Finding 3-Digit Even Numbers — LeetCode 2094 Python Solution
EasyRecursionArrayHash TableEnumerationSorting
- Problem
- #2094
- Pattern
- Sorting
- Reading time
- 2 min
- Source
- leetcode.com
The problem
You are given an integer array digits, where each element is a digit. The array may contain duplicates.
Example
- 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.
Python solution
Python
class Solution:
def findEvenNumbers(self, digits: List[int]) -> List[int]:
cnt = Counter(digits)
ans = []
for x in range(100, 1000, 2):
cnt1 = Counter()
y = x
while y:
y, v = divmod(y, 10)
cnt1[v] += 1
if all(cnt[i] >= cnt1[i] for i in range(10)):
ans.append(x)
return ansComplexity
| Measure | Complexity |
|---|---|
| Time | O(k \times 10^k), where k is the number of digits of the target even number, which is 3 in this problem |
| Space | O(1) auxiliary |
Pattern: Sorting
Spend O(n log n) once to buy an ordering that makes the rest of the problem trivial. LeetCode 2094. Finding 3-Digit Even Numbers is filed here because LeetCode tags it Sorting, which is the vocabulary this hub collects.
The sorting guide has the Python template for the pattern and the 401 LeetCode problems that use it.
Related problems
Frequently asked questions
- How hard is LeetCode 2094. Finding 3-Digit Even Numbers?
- LeetCode 2094. Finding 3-Digit Even Numbers is rated Easy on LeetCode.
- What is the time complexity of LeetCode 2094. Finding 3-Digit Even Numbers?
- The Python solution on this page runs in O(k \times 10^k), where k is the number of digits of the target even number, which is 3 in this problem.
- What is the space complexity of LeetCode 2094. Finding 3-Digit Even Numbers?
- The Python solution on this page uses O(1) auxiliary space.
- What topics does LeetCode 2094. Finding 3-Digit Even Numbers cover?
- LeetCode 2094. Finding 3-Digit Even Numbers is tagged Recursion, Array, Hash Table, Enumeration and Sorting on LeetCode.