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Program to group a set of points into k different groups in Python
Suppose we have a list of points and a number k. The points are in the form (x, y) representing Cartesian coordinates. We can group any two point p1 and p2 if the Euclidean distance between them is <= k, we have to find total number of disjoint groups.
So, if the input is like points = [[2, 2],[3, 3],[4, 4],[11, 11],[12, 12]], k = 2, then the output will be 2, as it can make two groups: ([2,2],[3,3],[4,4]) and ([11,11],[12,12])
To solve this, we will follow these steps −
Define a function dfs() . This will take i
if i is in seen, then
return
insert i into seen
for each nb in adj[i], do
dfs(nb)
From the main method, do the following−
adj := a map
n := size of points
for j in range 0 to n, do
for i in range 0 to j, do
p1 := points[i]
p2 := points[j]
if Euclidean distance between p1 and p2 < k, then
insert j at the end of adj[i]
insert i at the end of adj[j]
seen := a new set
ans := 0
for i in range 0 to n, do
if i not seen, then
ans := ans + 1
dfs(i)
return ans
Let us see the following implementation to get better understanding −
Example
from collections import defaultdict class Solution: def solve(self, points, k): adj = defaultdict(list) n = len(points) for j in range(n): for i in range(j): x1, y1 = points[i] x2, y2 = points[j] if (x1 - x2) ** 2 + (y1 - y2) ** 2 <= k ** 2: adj[i].append(j) adj[j].append(i) seen = set() def dfs(i): if i in seen: return seen.add(i) for nb in adj[i]: dfs(nb) ans = 0 for i in range(n): if i not in seen: ans += 1 dfs(i) return ans ob = Solution() points = [ [2, 2], [3, 3], [4, 4], [11, 11], [12, 12] ] k = 2 print(ob.solve(points, k))
Input
[[2, 2],[3, 3],[4, 4],[11, 11],[12, 12]],2
Output
2
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