Find the Maximum Number of Marked Indices | Two Pointers | MUST DO | TRICKY | OA

https://leetcode.com/problems/find-the-maximum-number-of-marked-indices/

You are given a 0-indexed integer array nums.
Initially, all of the indices are unmarked. You are allowed to make this operation any number of times:Pick two different unmarked indices i and j such that 2 * nums[i] <= nums[j], then mark i and j.
Return the maximum possible number of marked indices in nums using the above operation any number of times.

Example 1:
Input: nums = [3,5,2,4]
Output: 2
Explanation: In the first operation: pick i = 2 and j = 1, the operation is allowed because 2 * nums[2] <= nums[1]. Then mark index 2 and 1.
It can be shown that there's no other valid operation so the answer is 2.

Example 2:
Input: nums = [9,2,5,4]
Output: 4
Explanation: In the first operation: pick i = 3 and j = 0, the operation is allowed because 2 * nums[3] <= nums[0]. Then mark index 3 and 0.
In the second operation: pick i = 1 and j = 2, the operation is allowed because 2 * nums[1] <= nums[2]. Then mark index 1 and 2.
Since there is no other operation, the answer is 4.

Example 3:
Input: nums = [7,6,8]
Output: 0
Explanation: There is no valid operation to do, so the answer is 0.

Constraints:
1 <= nums.length <= 10^5
1 <= nums[i] <= 10^9
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class Solution {
public:
    int maxNumOfMarkedIndices(vector<int>& nums) {
        int n=nums.size();
        sort(nums.begin(),nums.end());
        int i=0;
        int j=n/2+n%2;
        int res=0;
        while(j<n){
            if(2*nums[i]<=nums[j]){
                res++;
                i++;
            }
            j++;
        }
        return res*2;
    }
};