Today I have done leetcode's September LeetCoding Challenge with cpp.

September LeetCoding Challenge 28

Description

Sort Array By Parity II

Given an array of integers nums, half of the integers in nums are odd, and the other half are even.

Sort the array so that whenever nums[i] is odd, i is odd, and whenever nums[i] is even, i is even.

Return any answer array that satisfies this condition.

Example 1:

Input: nums = [4,2,5,7]
Output: [4,5,2,7]
Explanation: [4,7,2,5], [2,5,4,7], [2,7,4,5] would also have been accepted.

Example 2:

Input: nums = [2,3]
Output: [2,3]

Constraints:

• 2 <= nums.length <= 2 * 104
• nums.length is even.
• Half of the integers in nums are even.
• 0 <= nums[i] <= 1000

Follow Up: Could you solve it in-place?

Solution

auto speedup = [](){
  cin.tie(nullptr);
  cout.tie(nullptr);
  ios::sync_with_stdio(false);
  return 0;
}();
class Solution {
public:
  vector<int> sortArrayByParityII(vector<int>& nums) {
    int even = 0;
    int odd = 1;
    int len = nums.size();
    while(even < len) {
      while(even < len && ~nums[even] & 1) even += 2;
      while(odd < len && nums[odd] & 1) odd += 2;
      if(even >= len) break;
      swap(nums[odd], nums[even]);
      odd += 2;
      even += 2;
    }
    return nums;
  }
};

// Accepted
// 61/61 cases passed (12 ms)
// Your runtime beats 97.83 % of cpp submissions
// Your memory usage beats 30.87 % of cpp submissions (21.5 MB)