LeetCode 3470: Permutations IV Solution

Master LeetCode problem 3470 (Permutations IV), 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.

3470. Permutations IV

Hard

Array Math Combinatorics Enumeration

Problem Explanation

Explanation:

To solve this problem, we can use backtracking to generate all possible alternating permutations of the first n positive integers. We can maintain a boolean array to keep track of which elements have been used in the current permutation. At each step, we try to add the next element, ensuring that it alternates in parity with the previous element.

We can sort the permutations in lexicographical order and return the k-th permutation if it exists.

Algorithm:

Create a boolean array to mark which elements have been used.
Use backtracking to generate all possible alternating permutations.
Sort the permutations lexicographically.
Return the k-th permutation if it exists.

Time Complexity: The time complexity of this approach is O(n!) where n is the number of elements.

Space Complexity: The space complexity is O(n) to store the boolean array and O(n!) for the permutations generated.

Solution Code

class Solution {
    public List<Integer> getPermutation(int n, long k) {
        List<Integer> result = new ArrayList<>();
        int[] nums = new int[n];
        for (int i = 0; i < n; i++) {
            nums[i] = i + 1;
        }
        boolean[] used = new boolean[n];
        backtrack(nums, used, new ArrayList<>(), result);

        if (k > result.size()) {
            return new ArrayList<>();
        }
        return result.get((int) k - 1);
    }

    private void backtrack(int[] nums, boolean[] used, List<Integer> temp, List<List<Integer>> result) {
        if (temp.size() == nums.length) {
            result.add(new ArrayList<>(temp));
        } else {
            for (int i = 0; i < nums.length; i++) {
                if (!used[i] && (temp.isEmpty() || (temp.get(temp.size() - 1) % 2 != nums[i] % 2))) {
                    used[i] = true;
                    temp.add(nums[i]);
                    backtrack(nums, used, temp, result);
                    temp.remove(temp.size() - 1);
                    used[i] = false;
                }
            }
        }
    }
}

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