "Sorted array is \n";
    printArray(arr);
    return 0;
}

JavaScript

function heapSort(arr) {
  let n = arr.length;
  for (let i = Math.floor(n / 2) - 1; i >= 0; i--)
    heapify(arr, n, i);
  for (let i = n - 1; i > 0; i--) {
    let temp = arr[0];
    arr[0] = arr[i];
    arr[i] = temp;
    heapify(arr, i, 0);
  }
}

function heapify(arr, n, i) {
  let largest = i;
  let l = 2 * i + 1;
  let r = 2 * i + 2;
  if (l < n && arr[l] > arr[largest])
    largest = l;
  if (r < n && arr[r] > arr[largest])
    largest = r;
  if (largest !== i) {
    let swap = arr[i];
    arr[i] = arr[largest];
    arr[largest] = swap;
    heapify(arr, n, largest);
  }
}

const arr = [12, 11, 13, 5, 6, 7];
heapSort(arr);
console.log("Sorted array is");
console.log(arr);

#include <stdio.h>

void swap(int *a, int *b) {
    int t = *a;
    *a = *b;
    *b = t;
}

void heapify(int arr[], int n, int i) {
    int largest = i;
    int l = 2 * i + 1;
    int r = 2 * i + 2;
    if (l < n && arr[l] > arr[largest])
        largest = l;
    if (r < n && arr[r] > arr[largest])
        largest = r;
    if (largest != i) {
        swap(&arr[i], &arr[largest]);
        heapify(arr, n, largest);
    }
}

void heapSort(int arr[], int n) {
    for (int i = n / 2 - 1; i >= 0; i--)
        heapify(arr, n, i);
    for (int i = n - 1; i > 0; i--) {
        swap(&arr[0], &arr[i]);
        heapify(arr, i, 0);
    }
}

void printArray(int arr[], int n) {
    for (int i = 0; i < n; ++i)
        printf("%d ", arr[i]);
    printf("\n");
}

int main() {
    int arr[] = {12, 11, 13, 5, 6, 7};
    int n = sizeof(arr) / sizeof(arr[0]);
    heapSort(arr, n);
    printf("Sorted array is \n");
    printArray(arr, n);
}

using System;

class HeapSort {
    public void Sort(int[] arr) {
        int n = arr.Length;
        for (int i = n / 2 - 1; i >= 0; i--)
            Heapify(arr, n, i);
        for (int i = n - 1; i > 0; i--) {
            int temp = arr[0];
            arr[0] = arr[i];
            arr[i] = temp;
            Heapify(arr, i, 0);
        }
    }

    void Heapify(int[] arr, int n, int i) {
        int largest = i;
        int l = 2 * i + 1;
        int r = 2 * i + 2;
        if (l < n && arr[l] > arr[largest])
            largest = l;
        if (r < n && arr[r] > arr[largest])
            largest = r;
        if (largest != i) {
            int swap = arr[i];
            arr[i] = arr[largest];
            arr[largest] = swap;
            Heapify(arr, n, largest);
        }
    }

    static void PrintArray(int[] arr) {
        foreach (int i in arr)
            Console.Write(i + " ");
        Console.WriteLine();
    }

    public static void Main() {
        int[] arr = {12, 11, 13, 5, 6, 7};
        HeapSort ob = new HeapSort();
        ob.Sort(arr);
        Console.WriteLine("Sorted array is");
        PrintArray(arr);
    }
}

In Python, heap sort is achieved through recursive heapify functions alongside a primary sorting loop. The algorithm builds a max heap and sorts by swapping and heapifying top elements.

Complexity

Time Complexity: O(n log n) due to heap operations for sorting.
Space Complexity: O(1), as the algorithm stores objects in place.

