Leetcode 3321. Find X-Sum of All K-Long Subarrays II

Leetcode 3321. Find X-Sum of All K-Long Subarrays II
- 1. 解题思路
- 2. 代码实现

题目链接：3321. Find X-Sum of All K-Long Subarrays II

1. 解题思路

这一题同样虽然是一道hard的题目，但也是比较常规的，就是通过一个滑动窗口不断地维护当前长度为k的滑动窗口内所有数字的出现次数，进而维护一个按照出现次数和大小从大到小排列的数组，最后使用这个数组维护top x的频次的数字的总和即可。

2. 代码实现

给出python代码实现如下：

class Solution:def findXSum(self, nums: List[int], k: int, x: int) -> List[int]:n = len(nums)cnt = Counter(nums[:k])q = sorted([(-v, -k) for k, v in cnt.items()])s = sum(it[0] * it[1] for it in q[:x])ans = [s]for i in range(k, n):key, val = nums[i-k], cnt[nums[i-k]]idx = bisect.bisect_left(q, (-val, -key))q.pop(idx)if idx < x:s -= val * keyif x <= len(q):s += q[x-1][0] * q[x-1][1]cnt[key] -= 1if cnt[key] > 0:bisect.insort(q, (-val+1, -key))nidx = bisect.bisect_left(q, (-val+1, -key))if nidx < x:s += key * (val-1)if x < len(q):s -= q[x][0] * q[x][1]key, val = nums[i], cnt[nums[i]]cnt[key] += 1if val == 0:bisect.insort(q, (-val-1, -key))idx = bisect.bisect_left(q, (-val-1, -key))if idx < x:s += (val+1) * keyif x < len(q):s -= q[x][0] * q[x][1]else:idx = bisect.bisect_left(q, (-val, -key))q.pop(idx)if idx < x:s -= val * keyif x <= len(q):s += q[x-1][0] * q[x-1][1]bisect.insort(q, (-val-1, -key))idx = bisect.bisect_left(q, (-val-1, -key))if idx < x:s += (val+1) * keyif x < len(q):s -= q[x][0] * q[x][1]ans.append(s)return ans