## 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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``````#include <iostream>
#include<bits/stdc++.h>
using namespace std;

void dfs(vector<set<int>>& graph, int node, int& result, vector<bool> visited) {
if(visited[node]) return;

visited[node] = true;
set<int> temp(graph[node]);
for(auto neighbor : temp) {
if(graph[node].find(neighbor) == graph[node].end())
continue;
graph[node].erase(neighbor);
graph[neighbor].erase(node);
if(visited[neighbor]) {
cout << "Node " << node << " Neighbor " << neighbor << endl;
result++;
}
else {
dfs(graph, neighbor, result, visited);
}
}
}

int main() {
vector<string> input;
for (std::string line; std::getline(std::cin, line);) {
input.push_back(line);
}

//construct the graph
int columns = input[0].size() + 1;
int rows = input.size() + 1;

int lastRowStartIndex = (rows-1)*columns;
vector<set<int>> graph(rows*columns, set<int>());
for(int i = 0; i < columns-1; i++) {
graph[i].insert(i+1);
graph[i+1].insert(i);

graph[i + lastRowStartIndex].insert(i + lastRowStartIndex + 1);
graph[i + lastRowStartIndex + 1].insert(i + lastRowStartIndex);
}

for(int i = 0; i < lastRowStartIndex; i+=columns) {
graph[i].insert(i+columns);
graph[i+columns].insert(i);

graph[i + columns - 1].insert(i+2*columns - 1);
graph[i + 2*columns - 1].insert(i + columns - 1);
}

int currentNode = 0;
for(string line : input) {
// cout << line << endl;
for(char c : line) {
if(c == '\\') {
graph[currentNode].insert(currentNode + columns + 1);
graph[currentNode + columns + 1].insert(currentNode);
}
else {
graph[currentNode+1].insert(currentNode + columns);
graph[currentNode + columns].insert(currentNode+1);
}
currentNode++;
}
currentNode++;
}

int result = 0;
vector<bool> visited(rows*columns, false);
// cout << "Everything is fine" << endl;
dfs(graph, 0, result, visited);

cout << result << endl;
return 0;
}``````

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``````#include <iostream>
#include<bits/stdc++.h>
using namespace std;

void dfs(vector<set<int>>& graph, int node, int& result, vector<bool> visited) {
if(visited[node]) return;

visited[node] = true;
set<int> temp(graph[node]);
for(auto neighbor : temp) {
if(graph[node].find(neighbor) == graph[node].end())
continue;
graph[node].erase(neighbor);
graph[neighbor].erase(node);
if(visited[neighbor]) {
cout << "Node " << node << " Neighbor " << neighbor << endl;
result++;
}
else {
dfs(graph, neighbor, result, visited);
}
}
}

int main() {
vector<string> input;
for (std::string line; std::getline(std::cin, line);) {
input.push_back(line);
}

//construct the graph
int columns = input[0].size() + 1;
int rows = input.size() + 1;

int lastRowStartIndex = (rows-1)*columns;
vector<set<int>> graph(rows*columns, set<int>());
for(int i = 0; i < columns-1; i++) {
graph[i].insert(i+1);
graph[i+1].insert(i);

graph[i + lastRowStartIndex].insert(i + lastRowStartIndex + 1);
graph[i + lastRowStartIndex + 1].insert(i + lastRowStartIndex);
}

for(int i = 0; i < lastRowStartIndex; i+=columns) {
graph[i].insert(i+columns);
graph[i+columns].insert(i);

graph[i + columns - 1].insert(i+2*columns - 1);
graph[i + 2*columns - 1].insert(i + columns - 1);
}

int currentNode = 0;
for(string line : input) {
// cout << line << endl;
for(char c : line) {
if(c == '\\') {
graph[currentNode].insert(currentNode + columns + 1);
graph[currentNode + columns + 1].insert(currentNode);
}
else {
graph[currentNode+1].insert(currentNode + columns);
graph[currentNode + columns].insert(currentNode+1);
}
currentNode++;
}
currentNode++;
}

int result = 0;
vector<bool> visited(rows*columns, false);
// cout << "Everything is fine" << endl;
dfs(graph, 0, result, visited);

cout << result << endl;
return 0;
}``````

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