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Approach 5: Default Approach

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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Complexity comparison

Approach	Complexity
Backtracking Solution	Time Complexity: O(n!) due to the combinatorial nature of backtracking. Space Complexity: O(n) for the result and used arrays.
Iterative Construction	Time Complexity: O(n^2) due to two nested loops traversing the positions. Space Complexity: O(n) for storing the result sequence.
Approach 1: Using Divide and Conquer	Time Complexity: O(n log n) because of recursive splitting followed by merging. Space Complexity: O(n) due to extra space for left and right parts during merge.
Approach 2: Using a Heap-based Sort	Time Complexity: O(n log n) due to heap operations for sorting. Space Complexity: O(1), as the algorithm stores objects in place.
Default Approach	–

Complexity analysis

Approach	Time	Space	When to Use
Backtracking with Greedy Placement	O(n!)	O(n)	Best general solution. Guarantees lexicographically largest sequence and commonly expected in interviews.
Iterative Construction	O(n^2)	O(n)	Useful when avoiding recursion or when implementing incremental construction logic.
Divide and Conquer Placement	O(n!)	O(n)	Conceptual recursive decomposition of the array into smaller segments.
Heap-based Ordering	O(n log n)	O(n)	When prioritizing larger numbers efficiently using a heap structure.

Frequently asked questions

Is Construct the Lexicographically Largest Valid Sequence easy or hard?

The problem is rated Medium on LeetCode but can feel closer to Hard if you are new to backtracking. The challenge comes from enforcing the distance constraint while also maximizing lexicographic order. Once you recognize the backtracking pattern and greedy ordering, the implementation becomes straightforward.

Construct the Lexicographically Largest Valid Sequence Python/Java solution

In Python or Java, the solution usually implements recursive backtracking. Iterate through positions in the array and attempt to place numbers from n down to 1 while checking the distance constraint. If placement fails later, revert the changes and try the next candidate number.

How to solve Construct the Lexicographically Largest Valid Sequence in O(n)?

A strict O(n) solution generally isn't achievable because the problem requires exploring multiple placements to satisfy distance constraints. Practical solutions rely on backtracking with pruning, which behaves efficiently for the small constraints used in the problem. The algorithm prioritizes larger numbers first so the first valid result is already the lexicographically largest.

What is the best approach for Construct the Lexicographically Largest Valid Sequence?

The best approach is backtracking with greedy placement. Fill the array from left to right and always try larger numbers first to maximize lexicographic order. When placing a number k greater than 1, ensure positions i and i + k are both empty before placing it. This approach explores valid configurations while pruning invalid states early.

Is Construct the Lexicographically Largest Valid Sequence asked at Google/Amazon/Meta?

Problems involving constraint placement and backtracking patterns appear frequently in interviews at companies like Google, Amazon, and Meta. Variations of distance-constrained array construction test recursion, pruning strategies, and greedy ordering. This problem is a good example of combining backtracking with lexicographic optimization.

What data structure is used in Construct the Lexicographically Largest Valid Sequence?

The main structure is an array of size 2n - 1 that stores the final sequence. Additional structures typically include a boolean visited array to track which numbers have been used and recursion for backtracking. Some variations also use a heap to prioritize larger numbers first.

What is the time complexity of Construct the Lexicographically Largest Valid Sequence?

The typical backtracking solution runs in O(n!) time in the worst case because it explores permutations of placements. However, strong pruning reduces the actual search space significantly. The space complexity is O(n) due to the recursion stack and the result array.

Construct the Lexicographically Largest Valid Sequence solution and explanation

The problem

Approach overview

Approach 1: Backtracking Solution

Approach 2: Iterative Construction

Approach 3: Approach 1: Using Divide and Conquer

Approach 4: Approach 2: Using a Heap-based Sort

Approach 5: Default Approach

Complexity comparison

Complexity analysis

Frequently asked questions

Comments

Construct the Lexicographically Largest Valid Sequence solution and explanation

The problem

Approach overview

Approach 1: Backtracking Solution

Approach 2: Iterative Construction

Approach 3: Approach 1: Using Divide and Conquer

Approach 4: Approach 2: Using a Heap-based Sort

Approach 5: Default Approach

Complexity comparison

Complexity analysis

Frequently asked questions

Comments

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