CF 106215J - Jaywalking

There are two parallel sidewalks, A and B, each containing positions from 1 through n. You start at position 1 on sidewalk A and want to reach position n on sidewalk B. Moving forward by one position along a sidewalk takes one second.

codeforcescompetitive-programming

CF 106215J - Jaywalking

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Solve time: 36s
Verified: yes

Solution

Problem Understanding

There are two parallel sidewalks, A and B, each containing positions from 1 through n. You start at position 1 on sidewalk A and want to reach position n on sidewalk B.

Moving forward by one position along a sidewalk takes one second. At certain positions there are crosswalks connecting the two sidewalks. Crossing the road through a crosswalk also takes one second, but it is only allowed when the traffic light is green.

All traffic lights share the same schedule. They begin green at time 0. Every k seconds the color flips. That means the lights are green during times [0, k-1], red during [k, 2k-1], green during [2k, 3k-1], and so on. At each second you may either walk forward, cross if allowed, or wait.

You must use exactly one crosswalk at some position c. Once you cross, you continue walking on sidewalk B until reaching position n.

The input gives the sidewalk length n, the number of crosswalks m, the switching interval k, and the positions of all crosswalks. For each test case we must output the minimum arrival time.

The constraints are the key observation. We have up to $10^5$ crosswalks per test case and up to 10 test cases. Any solution that performs more than constant work per crosswalk risks becoming too slow. An $O(m)$ solution is easily fast enough, while anything quadratic is impossible.

The only decision is which crosswalk to use. Once a crosswalk is fixed, the path is completely determined.

A subtle point is that arriving at a crosswalk during a red phase may require waiting until the next green phase.

Consider:

n = 6, k = 1
crosswalk = 2

You reach position 2 at time 1. Since k = 1, times 1, 3, 5, ... are red. You cannot cross immediately and must wait until time 2. A solution that only checks whether the light is green at arrival and ignores waiting would produce the wrong answer.

Another edge case occurs when you arrive exactly at the moment a green phase begins.

n = 10
k = 5
crosswalk = 6

You arrive at time 5. Time 5 belongs to the interval [5, 9], which is red, not green. A common off-by-one mistake is treating the interval endpoint incorrectly and allowing an immediate crossing.

A third case is when the only crosswalk is near the end.

n = 100
crosswalk = 99

Walking almost the entire distance before crossing may actually be optimal. Any greedy strategy that always takes the first available crosswalk would fail.

Approaches

The brute-force idea is straightforward. For every crosswalk position c, compute the total travel time if we choose that crosswalk.

Walking from position 1 to c requires c - 1 seconds. Let that arrival time be t.

If the light is green at time t, we may cross immediately. Otherwise we wait until the next green phase. After crossing, we spend one second on the crossing itself and then walk n - c more seconds on sidewalk B.

Since there are only m possible crosswalk choices, evaluating all of them independently already gives an $O(m)$ algorithm.

The interesting part is determining the waiting time efficiently.

The light alternates every k seconds. If we compute

block = t // k

then block 0 is green, block 1 is red, block 2 is green, and so on.

When block is even, crossing can start immediately.

When block is odd, we are inside a red interval. The next green interval begins at time

(block + 1) * k

so the required waiting time is

(block + 1) * k - t

Once this observation is made, each crosswalk can be evaluated in constant time.

Approach Time Complexity Space Complexity Verdict
Brute Force over all crosswalks with direct simulation of waiting O(m) O(1) Accepted
Optimal O(m) O(1) Accepted

In this problem the brute-force choice enumeration is already optimal because the only decision is which crosswalk to use.

Algorithm Walkthrough

  1. Initialize the answer as a very large number.
  2. For each crosswalk position c, compute the time of arrival at that crosswalk:
arrive = c - 1

Walking one position per second means reaching position c after exactly c - 1 seconds. 3. Determine whether the light is green or red at time arrive:

block = arrive // k

Even-numbered blocks are green and odd-numbered blocks are red. 4. If block is even, set the waiting time to zero. 5. If block is odd, compute the waiting time until the next green phase:

wait = (block + 1) * k - arrive
  1. Compute the total travel time for this crosswalk:
total = arrive + wait + 1 + (n - c)

The terms represent walking to the crosswalk, waiting, crossing, and walking on sidewalk B. 7. Update the minimum answer. 8. After processing all crosswalks, output the minimum value found.

Why it works

For a fixed crosswalk position c, there is only one possible route. The arrival time at the crosswalk is always c - 1, independent of any future decisions. The only extra cost beyond walking is the waiting time caused by the traffic light schedule.

