## Uber Interview Question for Software Engineers

Country: United States
Interview Type: In-Person

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4/5 is a graph problem - similar to finding the number of connected components in graph.
DFS solution:
In this graph every node has at most 2 edges. Every position (x, y) has 2 nodes. If it's a '/' in (x, y) and current node is upper half of (x,y), the next two nodes to search is right half of (x - 1, y) and lower half of (x, y - 1).
Other than DFS, union find and BFS will work as well.

``````public int segmentCount(char[][] m) {
int len = m[0].length;
boolean[] upperHalf = new boolean[m.length * len];
boolean[] lowerHalf = new boolean[m.length * len];

int count = 0;
for(int i = 0; i < m.length; i++) {
for(int j = 0; j < len; j++) {
if(!upperHalf[i*len + j]) {
count++;
dfs(m, upperHalf, lowerHalf, i, j, 0);
}
if(!lowerHalf[i*len + j]) {
count++;
dfs(m, upperHalf, lowerHalf, i, j, 1);
}
}
}
return count;
}
//upper:0, lower:1, left:2, right:3
private void dfs(char[][] m, boolean[] upperHalf, boolean[] lowerHalf, int x, int y, int position) {
if(x < 0 || x == m.length || y == m[0].length || y < 0) {
return;
}
if((position == 2 && m[x][y] == '\\') || (position == 3 && m[x][y] == '/')) position = 1;
if((position == 2 && m[x][y] == '/') || position == 3 && m[x][y] == '\\') position = 0;
int id = x * m[0].length + y;
if((position == 0 && upperHalf[id]) || (position == 1 && lowerHalf[id])) { //if visited
return;
}
if(position == 0) upperHalf[id] = true;
else lowerHalf[id] = true;
if(position == 0 && m[x][y] == '\\') {
dfs(m, upperHalf, lowerHalf, x, y + 1, 2); //go right
dfs(m, upperHalf, lowerHalf, x - 1, y, 1); //go up
}
if(position == 0 && m[x][y] == '/') {
dfs(m, upperHalf, lowerHalf, x, y - 1, 3); //go left
dfs(m, upperHalf, lowerHalf, x - 1, y, 1); //go up
}
if(position == 1 && m[x][y] == '\\') {
dfs(m, upperHalf, lowerHalf, x, y - 1, 3); //go left
dfs(m, upperHalf, lowerHalf, x + 1, y, 0); //go down
}
if(position == 1 && m[x][y] == '/') {
dfs(m, upperHalf, lowerHalf, x, y + 1, 2); //go right
dfs(m, upperHalf, lowerHalf, x + 1, y, 0); //go down
}
}``````

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Could you please elaborate a bit. I find it difficult to visualize how this problem is similar to connected components problem ?

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Could you please elaborate a bit. I find it difficult to visualize how this problem is similar to connected component problem So how the DFS is applied.

``````Does the following diagram divides into 12 graphically pieces or just 4 ? I guess it must be 4.

/ / /
/ / /
/ / /``````

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