力扣2968.执行操作使频率分数最大
方法一：滑窗 + 前缀和
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求前缀和数组s
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求一个数组补齐到中位数的差值
- 枚举右端点
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class Solution {public:int maxFrequencyScore(vector<int>& nums, long long k) {int res=0,n = nums.size();sort(nums.begin(),nums.end());vector<long long> s(n + 1, 0);for (int i = 0; i < n; i++) {s[i + 1] = s[i] + nums[i];}auto distance_sum = [&](int l, int i, int r) -> long long {long long left = (long long) nums[i] * (i - l) - (s[i] - s[l]);long long right = s[r + 1] - s[i + 1] - (long long) nums[i] * (r - i);return left + right;};for(int i=0,j=0;i<n;i++){while(distance_sum(j,(i+j)/2,i) > k)j ++;res = max(res,i-j+1);}return res;}};
方法二：滑窗 + 贡献法
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考虑每个元素对答案的贡献
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发现每枚举到一个数时 + nums[i] – nums[i + j / 2]
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同理 当left++时为 + nums[j] - nums[(i+j+1)/2]
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class Solution {public:int maxFrequencyScore(vector<int>& nums, long long k) {int res=0,n = nums.size();long long s = 0;sort(nums.begin(),nums.end());for(int i=0,j=0;i<n;i++){s += nums[i] - nums[(i+j)/2];while(s > k){s += nums[j++] - nums[(i+j)/2];}res = max(res,i-j+1);}return res;}};
