Description
You are given an integer array of unique positive integers nums. Consider the following graph:
- There are
nums.length nodes, labeled nums[0] to nums[nums.length - 1],
- There is an undirected edge between
nums[i] and nums[j] if nums[i] and nums[j] share a common factor greater than 1.
Return the size of the largest connected component in the graph.
Example 1:
Input: nums = [4,6,15,35]
Output: 4
Example 2:
Input: nums = [20,50,9,63]
Output: 2
Example 3:
Input: nums = [2,3,6,7,4,12,21,39]
Output: 8
Constraints:
1 <= nums.length <= 2 * 1041 <= nums[i] <= 105- All the values of
nums are unique.
Solutions
Solution 1
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26 | class UnionFind:
def __init__(self, n):
self.p = list(range(n))
def union(self, a, b):
pa, pb = self.find(a), self.find(b)
if pa != pb:
self.p[pa] = pb
def find(self, x):
if self.p[x] != x:
self.p[x] = self.find(self.p[x])
return self.p[x]
class Solution:
def largestComponentSize(self, nums: List[int]) -> int:
uf = UnionFind(max(nums) + 1)
for v in nums:
i = 2
while i <= v // i:
if v % i == 0:
uf.union(v, i)
uf.union(v, v // i)
i += 1
return max(Counter(uf.find(v) for v in nums).values())
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52 | class UnionFind {
int[] p;
UnionFind(int n) {
p = new int[n];
for (int i = 0; i < n; ++i) {
p[i] = i;
}
}
void union(int a, int b) {
int pa = find(a), pb = find(b);
if (pa != pb) {
p[pa] = pb;
}
}
int find(int x) {
if (p[x] != x) {
p[x] = find(p[x]);
}
return p[x];
}
}
class Solution {
public int largestComponentSize(int[] nums) {
int m = 0;
for (int v : nums) {
m = Math.max(m, v);
}
UnionFind uf = new UnionFind(m + 1);
for (int v : nums) {
int i = 2;
while (i <= v / i) {
if (v % i == 0) {
uf.union(v, i);
uf.union(v, v / i);
}
++i;
}
}
int[] cnt = new int[m + 1];
int ans = 0;
for (int v : nums) {
int t = uf.find(v);
++cnt[t];
ans = Math.max(ans, cnt[t]);
}
return ans;
}
}
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