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

Espresso Macchiato2024-10-14 9:09

[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](#1. 解题思路)
- [2. 代码实现](#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：

python 复制代码

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 * key
                if x <= len(q):
                    s += q[x-1][0] * q[x-1][1]

            cnt[key] -= 1
            if 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] += 1
            if val == 0:
                bisect.insort(q, (-val-1, -key))
                idx = bisect.bisect_left(q, (-val-1, -key))
                if idx < x:
                    s += (val+1) * key
                    if 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 * key
                    if 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) * key
                    if x < len(q):
                        s -= q[x][0] * q[x][1]

            ans.append(s)
        return ans

提交代码评测得到：耗时3055ms，占用内存42.3MB。