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Leetcode 3624. Number of Integers With Popcount-Depth Equal to K II

news 2025/7/22 6:32:16

Leetcode 3624. Number of Integers With Popcount-Depth Equal to K II
- 1. 解题思路
- 2. 代码实现

题目链接：3624. Number of Integers With Popcount-Depth Equal to K II

1. 解题思路

这一题就是一个segment tree的变体，唯一的区别在于其操作不同于一般的简单求和或者求最大最小值，而是返回一个合并后的Counter即可。

而关于segment tree的相关内容，网络上实在是有很多了，我自己反正也有一篇博客《经典算法：Segment Tree》作为备忘，这里就不过多展开了，有兴趣的读者自己查查了解下即可。

2. 代码实现

给出python代码实现如下：

class SegmentTree:def __init__(self, arr):self.length = len(arr)self.tree = self.build(arr)def feature_func(self, *args):ret = defaultdict(int)for cnt in args:for k, v in cnt.items():ret[k] += vreturn retdef build(self, arr):n = len(arr)tree = [defaultdict(int) for _ in range(2*n)]for i in range(n):tree[i+n][arr[i]] += 1for i in range(n-1, 0, -1):tree[i] = self.feature_func(tree[i<<1], tree[(i<<1) | 1])return treedef update(self, idx, val):idx = idx + self.lengthself.tree[idx] = defaultdict(int)self.tree[idx][val] += 1while idx > 1:self.tree[idx>>1] = self.feature_func(self.tree[idx], self.tree[idx ^ 1])idx = idx>>1returndef query(self, lb, rb):lb += self.length rb += self.lengthnodes = []while lb < rb:if lb & 1 == 1:nodes.append(self.tree[lb])lb += 1if rb & 1 == 0:nodes.append(self.tree[rb])rb -= 1lb = lb >> 1rb = rb >> 1if lb == rb:nodes.append(self.tree[rb])return self.feature_func(*nodes)class Solution:def popcountDepth(self, nums: List[int], queries: List[List[int]]) -> List[int]:def fn(num):if num == 1:return 0num = Counter(bin(num)[2:])["1"]return 1 + fn(num)nums = [fn(num) for num in nums]segment_tree = SegmentTree(nums)ans = []for query in queries:if len(query) == 3:_, idx, val = querysegment_tree.update(idx, fn(val))else:_, l, r, k = querycnt = segment_tree.query(l, r)ans.append(cnt[k])return ans