难度 中等
题目 Leetcode:
1838. Frequency of the Most Frequent Element
The frequency of an element is the number of times it occurs in an array.
You are given an integer array nums and an integer k. In one operation, you can choose an index of nums and increment the element at that index by 1.
Return the maximum possible frequency of an element after performing at most k operations.
Constraints
1 <= nums.length <= 105
1 <= nums[i] <= 105
1 <= k <= 105
Keyword
maximum possible frequency 最大可能频数
题目解析
本题大意是对给定数组内的元素操作至多k次,使得数组内相同的元素尽可能的多,即最大可能频数最大,一次操作就是选中一个元素使其的值+1;
这种题拿到第一反应就是先排序,这题用到了滑动窗口的思想,简单来说就是先分别定义left和right两个值作为滑动窗口的两端,而这题k的作用本质来说就等于说是一个条件,在滑动窗口过程中寻找k足够填满窗口的情况,然后记录当前频数是否是当前的最大可能频数。
解析完毕,以下位参考代码
1 class Solution { 2 public: 3 int maxFrequency(vector<int>& nums, int k) { 4 sort(nums.begin(),nums.end()); 5 int length=nums.size(); 6 long long ans=0,l=0,res=1; 7 for(int i=1;i<length;i++) 8 { 9 ans+=(long long)(nums[i]-nums[i-1])*(i-l); 10 if(ans>k)ans-=nums[i]-nums[l++]; 11 res=max(res,i-l+1); 12 } 13 return res; 14 } 15 };