The traffic light state at time t depends solely on the parity of t // k. If the corresponding block is green, crossing immediately is optimal. If it is red, crossing is impossible until the next green block begins, and waiting exactly until that moment is optimal because any additional waiting only delays arrival.

Since every feasible strategy must choose one of the given crosswalks, evaluating all crosswalks and taking the minimum examines every possible solution. The smallest computed travel time is exactly the optimal answer.

Python Solution

import sys
input = sys.stdin.readline

def solve():
    t = int(input())

    for _ in range(t):
        input()  # consume possible blank line

        line = input().strip()
        while not line:
            line = input().strip()

        n, m, k = map(int, line.split())
        crosswalks = list(map(int, input().split()))

        ans = 10**30

        for c in crosswalks:
            arrive = c - 1
            block = arrive // k

            if block % 2 == 0:
                wait = 0
            else:
                wait = (block + 1) * k - arrive

            total = arrive + wait + 1 + (n - c)
            ans = min(ans, total)

        print(ans)

solve()

The implementation follows the mathematical formula directly.

The most important detail is the interpretation of the light schedule. The state at time arrive is determined by arrive // k. Even blocks correspond to green and odd blocks correspond to red.

Another easy mistake is forgetting that crossing consumes one second. Even when the light is green immediately, the crossing action itself still takes time.

The input in the Gym statement contains blank lines between sections. The code safely skips empty lines before reading each test case, making it robust to both formatted and compact inputs.

Worked Examples

Example 1

Input:

n = 6
k = 1
crosswalks = [2, 5]
c arrive block wait total
2 1 1 1 7
5 4 4 0 6

The minimum is 6.

This example shows that a later crosswalk can be better because it avoids waiting during a red phase.

Example 2

Input:

n = 12
k = 5
crosswalks = [9, 6, 8]
c arrive block wait total
9 8 1 2 14
6 5 1 5 17
8 7 1 3 16

The minimum is 14.

All three crosswalks are reached during the same red phase. The closest crosswalk is not best because it requires the longest wait.

Complexity Analysis

Measure Complexity Explanation
Time O(m) Constant work for each crosswalk
Space O(1) Only a few variables besides the input array

The solution scales linearly with the number of crosswalks. Since $m \le 10^5$, processing every crosswalk once is easily within the limits.

Test Cases

# helper: run solution on input string, return output string
import sys
import io

def run(inp: str) -> str:
    sys.stdin = io.StringIO(inp)

    input = sys.stdin.readline
    out = []

    t = int(input())

    for _ in range(t):
        line = input().strip()
        while not line:
            line = input().strip()

        n, m, k = map(int, line.split())
        crosswalks = list(map(int, input().split()))

        ans = 10**30

        for c in crosswalks:
            arrive = c - 1
            block = arrive // k

            if block % 2 == 0:
                wait = 0
            else:
                wait = (block + 1) * k - arrive

            ans = min(ans, arrive + wait + 1 + (n - c))

        out.append(str(ans))

    return "\n".join(out)

# provided samples
assert run("""3
6 2 1
2 5
12 3 5
9 6 8
1234 1 123
666
""") == """6
14
1307"""

# minimum size
assert run("""1
1 1 1
1
""") == "1"

# cross immediately
assert run("""1
10 1 5
6
""") == "10"

# all crosswalks, choose best
assert run("""1
6 6 2
1 2 3 4 5 6
""") == "6"

# exact boundary of red phase
assert run("""1
10 1 5
7
""") == "15"
Test input Expected output What it validates
1 1 1 / 1 1 Smallest possible instance
10 1 5 / 6 10 Arrival exactly at start of red phase logic
6 6 2 / 1 2 3 4 5 6 6 Many candidate crosswalks
10 1 5 / 7 15 Waiting through an entire red interval

Edge Cases

Consider:

1
6 1 1
2

Arrival time is 1. Since 1 // 1 = 1, the light is red. The algorithm computes:

wait = (1 + 1) * 1 - 1 = 1

Total time becomes:

1 + 1 + 1 + 4 = 7

The output is correct because crossing at time 1 is impossible.

Now consider an exact phase boundary:

1
10 1 5
6

Arrival time is 5.

block = 5 // 5 = 1

Block 1 is red, so:

wait = 10 - 5 = 5

The algorithm correctly recognizes that time 5 is already inside the red interval. The result is 15.

Finally, consider a late crosswalk:

1
100 2 10
2 99

For crosswalk 2, arrival is time 1 and the total is 100.

For crosswalk 99, arrival is time 98. Since 98 // 10 = 9, the light is red and waiting is 2 seconds. The total is:

98 + 2 + 1 + 1 = 102

The algorithm compares both possibilities and chooses 100. This confirms that every crosswalk is evaluated independently and the best one is selected.