Solution For Construct The Lexicographically Largest Valid Sequence

Problem statement:

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

1. The integer 1 occurs once in the sequence.
2. Each integer between 2 and n occurs twice in the sequence.
3. For every integer i between 2 and n, the distance between the two occurrences of i is exactly i.
4. 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:

1. Create an empty result array of size 2n-1.
2. Initialize the first element of the result array to 1.
3. Initialize two pointers l and r to 0 and 1 respectively.
4. 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.
5. 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 IList LargestDivisibleSubset(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;
    }
}