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Largest Component Size by Common Factor in C++
Suppose we have an array A of unique positive integers, now consider the following graph −
There are length of A number of nodes, these are labelled A[0] to A[size of A - 1]; There is an edge between A[i] and A[j] when A[i] and A[j] share a common factor greater than 1. We have to find the size of the largest connected component in the graph.
So, if the input is like [4,6,15,35], then the output will be 4
To solve this, we will follow these steps −
Define an array parent
Define an array rank
Define an array rank
if parent[x] is same as -1, then −
return x
return parent[x] = getParent(parent[x])
Define a function unionn(), this will take x, y,
parX := getParent(x)
parY := getParent(y)
if parX is same as parY, then −
return
if rank[parX] >= rank[parY], then −
rank[parX] := rank[parX] + rank[parY]
parent[parY] := parX
Otherwise
rank[parY] := rank[parY] + rank[parX]
parent[parX] := parY
From the main method do the following −
ret := 0, n := size of A
parent := Define an array of size n fill this with -1
rank := Define an array of size n fill this with 1
Define one map m
for initialize i := 0, when i < n, update (increase i by 1), do −
x := A[i]
for initialize j := 2, when j * j <= x, update (increase j by 1), do −
if x mod j is same as 0, then &minsu;
if j is in m, then −
unionn(m[j], i)
Otherwise
m[j] := i
if (x / j) is in m, then −
unionn(m[x / j], i)
Otherwise
m[x / j] = i
if x is in m, then −
unionn(m[x], i)
Otherwise
m[x] := i
ret := maximum of ret and rank[getParent(i)]
return ret
Let us see the following implementation to get better understanding −
Example
#include <bits/stdc++.h>
using namespace std;
class Solution {
public:
vector<int> parent;
vector<int> rank;
int getParent(int x){
if (parent[x] == -1)
return x;
return parent[x] = getParent(parent[x]);
}
void unionn(int x, int y){
int parX = getParent(x);
int parY = getParent(y);
if (parX == parY)
return;
if (rank[parX] >= rank[parY]) {
rank[parX] += rank[parY];
parent[parY] = parX;
} else {
rank[parY] += rank[parX];
parent[parX] = parY;
}
}
int largestComponentSize(vector<int>& A) {
int ret = 0;
int n = A.size();
parent = vector<int>(n, -1);
rank = vector<int>(n, 1);
unordered_map<int, int> m;
for (int i = 0; i < n; i++) {
int x = A[i];
for (int j = 2; j * j <= x; j++) {
if (x % j == 0) {
if (m.count(j)) {
unionn(m[j], i);
} else {
m[j] = i;
}
if (m.count(x / j)) {
unionn(m[x / j], i);
} else {
m[x / j] = i;
}
}
}
if (m.count(x)) {
unionn(m[x], i);
} else {
m[x] = i;
}
ret = max(ret, rank[getParent(i)]);
}
return ret;
}
};
main(){
Solution ob;
vector<int> v = {4,6,15,35};
cout << (ob.largestComponentSize(v));
}Input
{4,6,15,35}Output
4
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