✅Yet Another Array Rearrangement Problem (Contest)

Yet Another Array Rearrangement Problem (Contest) easy Time Limit: 2 sec Memory Limit: 128000 kB

Problem Statement :

You are given an array A of size N. In one move, you can choose two indices i, j (1 <= i, j <= N) such that Ai+Aj is odd and swap Ai and Aj. Find the lexicographically smallest array you can obtain after any number of operations. Input The first line of the input contains a single integer N. The second line of the input contains N space seperated integers.

Constraints: 1 <= N <= 105 1 <= Ai <= 109 Output Print N space seperated integers, the lexicographically smallest array you can obtain after any number of operations. Example Sample Input: 3 5 2 8

Sample Output: 2 5 8

Explaination: We can swap 5 and 2 as their sum is odd.

link:https://my.newtonschool.co/playground/code/j5up4bwqsy1v/

```cpp
#include <bits/stdc++.h> // header file includes every Standard library
using namespace std;
#define int long long
signed main() {
	// Your code here
    int n;
    cin >>n;
    vector<int>a(n);
    for(auto &i:a) {
        cin >> i;
    }
    int odd=0;
    for(auto i:a)
    {
        if(i%2)odd++;
    }
    if(odd!=0 and odd !=n){
        sort(a.begin(),a.end());
    }
    for(auto i:a) cout<<i<< ' ';
}
```