LOGIC : First we find out the numbers which is occuring in the array for more than twice and then we are replacing it with INT_MAX simultaneously and also counting how many INT_MAX we are doing, let's call it maxcount . At last, after traversing the array, we sort the array so, all the INT_MAX replaced at the end. We got our modified nums array, and the number of elements in the modified array is nums.size()-maxcount. That is it!!

NOTE -> We are converting more than twice numbers into INT_MAX because the constraint of nums[i] << INT_MAX.

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C++ Code:

class Solution {
public:
    int removeDuplicates(vector<int>& nums) {
        int x=0, count=0, maxcount=0;
        for(int i=0; i<nums.size(); i++)
        {
            if(x==nums[i])
                count++;
            else
            {
                x = nums[i];
                count=1;
            }
                
            
            if(count >= 3)
            {
                nums[i]=INT_MAX;
                maxcount++;
            } 
        }
        
        sort(nums.begin(), nums.end());
        
        return nums.size()-maxcount;
    }
};