README.md

2094. Finding 3-Digit Even Numbers

You are given an integer array digits, where each element is a digit. The array may contain duplicates.

You need to find all the unique integers that follow the given requirements:

• The integer consists of the concatenation of three elements from digits in any arbitrary order.
• The integer does not have leading zeros.
• The integer is even.

For example, if the given digits were [1, 2, 3], integers 132 and 312 follow the requirements.

Return a sorted array of the unique integers.

Example 1:

Input: digits = [2,1,3,0]
Output: 102,120,130,132,210,230,302,310,312,320]
Explanation: All the possible integers that follow the requirements are in the output array.
Notice that there are no odd integers or integers with leading zeros.

Example 2:

Input: digits = [2,2,8,8,2]
Output: [222,228,282,288,822,828,882]
Explanation: The same digit can be used as many times as it appears in digits.
In this example, the digit 8 is used twice each time in 288, 828, and 882.

Example 3:

Input: digits = [3,7,5]
Output: []
Explanation: No even integers can be formed using the given digits.

Constraints:

• 3 <= digits.length <= 100
• 0 <= digits[i] <= 9

Solutions (Rust)

1. Solution

impl Solution {
    pub fn find_even_numbers(digits: Vec<i32>) -> Vec<i32> {
        let mut digits = digits;
        let mut ret = vec![];
        digits.sort_unstable();

        for i in 0..digits.len() {
            for j in 0..digits.len() {
                for k in 0..digits.len() {
                    let x = digits[i] * 100 + digits[j] * 10 + digits[k];

                    if i != j && j != k && i != k && x % 2 == 0 && x > *ret.last().unwrap_or(&99) {
                        ret.push(x);
                    }
                }
            }
        }

        ret
    }
}