Technical Interview with an Intel Engineer

We go through a mock coding interview with an engineer from Intel. Older video from last year which we just recently uploaded.

[tommyw]

I believe this is essentially the traveling salesman problem.

[manyrandomthings]

Hey bro your explanation is good but font of your editor is too small to see while making video could you please make fonts larger

[himanshugupta]

Is 9 months enough for companies like amazon and microsoft i have just started learning dsa

[kevin_m]

This is my Java solution (also anyone know if this is a problem on LC?)
// each 0 in the input is updated with a value showing the
// distance to the closest -1. Note that -2 are walls
private static void findDistance(int[][] graph)
{
Queue<int[]> queue = new LinkedList<>();

// populate the queue with the positions of the -1's

for(int i=0; i<graph.length; i++)
{
for(int j=0; j<graph[0].length; j++)
{
if(graph[i][j] == -1)
{
int[] {i, j});
}
}
}

int distance = 1;
while(!queue.isEmpty())
{
int size = queue.size();

while(size-- != 0)
{
int[] pos = queue.remove();
int r = pos[0];
int c = pos[1];

// check each of the 4 directions
checkCell(graph, r-1, c, distance, queue);
checkCell(graph, r+1, c, distance, queue);
checkCell(graph, r, c-1, distance, queue);
checkCell(graph, r, c+1, distance, queue);

}

distance++;
}
}

private static void checkCell(int[][] graph, int r, int c, int distance, Queue<int[]> queue)
{
if(r < 0 || r >= graph.length || c < 0 || c >= graph[0].length || graph[r][c] < 0)
{
return;
}

if(graph[r][c] == 0 || distance < graph[r][c])
{
graph[r][c] = distance;
queue.add(new int[] {r, c});
}
}

[yurSJ]

hey abrar im a new engineer that wants to improve on my coding interview skills.how can i get better at this? i admire how u are able to finish and ask these complex questions! thanks