LeetCode 3269: Constructing Two Increasing Arrays Solution

Master LeetCode problem 3269 (Constructing Two Increasing Arrays), a hard challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

3269. Constructing Two Increasing Arrays

Hard

Array Dynamic Programming

Problem Explanation

Explanation:

Given a positive integer n, construct two arrays a and b such that:

The length of both arrays is n.
Both arrays a and b are strictly increasing (i.e., a[i] < a[i + 1] and b[i] < b[i + 1] for all valid i).
The sum of a and b is the maximum possible.

To achieve this, we can construct the arrays a and b in such a way that they are mirror images of each other. We can start with the maximum possible value for a[0] and b[0]. Then, we can iteratively decrease the values for a[i] and b[i] while increasing the values for a[i+1] and b[i+1]. This way, we can ensure that both arrays are strictly increasing and their sum is maximum.

Solution Code

class Solution {
    public int[][] constructDistancedSequence(int n) {
        int[] result = new int[2 * n - 1];
        boolean[] visited = new boolean[n + 1];
        backtrack(result, visited, 0);
        
        int[][] res = new int[2][n];
        for (int i = 0; i < 2 * n - 1; i++) {
            res[i / n][i % n] = result[i];
        }
        return res;
    }
    
    private boolean backtrack(int[] result, boolean[] visited, int index) {
        if (index == result.length) {
            return true;
        }
        
        if (result[index] != 0) {
            return backtrack(result, visited, index + 1);
        }
        
        for (int i = visited.length - 1; i > 0; i--) {
            if (visited[i]) continue;
            
            if (i == 1) {
                result[index] = i;
                visited[i] = true;
                if (backtrack(result, visited, index + 1)) {
                    return true;
                }
                result[index] = 0;
                visited[i] = false;
            } else if (index + i < result.length && result[index + i] == 0) {
                result[index] = result[index + i] = i;
                visited[i] = true;
                if (backtrack(result, visited, index + 1)) {
                    return true;
                }
                result[index] = result[index + i] = 0;
                visited[i] = false;
            }
        }
        
        return false;
    }
}

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