Samsung Interview Question for Backend Developers


Country: India
Interview Type: Written Test




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of 0 vote

#include<iostream>
using namespace std;

void findans(double **graph, int nodes, int time, int start, double p, double answer[])
{
    
    if(time <= 0)
    {
        answer[start] += p;
        return;
    }

    for(int i=1;i<=nodes;i++)
        if(graph[start][i] != 0)
        {
            p *= graph[start][i];
            findans(graph, nodes, time-10 , i , p ,answer);
            p /= graph[start][i];
        }
}

int main()
{
    int t;
    cin>>t;
    while(t--)
    {
        int nodes,edges,time;
        cin>>nodes>>edges>>time;
        double **graph = new double*[nodes];
        for(int i=1;i<=nodes;i++)
        {
            graph[i] = new double[nodes];
            for(int j=1;j<=nodes;j++)
                graph[i][j] = 0;
        }

        for(int i=0;i<edges;i++)
        {   
            int start, end;
            double prob;
            cin>>start>>end>>prob;
            graph[start][end] = prob;
        }
        double answer[nodes] = {0.0};
        findans(graph, nodes, time, 1, 1.0, answer);
        double final_prob = -1; int final_div = 0;
        for(int i=1;i<=nodes;i++)
        {
            if(answer[i]>final_prob)
            {
                final_prob = answer[i];
                final_div = i;
            }
        }
        cout<<final_div<<" "<<final_prob<<endl;
    }
    return 0;
}

- Samurai July 26, 2019 | Flag Reply
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#include<bits/stdc++.h>
using namespace std;
//dt duration to stop in a division
void bfs(vector<pair<int,double> > g[],int n,int dt,int lm)
{
   queue<int> que;
   int division=1;
   double maxprob=0;
   double prob[n];
   bool visited[n];
   for(int i=0;i<n;i++)
    {
        prob[i]=0;
        visited[i]=false;
    }
    que.push(0);
    prob[0]=1;
    visited[0]=true;
    int limit=0;
    while(!que.empty())
    {
        int size=que.size();
        division=1,maxprob=0;
        while(size--)
        {
              int u=que.front();
              que.pop();
              for(int i=0;i<g[u].size();i++)
               {
                   int v=g[u][i].first;
                   double w=g[u][i].second;
                   if(u!=v)
                   {
                   prob[v]+=prob[u]*w;
                   if(maxprob<prob[v])
                     {
                      maxprob=prob[v];
                      division=v+1;
                     }
                      
                   }
                   if(visited[v]==false)
                    {
                     que.push(v);
                     visited[v]=true;
                     }
               }
        }
        
        limit+=dt;
        if(limit>=lm)
         break;
    }
    cout<<std::setprecision(6)<<std::fixed;
    cout<<division<<" "<<maxprob;
   
}
int main()
{
    int t,n,e,ed;
    int sd=10;
    while(cin>>t)
    {
         cin>>n>>e>>ed;
        vector<pair<int,double> > g[n];
        int x,y;
        double p;
        for(int i=0;i<e;i++)
        {
             cin>>x>>y>>p;
            g[x-1].push_back(make_pair(y-1,p));
        }
        cout<<t<<" ";
        bfs(g,n,sd,ed);
        cout<<"\n";
    }
}

- Anonymous September 10, 2019 | Flag Reply
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of 0 vote

can someone tell how probability is 0.774000 after 40 mins ?

- Anonymous September 27, 2019 | Flag Reply
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of 0 votes

you find the probability of reaching every node after 40 mins by multiplying each probability in the path. Now there can be multiple paths by which end node is the same. Just add the probability for those paths for each end point.
Later find the max from all end nodes.

- pre October 12, 2019 | Flag


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