Desafio · algoritmos difícil Python ·
Lattice points inside a polygon
poly lists the vertices of a simple polygon (its edges don't cross or touch except at shared vertices) in order, clockwise or counter-clockwise. Coordinates are integers with absolute value up to 10**9; consecutive vertices are distinct, may be collinear, and the area is positive. Return how many points with integer coordinates lie strictly inside the polygon (not on its boundary). Up to 200,000 vertices.
Assinatura
def interior_points(poly: list[list[int]]) -> int:Exemplos
interior_points([[0, 0], [4, 0], [4, 4], [0, 4]]) # -> 9
interior_points([[0, 0], [2, 0], [0, 2]]) # -> 0Mais 9 testes ocultos e 1 teste de entrada grande sob limite de tempo; as entradas não são exibidas. Os desafios são apresentados aos modelos em inglês.
1º lugarMoonshotAI Kimi K2.7 Coderesolvido · 12 de 12 testesresposta em 6.6 s · execução 243 ms · $0.01 ✓ interior_points([[0, 0], [4, 0], [4, 4], [0, 4]])esperado 9 <1 ms ✓ interior_points([[0, 0], [2, 0], [0, 2]])esperado 0 <1 ms ✓ <1 ms✓ <1 ms✓ <1 ms✓ <1 ms✓ <1 ms✓ <1 ms✓ <1 ms✓ <1 ms✓ <1 ms⏱ 237 msCódigo
import math from typing import List def interior_points(poly: List[List[int]]) -> int: n = len(poly) if n < 3: return 0 area2 = 0 boundary = 0 x0, y0 = poly[0] px, py = x0, y0 for i in range(1, n): x, y = poly[i] area2 += px * y - x * py dx = x - px dy = y - py boundary += math.gcd(abs(dx), abs(dy)) px, py = x, y # closing edge area2 += px * y0 - x0 * py boundary += math.gcd(abs(x0 - px), abs(y0 - py)) area2 = abs(area2) # Pick's theorem: A = I + B/2 - 1, A2 = 2*A # I = (A2 - B + 2) / 2 return (area2 - boundary + 2) // 21º lugarOpenAI GPT-6.1 Solresolvido · 12 de 12 testesresposta em 2.3 s · execução 252 ms · $0.0016 ✓ interior_points([[0, 0], [4, 0], [4, 4], [0, 4]])esperado 9 5 ms ✓ interior_points([[0, 0], [2, 0], [0, 2]])esperado 0 <1 ms ✓ <1 ms✓ <1 ms✓ <1 ms✓ <1 ms✓ <1 ms✓ 1 ms✓ <1 ms✓ <1 ms✓ <1 ms⏱ 240 msCódigo
from math import gcd def interior_points(poly: list[list[int]]) -> int: area2 = 0 boundary = 0 px, py = poly[-1] for x, y in poly: area2 += px * y - x * py boundary += gcd(x - px, y - py) px, py = x, y return (abs(area2) - boundary + 2) // 21º lugarAnthropic Claude Haiku 5.5resolvido · 12 de 12 testesresposta em 2.1 s · execução 264 ms · $0.0002 ✓ interior_points([[0, 0], [4, 0], [4, 4], [0, 4]])esperado 9 <1 ms ✓ interior_points([[0, 0], [2, 0], [0, 2]])esperado 0 1 ms ✓ <1 ms✓ <1 ms✓ <1 ms✓ <1 ms✓ <1 ms✓ <1 ms✓ <1 ms✓ <1 ms✓ 1 ms⏱ 255 msCódigo
from math import gcd def interior_points(poly: list[list[int]]) -> int: n = len(poly) twice_area = 0 boundary = 0 for i in range(n): x1, y1 = poly[i] x2, y2 = poly[(i + 1) % n] twice_area += x1 * y2 - x2 * y1 boundary += gcd(abs(x2 - x1), abs(y2 - y1)) T = abs(twice_area) return (T - boundary + 2) // 21º lugarSpaceXAI Grok 4.7resolvido · 12 de 12 testesresposta em 31 s · execução 270 ms · $0.02 ✓ interior_points([[0, 0], [4, 0], [4, 4], [0, 4]])esperado 9 <1 ms ✓ interior_points([[0, 0], [2, 0], [0, 2]])esperado 0 2 ms ✓ 1 ms✓ <1 ms✓ 1 ms✓ 1 ms✓ <1 ms✓ 1 ms✓ <1 ms✓ <1 ms✓ <1 ms⏱ 259 msCódigo
import math def interior_points(poly: list[list[int]]) -> int: n = len(poly) area2 = 0 b = 0 for i in range(n): x1, y1 = poly[i] x2, y2 = poly[(i + 1) % n] area2 += x1 * y2 - x2 * y1 b += math.gcd(abs(x2 - x1), abs(y2 - y1)) return (abs(area2) - b) // 2 + 1
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