Leetcode #2158: Amount of New Area Painted Each Day
In this guide, we solve Leetcode #2158 Amount of New Area Painted Each Day in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
There is a long and thin painting that can be represented by a number line. You are given a 0-indexed 2D integer array paint of length n, where paint[i] = [starti, endi].
Quick Facts
- Difficulty: Hard
- Premium: Yes
- Tags: Segment Tree, Array, Ordered Set
Intuition
The constraints allow a direct scan that keeps only the essential state.
By translating the requirements into a clean loop, the logic stays easy to reason about.
Approach
Iterate through the data once, updating the state needed to compute the answer.
Return the final state after the traversal is complete.
Steps:
- Parse the input.
- Iterate and update state.
- Return the computed answer.
Example
Input: paint = [[1,4],[4,7],[5,8]]
Output: [3,3,1]
Explanation:
On day 0, paint everything between 1 and 4.
The amount of new area painted on day 0 is 4 - 1 = 3.
On day 1, paint everything between 4 and 7.
The amount of new area painted on day 1 is 7 - 4 = 3.
On day 2, paint everything between 7 and 8.
Everything between 5 and 7 was already painted on day 1.
The amount of new area painted on day 2 is 8 - 7 = 1.
Python Solution
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()
ans = []
for i, (start, end) in enumerate(paint):
l, r = start + 1, end
v = tree.query(l, r)
ans.append(r - l + 1 - v)
tree.modify(l, r, 1)
return ans
Complexity
The time complexity is O(n). The space complexity is O(1) to O(n).
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.