## Samsung Interview Question

Software Engineers**Country:**India

**Interview Type:**Written Test

{

var min ;

S ---> W1 ---> W2 ----> W3 ----> ... ---> WN -----> D

1. take all permutations (recursion) of Wi to reach D .

2. update the distance( manhattan ) for the trip and check if value of the trip is < min

update min ;

print min ;

}

{take for both starting and ending points of wormholes as mentioned in the statement for the logic ..}

#include<bits/stdc++.h>

using namespace std;

vector<pair<int,int>>entry;

vector<pair<int,int>>exit1;

vector<int>cost;

int main()

{

int n;

cin>>n;

int x1,y1,x2,y2,d;

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

{

cin>>x1>>y1>>x2>>y2>>d;

entry.push_back({x1,y1});

exit1.push_back({x2,y2});

cost.push_back(d);

}

int sx=-1,sy=-1,ex=-1,ey=-1;

cin>>sx>>sy>>ex>>ey;

entry.push_back({sx,sy});

entry.push_back({ex,ey});

exit1.push_back({sx,sy});

exit1.push_back({ex,ey});

cost.push_back(0);

cost.push_back(0);

int dis[n+2][n+2];

memset(dis,0,sizeof(dis));

for(int i=0;i<n+2;i++)

{

for(int j=0;j<n+2;j++)

{

if(i==j)

continue;

int d1=abs(exit1[i].first-entry[j].first)+abs(exit1[i].second-entry[j].second)+cost[i];

int d2=abs(exit1[i].first-exit1[j].first)+abs(exit1[i].second-exit1[j].second)+cost[i];

d1=min(d1,d2);

dis[i][j]=d;

}

}

for(int i=0;i<n+2;i++)

{

for(int j=0;j<n+2;j++)

{

cout<<dis[i][j]<<" ";

}

cout<<endl;

}

for(int k=0;k<n+2;k++)

{

for(int i=0;i<n+2;i++)

{

for(int j=0;j<n+2;j++)

{

if(dis[i][k]+dis[k][j]<dis[i][j])

{

dis[i][j]=dis[i][k]+dis[k][j];

}

}

}

}

for(int i=0;i<n+2;i++)

{

for(int j=0;j<n+2;j++)

{

cout<<dis[i][j]<<" ";

}

cout<<endl;

}

cout<<dis[n][n+1];

return 0;

}

I think this might work

```
#include<bits/stdc++.h>
using namespace std;
vector<pair<int,int>>entry;
vector<pair<int,int>>exit1;
vector<int>cost;
int main()
{
int n;
cin>>n;
int x1,y1,x2,y2,d;
for(int i=0;i<n;i++)
{
cin>>x1>>y1>>x2>>y2>>d;
entry.push_back({x1,y1});
exit1.push_back({x2,y2});
cost.push_back(d);
}
int sx=-1,sy=-1,ex=-1,ey=-1;
cin>>sx>>sy>>ex>>ey;
entry.push_back({sx,sy});
entry.push_back({ex,ey});
exit1.push_back({sx,sy});
exit1.push_back({ex,ey});
cost.push_back(0);
cost.push_back(0);
int dis[n+2][n+2];
memset(dis,0,sizeof(dis));
for(int i=0;i<n+2;i++)
{
for(int j=0;j<n+2;j++)
{
if(i==j)
continue;
int d1=abs(exit1[i].first-entry[j].first)+abs(exit1[i].second-entry[j].second)+cost[i];
int d2=abs(exit1[i].first-exit1[j].first)+abs(exit1[i].second-exit1[j].second)+cost[i];
d1=min(d1,d2);
dis[i][j]=d;
}
}
for(int i=0;i<n+2;i++)
{
for(int j=0;j<n+2;j++)
{
cout<<dis[i][j]<<" ";
}
cout<<endl;
}
for(int k=0;k<n+2;k++)
{
for(int i=0;i<n+2;i++)
{
for(int j=0;j<n+2;j++)
{
if(dis[i][k]+dis[k][j]<dis[i][j])
{
dis[i][j]=dis[i][k]+dis[k][j];
}
}
}
}
for(int i=0;i<n+2;i++)
{
for(int j=0;j<n+2;j++)
{
cout<<dis[i][j]<<" ";
}
cout<<endl;
}
cout<<dis[n][n+1];
return 0;
}
```

```
int d1=abs(exit1[i].first-entry[j].first)+abs(exit1[i].second-entry[j].second)+cost[i];
int d2=abs(exit1[i].first-exit1[j].first)+abs(exit1[i].second-exit1[j].second)+cost[i];
d1=min(d1,d2);
dis[i][j]=d;
```

