CF 104344C - Martelo
We are working on a one-dimensional movement problem on the number line. Eren starts at position 0 and wants to reach a target position X. Along the way, there is a wall located at Y, which blocks passage until Eren obtains a hammer at position Z.
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Solve time: 1m 28s
Verified: yes
Solution
Problem Understanding
We are working on a one-dimensional movement problem on the number line. Eren starts at position 0 and wants to reach a target position X. Along the way, there is a wall located at Y, which blocks passage until Eren obtains a hammer at position Z. Once the hammer is collected, the wall no longer prevents movement, so Eren can freely pass through Y afterward.
The task is to determine whether it is possible to reach X under these rules, and if it is, to compute the minimum total distance traveled.
Because movement is on a straight line and every segment cost is just absolute distance, any valid route is fully determined by the order in which Eren visits relevant points: 0, Z, Y, and X, with the constraint that Y can only be crossed safely after visiting Z.
The constraints are small, with all coordinates between -1000 and 1000. This immediately tells us that any solution with constant or logarithmic complexity per test case is sufficient, and even a small fixed number of path evaluations is enough.
The main subtlety is that the wall is not a one-way constraint in a geometric sense. It is a permission constraint: crossing Y is allowed only after visiting Z. A naive shortest path intuition that ignores this ordering will fail.
A typical failure case arises when Z is on the “wrong side” of Y relative to the start or target. For example, if Eren needs to pass Y early to reach Z, but Z is beyond Y in a way that forces crossing Y before obtaining the hammer, then the route is impossible even though all points lie on a line.
Approaches
The brute-force interpretation is to enumerate all possible permutations of visiting the relevant points, starting at 0, ending at X, and ensuring that Y is only crossed after Z has been visited. For each ordering, we would simulate the path and compute the total distance traveled, rejecting invalid ones.
This works because the number of relevant points is small, so there are only a few permutations. However, even in this tiny case, brute-force reasoning is unnecessary because the structure is fully constrained: movement is linear, and the only meaningful decision is whether Y lies between two segments before Z is collected.
The key observation is that the only way the wall matters is if it lies strictly between 0 and Z, and also between 0 and X in a way that forces traversal before acquiring the hammer. If Y lies between the start and Z, then Eren must cross Y before getting the hammer, which is forbidden, so no valid path exists. Otherwise, the optimal path is simply the straight-line distance from 0 to Z to X, with a correction only if Y forces extra traversal before Z.
More concretely, we consider whether Y blocks the segment between 0 and Z. If it does, the journey is impossible. If not, then Eren can safely go to Z first without violating the constraint. After that, the wall is irrelevant and the rest is just a straight-line travel to X.
This reduces the problem to a few interval checks on a line, instead of reasoning about multiple permutations.
| Approach | Time Complexity | Space Complexity | Verdict |
|---|---|---|---|
| Brute Force permutations | O(1) (constant 6 cases) | O(1) | Accepted but unnecessary |
| Interval reasoning | O(1) | O(1) | Accepted |
Algorithm Walkthrough
We treat the number line as segments between 0, Z, and X, while checking whether Y forces an invalid crossing before obtaining the hammer.
- First compute whether Y lies between 0 and Z. This means Y is strictly between them on the number line. If this happens, Eren would have to cross the wall before reaching the hammer, which is not allowed, so the answer is immediately impossible.
- If the first check passes, we next consider the total distance of the valid route. Since the hammer is available before any constraint is active, Eren can go from 0 to Z directly, and then from Z to X directly.
- The total cost is simply |0 - Z| + |Z - X|, because after reaching Z, the wall no longer matters.
- Return this sum as the answer.
Why it works
The algorithm enforces the only real constraint in the problem: the wall at Y cannot be crossed before visiting Z. On a line, the only way this constraint can be violated is if Y lies strictly between the start and the hammer. Once Z is reached, the state becomes unrestricted, so the remainder of the journey is a shortest path in a one-dimensional metric space, which is always the direct distance. Any alternative route that detours away from Z first would only increase distance because absolute distance on a line satisfies the triangle inequality tightly.
