2080. 区间内查询数字的频率

题目链接	2080. 区间内查询数字的频率
思路	二分法（upper_bound - lower_bound）
题解链接	简洁写法：统计位置+二分查找（Python/Java/C++/Go/JS/Rust）
关键点	预先处理得到每个值所处位置的列表
时间复杂度	\(O(n + m \log n)\)
空间复杂度	\(O(n)\)

代码实现：

class RangeFreqQuery:def __init__(self, arr: List[int]):positions = defaultdict(list)for i, v in enumerate(arr):positions[v].append(i)self.positions = positionsdef query(self, L: int, R: int, value: int) -> int:positions = self.positions[value]n = len(positions)def upper_bound():left, right = -1, nwhile left + 1 < right:mid = (left+right) // 2if positions[mid] > R:right = midelse:left = midreturn rightdef lower_bound():left, right = -1, nwhile left + 1 < right:mid = (left+right) // 2if positions[mid] < L:left = midelse:right = midreturn rightreturn upper_bound() - lower_bound()

Python-官方库

class RangeFreqQuery:def __init__(self, arr: List[int]):positions = defaultdict(list)for i, v in enumerate(arr):positions[v].append(i)self.positions = positionsdef query(self, L: int, R: int, value: int) -> int:positions = self.positions[value]return bisect_right(positions, R) - bisect_left(positions, L)