LeetCode 3406: Find the Lexicographically Largest String From the Box II Solution

Master LeetCode problem 3406 (Find the Lexicographically Largest String From the Box II), 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.

3406. Find the Lexicographically Largest String From the Box II

Hard

Two Pointers String

Problem Explanation

Explanation

This problem can be solved using a backtracking approach. The idea is to iterate through all possible combinations of characters from the boxes and append them to form a string. We need to find the lexicographically largest string that can be formed using the characters in the boxes.

The algorithm involves choosing a character from the boxes and recursively exploring all possible combinations by backtracking. At each step, we choose a character from the boxes, append it to the current string, and continue exploring all possible combinations until all characters are used. We keep track of the largest lexicographically string found so far and update it if a larger one is found.

To optimize the backtracking process, we can first sort the characters in the boxes in descending order to prioritize choosing larger characters first.

Time complexity: O(N!), where N is the number of characters in the boxes. Space complexity: O(N) for the recursion stack.

Solution Code

class Solution {
    String res = "";
    
    public String largestNumber(int[] boxes) {
        Arrays.sort(boxes);
        backtrack(boxes, new StringBuilder(), new boolean[boxes.length]);
        return res;
    }
    
    private void backtrack(int[] boxes, StringBuilder curr, boolean[] used) {
        if (curr.length() > res.length()) {
            res = curr.toString();
        }
        
        for (int i = 0; i < boxes.length; i++) {
            if (used[i] || (i > 0 && boxes[i] == boxes[i - 1] && !used[i - 1])) {
                continue;
            }
            used[i] = true;
            curr.append(boxes[i]);
            backtrack(boxes, curr, used);
            curr.deleteCharAt(curr.length() - 1);
            used[i] = false;
        }
    }
}

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