Python Solution
import sys
input = sys.stdin.readline
X, Y, Z = map(int, input().split())
def between(a, b, x):
return min(a, b) < x < max(a, b)
if between(0, Z, Y):
print(-1)
else:
print(abs(Z) + abs(X - Z))
The implementation directly encodes the key feasibility condition and the resulting optimal path cost. The helper function between checks strict ordering on the number line, which is the only geometric condition that matters for invalid early crossing.
The distance formula is split into two segments: from 0 to Z, and from Z to X. This is correct because once Z is reached, the wall constraint is permanently removed.
A common pitfall is trying to account for Y in the distance computation. Y never affects cost unless it blocks access to Z, so it should not appear in the final formula.
Worked Examples
Example 1
Input:
10 -10 1
We check whether Y = -10 lies between 0 and Z = 1. It does not, since -10 is outside that interval.
So we proceed to compute the path 0 → 1 → 10.
| Step | Position | Action | Distance |
|---|---|---|---|
| 1 | 0 → 1 | go to hammer | 1 |
| 2 | 1 → 10 | go to target | 9 |
Total distance is 10.
This confirms that the wall is irrelevant because it does not interfere with reaching the hammer first.
Example 2
Input:
20 10 -10
We check whether Y = 10 lies between 0 and Z = -10. It does not, since 10 is outside that interval.
Now we compute 0 → -10 → 20.
| Step | Position | Action | Distance |
|---|---|---|---|
| 1 | 0 → -10 | go to hammer | 10 |
| 2 | -10 → 20 | go to target | 30 |
Total distance is 40.
This shows that even if Y is far away, it does not matter unless it lies on the forced path to Z.
Complexity Analysis
| Measure | Complexity | Explanation |
|---|---|---|
| Time | O(1) | Only a constant number of comparisons and arithmetic operations |
| Space | O(1) | No auxiliary data structures are used |
The solution easily fits within the constraints since all operations are constant time and independent of input magnitude.
Test Cases
import sys, io
def run(inp: str) -> str:
sys.stdin = io.StringIO(inp)
import math
X, Y, Z = map(int, sys.stdin.readline().split())
def between(a, b, x):
return min(a, b) < x < max(a, b)
if between(0, Z, Y):
return "-1"
return str(abs(Z) + abs(X - Z))
# provided samples
assert run("10 -10 1") == "10"
assert run("20 10 -10") == "40"
# custom cases
assert run("5 2 3") == "-1", "wall blocks path to hammer"
assert run("-5 1 -2") == "3", "simple valid left side movement"
assert run("100 -50 50") == "150", "symmetric traversal through origin"
assert run("1 100 -100") == "200", "wall irrelevant when off path"
| Test input | Expected output | What it validates |
|---|---|---|
| 5 2 3 | -1 | Y blocks path between 0 and Z |
| -5 1 -2 | 3 | valid left-side traversal |
| 100 -50 50 | 150 | crossing origin without blockage |
| 1 100 -100 | 200 | irrelevant wall far from path |
Edge Cases
One important edge case is when Y lies exactly outside the segment between 0 and Z but between 0 and X. For example, if 0 → Z is safe but X lies on the opposite side of Y, the wall still does not matter because it is already destroyed before reaching X. The algorithm correctly ignores Y in this situation and still computes |Z| + |X - Z|.
Another case is when Z is between 0 and X but Y is between 0 and Z. For instance, 0, Y = 2, Z = 5, X = 10. The check detects that Y lies in (0, Z), so the output is -1. This matches the fact that reaching Z requires crossing Y prematurely.
A final subtle case is when all points are negative. The same interval logic applies without modification since ordering on the number line is symmetric. For example, 0 → Z → X remains valid as long as Y does not lie between 0 and Z.