Partition an Array into Even Values Followed by Odd Values
Company: Amazon
Role: Software Engineer
Category: Coding & Algorithms
Difficulty: easy
Interview Round: Other
### Problem
Rearrange every value of `nums` so all even integers appear before all odd integers. Preserve every occurrence. Any ordering within the even and odd portions is acceptable.
```cpp
class Solution {
public:
vector<int> sortArrayByParity(vector<int>& nums);
};
```
### Examples
**Example 1:**
```text
Input: nums = [5,0,3,2]
Output: [0,2,5,3]
Explanation: The two evens precede both odds;[2,0,3,5] is another valid answer.
```
**Example 2:**
```text
Input: nums = [7,1,9]
Output: [7,1,9]
Explanation: With no even values, the original order already satisfies the requirement.
```
### Constraints
- `1 <= nums.length <= 5000`.
- `0 <= nums[i] <= 5000`.
### Contract considerations
The result need not be numerically sorted; it must only place every even value before every odd value.
```hint Read the boundary condition
Maintain the index where the next even value belongs.
```
### What a Strong Answer Covers
The result need not be numerically sorted; it must only place every even value before every odd value.
### Follow-up Questions
Optional practice extension, separate from the original task: How could you preserve the original relative order within each parity group?
Overview: Partition an Array into Even Values Followed by Odd Values. Rearrange every value of nums so all even integers appear before all odd integers.
