Snap interview geometric algorithm question. This article demonstrates how to handle geometric problems on a 2D grid through the building perimeter calculation problem. oavoservice helps you master core geometric algorithm techniques.
📋 Problem Redefinition
Given a 2D grid where 1 represents a building and 0 represents empty space. Buildings may connect to form irregular shapes. Calculate the perimeter of the outer boundary of all buildings (only count edges touching empty space or the boundary).
Example:
Input: grid = [
[1,1,0],
[1,1,0],
[0,0,1]
]
Output: 12
🎯 Core Idea
- Boundary Identification - Find all edges touching empty space
- Deduplication - Avoid double counting shared edges
- Edge Cases - Handle grid boundaries
- Optimization - Reduce unnecessary traversals
💡 Solution Method (oavoservice Innovative Approach)
Method 1: Direct Counting
def calculate_perimeter(grid):
if not grid or not grid[0]:
return 0
rows, cols = len(grid), len(grid[0])
perimeter = 0
for i in range(rows):
for j in range(cols):
if grid[i][j] == 1:
# Check 4 directions
if i == 0 or grid[i-1][j] == 0:
perimeter += 1
if i == rows-1 or grid[i+1][j] == 0:
perimeter += 1
if j == 0 or grid[i][j-1] == 0:
perimeter += 1
if j == cols-1 or grid[i][j+1] == 0:
perimeter += 1
return perimeter
Time Complexity: O(m × n) Space Complexity: O(1)
Method 2: Mathematical Optimization
oavoservice optimization strategy:
def calculate_perimeter_optimized(grid):
rows, cols = len(grid), len(grid[0])
buildings = 0
shared_edges = 0
for i in range(rows):
for j in range(cols):
if grid[i][j] == 1:
buildings += 1
# Check right and bottom neighbors for shared edges
if j < cols - 1 and grid[i][j+1] == 1:
shared_edges += 1
if i < rows - 1 and grid[i+1][j] == 1:
shared_edges += 1
# Each building contributes 4 edges, subtract 2 for each shared edge
return buildings * 4 - shared_edges * 2
🚀 Extension Problems
1. 3D Surface Area
def calculate_surface_area_3d(grid):
"""
grid[i][j] represents height of building at that position
"""
if not grid:
return 0
rows, cols = len(grid), len(grid[0])
surface_area = 0
for i in range(rows):
for j in range(cols):
if grid[i][j] > 0:
# Top and Bottom
surface_area += 2
# 4 Sides
for di, dj in [(0,1), (0,-1), (1,0), (-1,0)]:
ni, nj = i + di, j + dj
neighbor_height = 0
if 0 <= ni < rows and 0 <= nj < cols:
neighbor_height = grid[ni][nj]
# Only the part higher than neighbor contributes to surface area
if grid[i][j] > neighbor_height:
surface_area += grid[i][j] - neighbor_height
return surface_area
💼 How oavoservice Helps
Geometric Thinking - Transforming problems into geometric calculations Optimization Strategy - Mathematical methods to reduce computation Extension Ability - 3D and complex scenarios Code Implementation - Clear logic structure
Contact oavoservice for professional geometric algorithm interview assistance!
Tags: #Snap #GeometricAlgorithm #GridTraversal #VOHelp #InterviewPrep #1point3acres
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