LeetCode 3483: Unique 3-Digit Even Numbers Solution

Master LeetCode problem 3483 (Unique 3-Digit Even Numbers), a easy 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.

3483. Unique 3-Digit Even Numbers

Easy

Array Hash Table Recursion Enumeration

Problem Explanation

Explanation:

To solve this problem, we can generate all possible permutations of the given digits and check if each permutation forms a valid 3-digit even number. We need to ensure that each digit is used at most once and there are no leading zeros in the resulting numbers.

Generate all permutations of the given digits using backtracking.
Check if each permutation represents a valid 3-digit even number.
Count the total number of valid 3-digit even numbers.

Time Complexity:

The time complexity of this approach is O(N!) where N is the number of digits in the input array.

Space Complexity:

The space complexity is O(N) for storing the temporary permutation and visited array.

Solution Code

class Solution {
    public int countUniqueThreeDigitEvenNumbers(int[] digits) {
        boolean[] visited = new boolean[digits.length];
        return backtrack(digits, visited, new ArrayList<>(), 0);
    }
    
    private int backtrack(int[] digits, boolean[] visited, List<Integer> curr, int count) {
        if (curr.size() == 3) {
            int num = curr.get(0) * 100 + curr.get(1) * 10 + curr.get(2);
            if (num % 2 == 0) {
                return 1;
            }
            return 0;
        }
        
        int total = 0;
        for (int i = 0; i < digits.length; i++) {
            if (!visited[i]) {
                visited[i] = true;
                curr.add(digits[i]);
                total += backtrack(digits, visited, curr, count);
                curr.remove(curr.size() - 1);
                visited[i] = false;
            }
        }
        
        return total;
    }
}

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