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

implSolution{pubfnfind_even_numbers(digits:Vec<i32>) -> Vec<i32>{letmut digits = digits;letmut ret = vec![]; digits.sort_unstable();for i in0..digits.len(){for j in0..digits.len(){for k in0..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 }}