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.


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;

    while(!q.empty()) {
        for(auto x: v[p.first]) {
            if(!visit[x]) {
                if(a[x] == id) {

int main()
    int n,m,i,x,y;

    for(i=0;i<m;++i) {

    for(i=0;i<n;++i) {

    int ans=1e9;
    for(i=0;i<n;++i) {
        if(a[i]==id) {
            ans=min(ans, c);
    if(ans==1e9) {

    return 0;

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