Sliding Window Maximum

/*
Logic:(similar to skyline problem)
We will have a max heap and a map which will store the window elements as keys.
first insert first window and find it's max and put into _max and answer vector.
whenver you get a new number,push it into the queue,update the max and update
the map.if max_element is same as the previous element before the start of window
let's say prev, then it is possible that prev is not in the new window so get the
maximum number present in the window using the max heap and map and update the
_max and push it into the answer vector.
*/
class Solution {
public:
    vector<int> maxSlidingWindow(vector<int>& nums, int k) {
        unordered_map<int,int> mp;
        priority_queue<int> q;
        vector <int> ans;
        int _max=nums[0],i=0;
        for(i=0;i<k;i++){
            _max=max(_max,nums[i]);
            mp[nums[i]]++;
            q.push(nums[i]);
        }
        ans.push_back(_max);
        for(;i<nums.size();i++){
            q.push(nums[i]);
            _max=max(_max,nums[i]);
            mp[nums[i]]++;
            if(nums[i-k]==_max){
                mp[nums[i-k]]--;
                while(!mp[q.top()]){
                    q.pop();
                }
                _max=q.top();
            }
            else{
                mp[nums[i-k]]--;
            }
            ans.push_back(_max);
        }
        return ans;
    }
};