**Find the nearest clone**

### Problem Statement :

In this challenge, there is a connected undirected graph where each of the nodes is a color. Given a color, find the shortest path connecting any two nodes of that color. Each edge has a weight of . If there is not a pair or if the color is not found, print -1. Function Description Complete the findShortest function in the editor below. It should return an integer representing the length of the shortest path between two nodes of the same color, or -1 if it is not possible. findShortest has the following parameter(s): g_nodes: an integer, the number of nodes g_from: an array of integers, the start nodes for each edge g_to: an array of integers, the end nodes for each edge ids: an array of integers, the color id per node val: an integer, the id of the color to match Input Format The first line contains two space-separated integers n and m, the number of nodes and edges in the graph. Each of the next m lines contains two space-separated integers g_from[ i ] and g_to[ i ] , the nodes connected by an edge. The next line contains n space-seperated integers, ids[ i ], representing the color id of each node from 1 to n. The last line contains the id of the color to analyze. Note: The nodes are indexed from 1 to n. Constraints 1 <= n <= 10^6 1 <= m <= 10^6 1 <= ids[ i ] <= 10^8 Output Format Print the single integer representing the smallest path length or -1.

### Solution :

` ````
Solution in C++ :
In C ++ :
#include <bits/stdc++.h>
#define MAX 1000003
using namespace std;
vector<int> v[MAX];
bool visit[MAX];
int a[MAX],c,id;
void bfs(int i)
{
memset(visit, false, sizeof visit);
queue<pair<int,int> > q;
pair<int,int> p;
q.push({i,0});
visit[i]=true;
while(!q.empty()) {
p=q.front();
q.pop();
for(auto x: v[p.first]) {
if(!visit[x]) {
if(a[x] == id) {
c=p.second+1;
return;
}
visit[x]=true;
q.push({x,p.second+1});
}
}
}
}
int main()
{
int n,m,i,x,y;
cin>>n>>m;
for(i=0;i<m;++i) {
cin>>x>>y;
x-=1,y-=1;
v[x].push_back(y);
v[y].push_back(x);
}
for(i=0;i<n;++i) {
cin>>a[i];
}
cin>>id;
int ans=1e9;
for(i=0;i<n;++i) {
if(a[i]==id) {
c=1e9;
bfs(i);
ans=min(ans, c);
}
}
if(ans==1e9) {
ans=-1;
}
cout<<ans<<endl;
return 0;
}
```

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