problem:1838. 最高频元素的频数

思路：滑动窗口维护窗口的操作数<=k

Accode:

class Solution {
public:
    int maxFrequency(vector<int>& nums, int k) {
        sort(nums.begin(),nums.end());
        long long sum = 0;
        int len = nums.size();
        int ans = 0;
        deque<int> deq;
        for(int i=0,j=0;i<len;i++){
            if(deq.size()) sum+=1LL*(nums[i]-nums[deq.back()])*(i-deq.front());
            while(deq.size() && sum>k){
                int curr = nums[deq.front()];
                deq.pop_front();
                sum-=nums[i]-curr;
            }
            deq.push_back(i);
            ans = max(ans,(int)deq.size());
        }
        return ans;
    }
};

时间复杂度：$o(n)$