- [Leetcode 3624. Number of Integers With Popcount-Depth Equal to K II](#Leetcode 3624. Number of Integers With Popcount-Depth Equal to K II)
- [1. 解题思路](#1. 解题思路)
- [2. 代码实现](#2. 代码实现)
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：
python
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] += v
return ret
def build(self, arr):
n = len(arr)
tree = [defaultdict(int) for _ in range(2*n)]
for i in range(n):
tree[i+n][arr[i]] += 1
for i in range(n-1, 0, -1):
tree[i] = self.feature_func(tree[i<<1], tree[(i<<1) | 1])
return tree
def update(self, idx, val):
idx = idx + self.length
self.tree[idx] = defaultdict(int)
self.tree[idx][val] += 1
while idx > 1:
self.tree[idx>>1] = self.feature_func(self.tree[idx], self.tree[idx ^ 1])
idx = idx>>1
return
def query(self, lb, rb):
lb += self.length
rb += self.length
nodes = []
while lb < rb:
if lb & 1 == 1:
nodes.append(self.tree[lb])
lb += 1
if rb & 1 == 0:
nodes.append(self.tree[rb])
rb -= 1
lb = lb >> 1
rb = rb >> 1
if 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 0
num = 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 = query
segment_tree.update(idx, fn(val))
else:
_, l, r, k = query
cnt = segment_tree.query(l, r)
ans.append(cnt[k])
return ans提交代码评测得到：耗时6716ms，占用内存94.28MB。