Leetcode #2249: Count Lattice Points Inside a Circle
In this guide, we solve Leetcode #2249 Count Lattice Points Inside a Circle 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
Given a 2D integer array circles where circles[i] = [xi, yi, ri] represents the center (xi, yi) and radius ri of the ith circle drawn on a grid, return the number of lattice points that are present inside at least one circle. Note: A lattice point is a point with integer coordinates.
Quick Facts
- Difficulty: Medium
- Premium: No
- Tags: Geometry, Array, Hash Table, Math, Enumeration
Intuition
Fast membership checks and value lookups are the heart of this problem, which makes a hash map the natural choice.
By storing what we have already seen (or counts/indexes), we can answer the question in one pass without backtracking.
Approach
Scan the input once, using the map to detect when the condition is satisfied and to update state as you go.
This keeps the solution linear while remaining easy to explain in an interview setting.
Steps:
- Initialize a hash map for seen items or counts.
- Iterate through the input, querying/updating the map.
- Return the first valid result or the final computed value.
Example
Input: circles = [[2,2,1]]
Output: 5
Explanation:
The figure above shows the given circle.
The lattice points present inside the circle are (1, 2), (2, 1), (2, 2), (2, 3), and (3, 2) and are shown in green.
Other points such as (1, 1) and (1, 3), which are shown in red, are not considered inside the circle.
Hence, the number of lattice points present inside at least one circle is 5.
Python Solution
class Solution:
def countLatticePoints(self, circles: List[List[int]]) -> int:
ans = 0
mx = max(x + r for x, _, r in circles)
my = max(y + r for _, y, r in circles)
for i in range(mx + 1):
for j in range(my + 1):
for x, y, r in circles:
dx, dy = i - x, j - y
if dx * dx + dy * dy <= r * r:
ans += 1
break
return ans
Complexity
The time complexity is O(n). The space complexity is 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.