Minimize Direction Violations in a Directed Road Network
Company: Google
Role: Software Engineer
Category: Coding & Algorithms
Difficulty: medium
Interview Round: Onsite
# Minimize Direction Violations in a Directed Road Network
There are `n` nodes numbered from `0` through `n - 1`. Every input edge `[u, v]` has an original direction from `u` to `v`, but you may physically traverse it in either direction.
- Traversing an edge in its original direction costs `0`.
- Traversing it against its original direction costs `1`.
Given `start` and `end`, return the minimum total cost of any route from `start` to `end`. Return `-1` if the two nodes are disconnected even when edges may be traversed both ways.
## Function Signature
```python
def minimum_direction_violations(
n: int,
edges: list[list[int]],
start: int,
end: int,
) -> int:
...
```
## Constraints
- `1 <= n <= 200_000`
- `0 <= len(edges) <= 300_000`
- `0 <= u, v < n` and `u != v`
- Multiple directed edges between the same pair of nodes are allowed.
- `0 <= start, end < n`
## Examples
```text
Input: n = 5, edges = [[0, 1], [2, 1], [2, 3], [4, 3]], start = 0, end = 4
Output: 2
```
```text
Input: n = 4, edges = [[0, 1], [1, 2]], start = 0, end = 3
Output: -1
```
```text
Input: n = 3, edges = [[0, 1], [1, 2]], start = 2, end = 0
Output: 2
```
Quick Answer: Find a route through a directed road network that minimizes travel against the original edge directions. Model forward and reverse traversal costs correctly, handle parallel edges and disconnected nodes, and choose a shortest-path approach that scales to hundreds of thousands of vertices and edges.
There are `n` intersections numbered `0` through `n - 1`, joined by one-way roads.
Each road is given as `edges[i] = [u, v]`, meaning the road's signposted direction runs
from `u` to `v`. You may still physically drive any road in either direction:
- driving a road in its signposted direction costs `0`;
- driving a road against its signposted direction costs `1` (one *direction violation*).
Given `start` and `end`, return the **minimum total number of direction violations** on
any route from `start` to `end`. Return `-1` when no route exists at all, even though
every road may be driven both ways.
### Details
- Reachability is decided by the *undirected* version of the graph, because every road is
drivable both ways. Only the cost depends on orientation.
- The edge list may contain parallel duplicates (`[u, v]` listed twice) and anti-parallel
pairs (both `[u, v]` and `[v, u]`). Every listed road is independently usable.
- If `start == end`, the answer is `0`.
### Output semantics
Return a single integer: the minimum violation count, or `-1` when `end` is unreachable
from `start`. Because the return value is one scalar, the answer is unique even when
several different routes achieve it, so there is no ordering or tie-breaking to define.
Every quantity fits in a signed 32-bit integer (the answer never exceeds `edges.length`),
so `int` is the correct type in Java and C++; no `long` / `long long` is required.
### Example 1
```text
Input: n = 5, edges = [[0, 1], [2, 1], [2, 3], [4, 3]], start = 0, end = 4
Output: 2
```
The route `0 -> 1 -> 2 -> 3 -> 4` costs `0 + 1 + 0 + 1 = 2`. Hops `1 -> 2` and `3 -> 4`
run against the signposted directions of `[2, 1]` and `[4, 3]`.
### Example 2
```text
Input: n = 4, edges = [[0, 1], [1, 2]], start = 0, end = 3
Output: -1
```
Node `3` touches no road, so it is unreachable in both the directed and the undirected
reading of the network.
### Example 3
```text
Input: n = 3, edges = [[0, 1], [1, 2]], start = 2, end = 0
Output: 2
```
Travelling `2 -> 1 -> 0` runs against both signposts, so each of the two roads costs `1`.
### Example 4
```text
Input: n = 5, edges = [[4, 0], [0, 1], [1, 2], [2, 3], [3, 4]], start = 0, end = 4
Output: 0
```
The single road `[4, 0]` reaches `end` in one hop but costs `1`. The four-hop route
`0 -> 1 -> 2 -> 3 -> 4` follows every signpost and costs `0`, so the answer is `0`.
Fewest roads is not the same objective as fewest violations.
Constraints
- 1 <= n <= 200000
- 0 <= edges.length <= 300000
- edges[i].length == 2
- 0 <= edges[i][0], edges[i][1] < n and edges[i][0] != edges[i][1]
- 0 <= start < n and 0 <= end < n
- Parallel duplicates and anti-parallel pairs are allowed in edges
- 0 <= answer <= edges.length, so every value fits in a signed 32-bit integer (use int in Java and C++; long / long long are not needed)
Examples
Input: (5, [[0, 1], [2, 1], [2, 3], [4, 3]], 0, 4)
Expected Output: 2
Input: (4, [[0, 1], [1, 2]], 0, 3)
Expected Output: -1
Hints
- Each road gives you two possible moves, not one: the signposted move and the reverse move. What weight does each of those two moves deserve?
- Plain breadth-first search answers 'fewest roads driven', which is a different question from 'fewest violations'. Example 4 is built to punish that confusion.
- Once every move weighs either 0 or 1, you do not need a full priority queue. Think about which end of a double-ended queue a free move belongs on so that the queue stays sorted by cost on its own.