850. Rectangle Area II

# Time:  O(nlogn)
# Space: O(n)

class SegmentTreeNode(object):
    def __init__(self, start, end):
        self.start, self.end = start, end
        self.total = self.count = 0
        self._left = self._right = None

    def mid(self):
        return (self.start+self.end) // 2

    def left(self):
        self._left = self._left or SegmentTreeNode(self.start, self.mid())
        return self._left

    def right(self):
        self._right = self._right or SegmentTreeNode(self.mid(), self.end)
        return self._right

    def update(self, X, i, j, val):
        if i >= j:
            return 0
        if self.start == i and self.end == j:
            self.count += val
        else:
            self.left().update(X, i, min(self.mid(), j), val)
            self.right().update(X, max(self.mid(), i), j, val)
        if self.count > 0:
            self.total = X[self.end]-X[self.start]
        else:
            self.total = self.left().total + self.right().total
        return self.total


class Solution(object):
    def rectangleArea(self, rectangles):
        """
        :type rectangles: List[List[int]]
        :rtype: int
        """
        OPEN, CLOSE = 1, -1
        events = []
        X = set()
        for x1, y1, x2, y2 in rectangles:
            events.append((y1, OPEN, x1, x2))
            events.append((y2, CLOSE, x1, x2))
            X.add(x1)
            X.add(x2)
        events.sort()
        X = sorted(X)
        Xi = {x: i for i, x in enumerate(X)}

        st = SegmentTreeNode(0, len(X)-1)
        result = 0
        cur_x_sum = 0
        cur_y = events[0][0]
        for y, typ, x1, x2 in events:
            result += cur_x_sum * (y-cur_y)
            cur_x_sum = st.update(X, Xi[x1], Xi[x2], typ)
            cur_y = y
        return result % (10**9+7)

Solution from kamyu104/LeetCode-Solutions · MIT