Why have you taken the minimum of both distances? According to question, the shortest path from source to destination may take the longer path to a wormhole so as to get shorter path overall.

Difficult assignment to make a closet loophole towards warmhole mystery closer than point for both blackhole rather than whitehole over the space shining with prior glamour of film academy appointment in many session, informer parking at the northport central complex with a limited value.

Transportation neither be intimated with public as well as private so that condonation of delay in transit be reinstatement on event of the day.

Difficult assignment to make a closet loophole towards warmhole mystery closer than point for both blackhole rather than whitehole over the space shining with prior glamour of film academy appointment in many session, informer parking at the northport central complex with a limited value.

Transportation neither be intimated with public as well as private so that condonation of delay in transit be reinstatement on event of the day.

```
#include <iostream>
using namespace std;
struct grid {
int x;
int y;
};
int main()
{
int n;
cin >> n;
grid s,d;
cin >> s.x >> s.y >> d.x >> d.y;
grid* vertex = new grid[2 * n + 2];
int* a = new int[n];
for (int i = 0;i < 2*n;i+=2) {
cin >> vertex[i].x >> vertex[i].y;
cin >> vertex[i+1].x >> vertex[i+1].y;
cin >> a[i / 2];
}
vertex[2 * n] = s;
vertex[2 * n + 1] = d;
int** graph = new int* [2 * n + 2];
for (int i = 0;i < 2 * n + 2;i++) {
graph[i] = new int[2 * n + 2];
}
int V = 2 * n + 2;
for (int i = 0;i < V;i++) {
for (int j = 0;j < V;j++) {
graph[i][j] = abs(vertex[i].x - vertex[j].x) + abs(vertex[i].y - vertex[j].y);
}
}
for (int i = 0;i < n;i++) {
if (a[i] < graph[2*i][2*i+1]) {
graph[2 * i][2 * i + 1] = a[i];
graph[2 * i + 1][2 * i] = a[i];
}
}
for (int i = 0;i < V;i++) {
for (int j = 0;j < V;j++) {
cout << graph[i][j] << " ";
}
cout << endl;
}
for (int k = 0;k < V;k++) {
for (int i = 0;i < V;i++) {
for (int j = 0;j < V;j++) {
if (graph[i][j] > graph[i][k] + graph[k][j]) {
graph[i][j] = graph[i][k] + graph[k][j];
}
}
}
}
cout << graph[2 * n][2 * n + 1];
return 0;
```

