# 2094. Finding 3-Digit Even Numbers

## Description

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

• class Solution {
public int[] findEvenNumbers(int[] digits) {
int[] counter = count(digits);
List<Integer> ans = new ArrayList<>();
for (int i = 100; i < 1000; i += 2) {
int[] t = new int[3];
for (int j = 0, k = i; k > 0; ++j) {
t[j] = k % 10;
k /= 10;
}
int[] cnt = count(t);
if (check(counter, cnt)) {
}
}
return ans.stream().mapToInt(Integer::valueOf).toArray();
}

private boolean check(int[] cnt1, int[] cnt2) {
for (int i = 0; i < 10; ++i) {
if (cnt1[i] < cnt2[i]) {
return false;
}
}
return true;
}

private int[] count(int[] nums) {
int[] counter = new int[10];
for (int num : nums) {
++counter[num];
}
return counter;
}
}

• class Solution {
public:
vector<int> findEvenNumbers(vector<int>& digits) {
vector<int> counter = count(digits);
vector<int> ans;
for (int i = 100; i < 1000; i += 2) {
vector<int> t(3);
for (int j = 0, k = i; k > 0; ++j) {
t[j] = k % 10;
k /= 10;
}
vector<int> cnt = count(t);
if (check(counter, cnt)) ans.push_back(i);
}
return ans;
}

vector<int> count(vector<int>& nums) {
vector<int> counter(10);
for (int num : nums) ++counter[num];
return counter;
}

bool check(vector<int>& cnt1, vector<int>& cnt2) {
for (int i = 0; i < 10; ++i)
if (cnt1[i] < cnt2[i])
return false;
return true;
}
};

• class Solution:
def findEvenNumbers(self, digits: List[int]) -> List[int]:
ans = []
counter = Counter(digits)
for i in range(100, 1000, 2):
t = []
k = i
while k:
t.append(k % 10)
k //= 10
cnt = Counter(t)
if all([counter[i] >= cnt[i] for i in range(10)]):
ans.append(i)
return ans


• func findEvenNumbers(digits []int) []int {
counter := count(digits)
var ans []int
for i := 100; i < 1000; i += 2 {
t := make([]int, 3)
k := i
for j := 0; k > 0; j++ {
t[j] = k % 10
k /= 10
}
cnt := count(t)
if check(counter, cnt) {
ans = append(ans, i)
}
}
return ans
}

func count(nums []int) []int {
counter := make([]int, 10)
for _, num := range nums {
counter[num]++
}
return counter
}

func check(cnt1, cnt2 []int) bool {
for i := 0; i < 10; i++ {
if cnt1[i] < cnt2[i] {
return false
}
}
return true
}

• function findEvenNumbers(digits: number[]): number[] {
let record = new Array(10).fill(0);
for (let digit of digits) {
record[digit]++;
}
let ans = [];
for (let i = 100; i < 1000; i += 2) {
if (check(record, String(i))) {
ans.push(i);
}
}
return ans;
}

function check(target: Array<number>, digits: string): boolean {
let record = new Array(10).fill(0);
for (let digit of digits) {
record[digit]++;
}

for (let i = 0; i < 10; i++) {
if (record[i] > target[i]) return false;
}
return true;
}