Blocked Pipeline - Google Top Interview Questions
Problem Statement :
You are given an integer n and a two-dimensional list of integers requests.
Consider a 2 x n matrix m where each cell can either be blocked or unblocked.
It starts off as completely unblocked.
Each element in requests contains [row, col, type] meaning that m[row][col] becomes blocked if type = 1 and it becomes unblocked if type = 0.
You want to process requests one by one and after processing each one check whether there is an unblocked path from left to right.
That is, whether you can start off either m[0][0] or m[1][0] and on each step either move right, up, or down and then end up at either m[0][n - 1] or m[1][n - 1].
Return the number of requests that result in such unblocked path after processing
Constraints
1 ≤ n ≤ 100,000
0 ≤ r ≤ 100,000 where r is the length of requests
Example 1
Input
n = 4
requests = [
[0, 2, 1],
[1, 3, 1],
[1, 3, 0]
]
Output
2
Explanation
After setting m[0][2] there is still a path from left to right. But if we block m[1][3], there is no longer a
path. After we unblock m[1][3], there is a path again. So the first and last request result in an unblocked
path.
Solution :
Solution in C++ :
int solve(int n, vector<vector<int>>& requests) {
int num_single = 0;
int num_pair = 0;
vector matrix(2, vector<bool>(n));
int res = 0;
for (auto&& req : requests) {
int row = req[0], col = req[1];
bool type = req[2];
if (type != matrix[row][col]) {
bool old_0 = matrix[0][col];
bool old_1 = matrix[1][col];
matrix[row][col] = type;
int prv = col - 1, nxt = col + 1;
if (type) {
// blocking
if (matrix[0][col] && matrix[1][col]) num_single++;
if (prv >= 0 && matrix[row ^ 1][prv]) num_pair++;
if (nxt < n && matrix[row ^ 1][nxt]) num_pair++;
} else {
// unblocking
if (old_0 && old_1) num_single--;
if (prv >= 0 && matrix[row ^ 1][prv]) num_pair--;
if (nxt < n && matrix[row ^ 1][nxt]) num_pair--;
}
}
if (!num_single && !num_pair) res++;
}
return res;
}
Solution in Java :
import java.util.*;
class Solution {
public int solve(int n, int[][] requests) {
int[][] a = new int[2][n];
// successfull requests
int ans = 0;
// free columns with no blocked cells in both rows
int freeCols = n;
// cross blocks like illustrated before
// 000x000
// 0000x00
int crossBlocks = 0;
for (int[] request : requests) {
int col = request[1];
int row = request[0];
int type = request[2];
if (type == 0) {
if (a[row][col] == 1) {
a[row][col] = 0;
if (a[1 - row][col] == 1) {
freeCols++;
}
if (col - 1 >= 0 && a[1 - row][col - 1] == 1) {
crossBlocks--;
}
if (col + 1 < n && a[1 - row][col + 1] == 1) {
crossBlocks--;
}
}
} else {
if (a[row][col] == 0) {
a[row][col] = 1;
if (a[1 - row][col] == 1) {
freeCols--;
}
if (col - 1 >= 0 && a[1 - row][col - 1] == 1) {
crossBlocks++;
}
if (col + 1 < n && a[1 - row][col + 1] == 1) {
crossBlocks++;
}
}
}
// Passage is possible only if all cols are free and there are not crossBlocks
if (freeCols == n && crossBlocks == 0) {
ans++;
}
}
return ans;
}
}
Solution in Python :
class Solution:
def solve(self, N, requests):
m = [[0 for _ in range(N)] for __ in range(2)]
neis = set()
ans = 0
for r, c, t in requests:
if t == 1:
m[r][c] = 1
for col in (-1, 1, 0):
if 0 <= (c + col) < N and m[(r + 1) % 2][c + col] == 1:
neis.add((r, c, (r + 1) % 2, (c + col)))
else:
m[r][c] = 0
for col in (-1, 1, 0):
if 0 <= (c + col) < N and m[(r + 1) % 2][c + col] == 1:
neis.discard((r, c, (r + 1) % 2, (c + col)))
neis.discard(((r + 1) % 2, (c + col), r, c))
if not neis and (not (m[-1][-1] & m[0][-1])):
ans += 1
return ans
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