}

```
#include <iostream>
using namespace std;
struct grid {
int x;
int y;
};
int main()
{
int n;
cin >> n;
grid s,d;
cin >> s.x >> s.y >> d.x >> d.y;
grid* vertex = new grid[2 * n + 2];
int* a = new int[n];
for (int i = 0;i < 2*n;i+=2) {
cin >> vertex[i].x >> vertex[i].y;
cin >> vertex[i+1].x >> vertex[i+1].y;
cin >> a[i / 2];
}
vertex[2 * n] = s;
vertex[2 * n + 1] = d;
int** graph = new int* [2 * n + 2];
for (int i = 0;i < 2 * n + 2;i++) {
graph[i] = new int[2 * n + 2];
}
int V = 2 * n + 2;
for (int i = 0;i < V;i++) {
for (int j = 0;j < V;j++) {
graph[i][j] = abs(vertex[i].x - vertex[j].x) + abs(vertex[i].y - vertex[j].y);
}
}
for (int i = 0;i < n;i++) {
if (a[i] < graph[2*i][2*i+1]) {
graph[2 * i][2 * i + 1] = a[i];
graph[2 * i + 1][2 * i] = a[i];
}
}
for (int i = 0;i < V;i++) {
for (int j = 0;j < V;j++) {
cout << graph[i][j] << " ";
}
cout << endl;
}
for (int k = 0;k < V;k++) {
for (int i = 0;i < V;i++) {
for (int j = 0;j < V;j++) {
if (graph[i][j] > graph[i][k] + graph[k][j]) {
graph[i][j] = graph[i][k] + graph[k][j];
}
}
}
}
cout << graph[2 * n][2 * n + 1];
return 0;
}
```

```
#include <iostream>
using namespace std;
struct grid {
int x;
int y;
};
int main()
{
int n;
cin >> n;
grid s,d;
cin >> s.x >> s.y >> d.x >> d.y;
grid* vertex = new grid[2 * n + 2];
int* a = new int[n];
for (int i = 0;i < 2*n;i+=2) {
cin >> vertex[i].x >> vertex[i].y;
cin >> vertex[i+1].x >> vertex[i+1].y;
cin >> a[i / 2];
}
vertex[2 * n] = s;
vertex[2 * n + 1] = d;
int** graph = new int* [2 * n + 2];
for (int i = 0;i < 2 * n + 2;i++) {
graph[i] = new int[2 * n + 2];
}
int V = 2 * n + 2;
for (int i = 0;i < V;i++) {
for (int j = 0;j < V;j++) {
graph[i][j] = abs(vertex[i].x - vertex[j].x) + abs(vertex[i].y - vertex[j].y);
}
}
for (int i = 0;i < n;i++) {
if (a[i] < graph[2*i][2*i+1]) {
graph[2 * i][2 * i + 1] = a[i];
graph[2 * i + 1][2 * i] = a[i];
}
}
for (int i = 0;i < V;i++) {
for (int j = 0;j < V;j++) {
cout << graph[i][j] << " ";
}
cout << endl;
}
for (int k = 0;k < V;k++) {
for (int i = 0;i < V;i++) {
for (int j = 0;j < V;j++) {
if (graph[i][j] > graph[i][k] + graph[k][j]) {
graph[i][j] = graph[i][k] + graph[k][j];
}
}
}
}
cout << graph[2 * n][2 * n + 1];
return 0;
```

}

```
#include<iostream>
using namespace std;
int ans,mask[10],w[10][5],n;
int distance(int sx,int sy,int dx,int dy) {
int xd=(sx>dx)?(sx-dx):(dx-sx);
int yd=(sy>dy)?(sy-dy):(dy-sy);
return (xd+yd);
}
void cal(int sx,int sy,int dx,int dy,int dis) {
ans=min(ans,distance(sx,sy,dx,dy)+dis);
for(int i=0;i<n;i++) {
if(mask[i]==0) {
mask[i]=1;
int temp=distance(sx,sy,w[i][0],w[i][1])+dis+w[i][4];
cal(w[i][2],w[i][3],dx,dy,temp);
temp=distance(sx,sy,w[i][2],w[i][3])+dis+w[i][4];
cal(w[i][0],w[i][1],dx,dy,temp);
mask[i]=0;
}
}
}
int main() {
int t;
cin>>t;
while(t--) {
cin>>n;
int sx,sy,dx,dy;
cin>>sx>>sy>>dx>>dy;
for(int i=0;i<n;i++) {
mask[i]=0;
for(int j=0;j<5;j++) {
cin>>w[i][j];
}
}
ans=999999;
cal(sx,sy,dx,dy,0);
cout<<ans<<endl;
}
return 0;
}
```

- Mei Li February 12, 2017