# Solution For Construct The Lexicographically Largest Valid Sequence

Problem statement:

Given an integer n, find a sequence that satisfies all of the following constraints:

- 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 sequence is lexicographically largest.

Return the lexicographically largest valid sequence. It is guaranteed that under the given constraints, there is always a valid sequence.

Solution:

To construct the lexicographically largest valid sequence, we can start with the largest numbers and work our way down. We can start with the largest number n and place it in the first position. Then, we can place the next largest number n-1 in the position that is i+d, where i is the current position and d is the distance required by the problem (which is i itself). Then we can place n-2 in the next position, and so on. In this way, we can construct a valid sequence that is also lexicographically largest.

Here is a step-by-step algorithm to solve this problem:

- Create an empty result array of size 2n-1.
- Initialize the first element of the result array to 1.
- Initialize two pointers l and r to 0 and 1 respectively.
- For each number i from n to 2, do the following:

a. If i is even, set result[r+i-1] = i, result[l+i-1] = i-1.

b. If i is odd, set result[r+i-1] = i-1, result[l+i-1] = i.

c. Increment l by i-1 and r by i. - Return the result array.

Explanation:

We start with the number n and place it in the first position (i.e., result[0] = n). Then, we need to place the number n-1 at a distance of 2 from the current position. Since the distance required is i itself (i.e., 2), we can place n-1 at index 1, which is i-1 away from the current position. Similarly, we can place n-2 at a distance of 3 from the current position, n-3 at a distance of 4, and so on. We keep track of two pointers l and r, which represent the left and right ends of the current sequence. By incrementing l by i-1 and r by i, we are ensuring that the next pair of values (i.e., i-1 and i) are placed at the required distance from each other.

For example, let’s say n=5. We start with the first element as 5 (i.e., result[0]=5). Then, we need to place 4 and 3 at a distance of 2 and 3 respectively from the first element. Therefore, we can place 4 at index 1 (i.e., result[1]=4) and 3 at index 4 (i.e., result[4]=3). Similarly, we can place 2 and 1 at indices 2 and 6 respectively.

The resulting array would be [5,4,2,3,1,4,5], which satisfies all the constraints of the problem and is also the lexicographically largest sequence possible.

Time Complexity:

The time complexity of the above algorithm is O(n), since we are iterating over all numbers from n to 2 and performing constant time operations inside the loop. Therefore, the algorithm is efficient and can handle large values of n without any issues.

## Step by Step Implementation For Construct The Lexicographically Largest Valid Sequence

public class Solution { public String largestValidSequence(String s) { StringBuilder sb = new StringBuilder(); int open = 0, close = 0; for (int i = 0; i < s.length(); i++) { char c = s.charAt(i); if (c == '(') { open++; } else { close++; } if (open == close) { sb.append(s.substring(0, i + 1)); sb.append(largestValidSequence(s.substring(i + 1))); break; } else if (open < close) { return ""; } } return sb.toString(); } }

Given a set of distinct positive integers, find the largest subset such that every pair (Si, Sj) of elements in this subset satisfies: Si % Sj = 0 or Sj % Si = 0. If there are multiple solutions, return any subset is fine. def largestDivisibleSubset(self, nums): if not nums: return [] nums.sort() dp = [[num] for num in nums] for i in range(len(nums)): for j in range(i): if nums[i] % nums[j] == 0 and len(dp[i]) < len(dp[j]) + 1: dp[i] = dp[j] + [nums[i]] return max(dp, key = len)

var constructLargest = function(s) { // sort the string in reverse order var sorted = s.split('').sort((a, b) => b.charCodeAt(0) - a.charCodeAt(0)); // initialize our stack and result string var stack = []; var result = ''; // iterate through the sorted string for (var i = 0; i < sorted.length; i++) { // if the current character is less than the last character in the stack // we pop it off and add it to the result string if (sorted[i] < stack[stack.length - 1]) { result += stack.pop(); } // otherwise, we push the current character onto the stack else { stack.push(sorted[i]); } } // return the result string + the remaining characters in the stack // (in reverse order) return result + stack.reverse().join(''); };

The following is a C++ solution for the leetcode problem construct-the-lexicographically-largest-valid-sequence: #include#include #include using namespace std; string constructLargestValidSequence(string s) { // sort the string in reverse lexicographical order sort(s.rbegin(), s.rend()); // initialize an empty string to store the largest valid sequence string largestValidSeq = ""; // go through each character in the sorted string for (char c : s) { // if the character is a '(', we need to find its corresponding ')' if (c == '(') { // search for the first ')' after the current '(', // and if found, add the two parentheses to the largest valid sequence int pos = s.find(')', largestValidSeq.size()); if (pos != string::npos) { largestValidSeq += c; largestValidSeq += s[pos]; } } // if the character is a ')', we can simply ignore it else if (c == ')') { continue; } // otherwise, we can add the character to the largest valid sequence else { largestValidSeq += c; } } return largestValidSeq; } int main() { string s = "()(()((("; cout << constructLargestValidSequence(s) << endl; return 0; }

Given a set of distinct positive integers, find the largest subset such that every pair (Si, Sj) of elements in this subset satisfies: Si % Sj = 0 or Sj % Si = 0. If there are multiple solutions, return any subset is fine. class Solution { public IListLargestDivisibleSubset(int[] nums) { // sort the array first Array.Sort(nums); // create an array to track the length of the largest divisible subset at each index int[] dp = new int[nums.Length]; // create an array to track the index of the previous element in the largest divisible subset at each index int[] prev = new int[nums.Length]; // initialize the first element in both arrays dp[0] = 1; prev[0] = -1; // track the overall largest divisible subset int maxIndex = 0; int max = 1; // iterate through the array, starting at the second element for (int i = 1; i < nums.Length; i++) { // iterate through all previous elements for (int j = 0; j < i; j++) { // check if the current element is divisible by the previous element if (nums[i] % nums[j] == 0 && dp[j] + 1 > dp[i]) { // update the length of the largest divisible subset at the current index dp[i] = dp[j] + 1; // update the index of the previous element in the largest divisible subset at the current index prev[i] = j; } } // check if the current largest divisible subset is larger than the overall largest divisible subset if (dp[i] > max) { // update the overall largest divisible subset max = dp[i]; maxIndex = i; } } // create a list to store the elements in the overall largest divisible subset List result = new List (); // iterate through the overall largest divisible subset, starting at the last element while (maxIndex != -1) { // add the current element to the list result.Add(nums[maxIndex]); // move to the previous element in the subset maxIndex = prev[maxIndex]; } // return the list of elements in the overall largest divisible subset return result; } }