2094. Finding 3-Digit Even Numbers LeetCode Solution
In this guide, you will get 2094. Finding 3-Digit Even Numbers LeetCode Solution with the best time and space complexity. The solution to Finding -Digit Even Numbers problem is provided in various programming languages like C++, Java, and Python. This will be helpful for you if you are preparing for placements, hackathons, interviews, or practice purposes. The solutions provided here are very easy to follow and include detailed explanations.
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.
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
Complexity Analysis
Time Complexity: O(10^3) = O(1)
Space Complexity: O(10 + |\texttt{digits}|)
2094. Finding 3-Digit Even Numbers LeetCode Solution in C++
class Solution {
public:
vector<int> findEvenNumbers(vector<int>& digits) {
vector<int> ans;
vector<int> count(10);
for (const int digit : digits)
++count[digit];
// Try to construct `abc`.
for (int a = 1; a <= 9; ++a)
for (int b = 0; b <= 9; ++b)
for (int c = 0; c <= 8; c += 2)
if (count[a] > 0 && count[b] > (b == a) &&
count[c] > (c == a) + (c == b))
ans.push_back(a * 100 + b * 10 + c);
return ans;
}
}; /* code provided by PROGIEZ */
2094. Finding 3-Digit Even Numbers LeetCode Solution in Java
class Solution {
public int[] findEvenNumbers(int[] digits) {
List<Integer> ans = new ArrayList<>();
int[] count = new int[10];
for (final int digit : digits)
++count[digit];
// Try to construct `abc`.
for (int a = 1; a <= 9; ++a)
for (int b = 0; b <= 9; ++b)
for (int c = 0; c <= 8; c += 2)
if (count[a] > 0 && count[b] > (b == a ? 1 : 0) &&
count[c] > (c == a ? 1 : 0) + (c == b ? 1 : 0))
ans.add(a * 100 + b * 10 + c);
return ans.stream().mapToInt(Integer::intValue).toArray();
}
} // code provided by PROGIEZ
2094. Finding 3-Digit Even Numbers LeetCode Solution in Python
class Solution:
def findEvenNumbers(self, digits: list[int]) -> list[int]:
ans = []
count = collections.Counter(digits)
# Try to construct `abc`.
for a in range(1, 10):
for b in range(0, 10):
for c in range(0, 9, 2):
if count[a] > 0 and count[b] > (
b == a) and count[c] > (
c == a) + (
c == b):
ans.append(a * 100 + b * 10 + c)
return ans # code by PROGIEZ
