2158. Amount of New Area Painted Each Day

Updated: 2024-03-12
2 min read
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LeetCode problem 2158

class Node:
    def __init__(self, l, r):
        self.left = None
        self.right = None
        self.l = l
        self.r = r
        self.mid = (l + r) >> 1
        self.v = 0
        self.add = 0


class SegmentTree:
    def __init__(self):
        self.root = Node(1, 10**5 + 10)

    def modify(self, l, r, v, node=None):
        if l > r:
            return
        if node is None:
            node = self.root
        if node.l >= l and node.r <= r:
            node.v = node.r - node.l + 1
            node.add = v
            return
        self.pushdown(node)
        if l <= node.mid:
            self.modify(l, r, v, node.left)
        if r > node.mid:
            self.modify(l, r, v, node.right)
        self.pushup(node)

    def query(self, l, r, node=None):
        if l > r:
            return 0
        if node is None:
            node = self.root
        if node.l >= l and node.r <= r:
            return node.v
        self.pushdown(node)
        v = 0
        if l <= node.mid:
            v += self.query(l, r, node.left)
        if r > node.mid:
            v += self.query(l, r, node.right)
        return v

    def pushup(self, node):
        node.v = node.left.v + node.right.v

    def pushdown(self, node):
        if node.left is None:
            node.left = Node(node.l, node.mid)
        if node.right is None:
            node.right = Node(node.mid + 1, node.r)
        if node.add:
            left, right = node.left, node.right
            left.v = left.r - left.l + 1
            right.v = right.r - right.l + 1
            left.add = node.add
            right.add = node.add
            node.add = 0


class Solution:
    def amountPainted(self, paint: List[List[int]]) -> List[int]:
        tree = SegmentTree()
        res = []
        for i, (start, end) in enumerate(paint):
            l, r = start + 1, end
            v = tree.query(l, r)
            res.append(r - l + 1 - v)
            tree.modify(l, r, 1)
        return res