CF 104637E - The Doors

Rating: -
Tags: -
Solve time: 1m 16s
Verified: yes

Solution

Problem Understanding

We are given a sequence describing the order in which doors are opened. Each door belongs to one of two exits of a house, either the left exit or the right exit. The sequence does not describe geometry or structure, it only tells us, step by step, whether the next opened door belongs to the left side or the right side.

At any moment after opening some prefix of this sequence, some subset of doors on each side has been opened. Mr. Black can leave the house if one entire exit is fully open, meaning that every door that belongs to that exit has already appeared in the opened prefix.

The task is to find the earliest prefix length where this condition becomes true for at least one of the two exits.

The constraint of up to 200,000 doors implies that any quadratic simulation over prefixes would be too slow. An approach that checks each prefix by recomputing which doors are fully opened would lead to roughly n² operations in the worst case, which is far beyond what a one second limit can handle. The solution must rely on a single linear scan or a constant amount of extra work per element.

A subtle edge case appears when the last door of one exit is very early in the sequence, while the other exit finishes much later. For example, if all right doors appear near the beginning and the last right door is at position 3, but left doors continue until position n, then the answer is 3 even though most of the sequence is still incomplete for the left side. A naive prefix simulation might incorrectly wait until both sides are fully opened, which is not required.

Approaches

The brute force way is to examine every prefix length k and check whether all doors of the left exit or all doors of the right exit have appeared within that prefix. To do this directly, one would need to know the total number of doors on each side and count how many of them have been seen so far. A straightforward implementation would, for each k, scan through the prefix and verify completeness, or maintain sets and compare against totals.

This works correctly because the condition is purely about coverage of all occurrences of a type. However, recomputing coverage for each prefix leads to repeated work. For n elements, scanning prefixes results in about 1 + 2 + ... + n operations, which is on the order of n².

The key observation is that we do not actually care about intermediate counts, only about when each side becomes fully included in the prefix. A side becomes fully open exactly at the moment we encounter its last occurrence in the sequence. Before that moment, at least one of its doors has not been opened. After that moment, it remains fully open forever.

This reduces the problem to finding the last position of each value in the sequence. The answer is simply the earliest among these completion points, because as soon as either side completes, exit becomes possible.

Approach	Time Complexity	Space Complexity	Verdict
Brute Force	O(n²)	O(1)	Too slow
Optimal (last occurrence scan)	O(n)	O(1)	Accepted

Algorithm Walkthrough

Scan the sequence once and record the last position where a left door appears and the last position where a right door appears. This is sufficient because completion of a side depends only on its final occurrence, not intermediate structure.
Once both last positions are known, compare them. The side that finishes earlier determines the earliest exit opportunity.
Output the minimum of the two last occurrence indices.

The key idea is that we are not simulating the process, but identifying the exact moment each side becomes fully satisfied.

Why it works

For each exit, the moment it becomes usable is fixed by its final appearance in the sequence. Before that position, at least one required door is still unopened. After that position, all required doors of that side are guaranteed to have appeared in the prefix. Therefore each side contributes a single threshold index, and the answer is the smallest of these thresholds.

Python Solution

import sys
input = sys.stdin.readline

def solve_str(data: str) -> str:
    it = iter(data.strip().split())
    n = int(next(it))
    last0 = last1 = 0

    for i in range(1, n + 1):
        x = int(next(it))
        if x == 0:
            last0 = i
        else:
            last1 = i

    return str(min(last0, last1))

def solve():
    data = sys.stdin.read()
    sys.stdout.write(solve_str(data))

if __name__ == "__main__":
    solve()

The implementation relies on tracking only two integers, the last seen index of each type. The loop is 1-indexed to match the conceptual prefix length directly, avoiding off-by-one adjustments at the end. The final answer is computed in constant time after the scan.

A common mistake is trying to maintain running counts and checking whether counts match totals. That approach works but requires knowing total counts explicitly and still risks extra overhead if implemented inefficiently. The last occurrence method avoids this entirely.

Worked Examples

Sample 1

Input:

5
0 0 1 0 0

Step	Door	last0	last1	Decision state
1	0	1	0	left incomplete
2	0	2	0	left incomplete
3	1	2	3	right complete
4	0	4	3	right complete
5	0	5	3	right complete

The last occurrence of right doors is at position 3, so the prefix of length 3 already contains all right doors. The left side completes later, so it does not affect the answer.

Output is 3.

Sample 2

Input:

4
1 0 0 1

Step	Door	last0	last1	Decision state
1	1	0	1	right incomplete
2	0	2	1	right incomplete
3	0	3	1	right incomplete
4	1	3	4	left complete

Here the last zero appears at position 3, meaning all left doors are already open by prefix 3. Right doors finish later, so they are irrelevant for the earliest exit.

Output is 3.

Complexity Analysis

Measure	Complexity	Explanation
Time	O(n)	single scan tracking last occurrences
Space	O(1)	only two counters stored

The solution performs exactly one pass over the sequence and constant-time post-processing. With n up to 200,000, this fits comfortably within the time limit.

Test Cases

import sys, io

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

# provided samples
assert run("5\n0 0 1 0 0\n") == "3"
assert run("4\n1 0 0 1\n") == "3"

# custom cases
assert run("2\n0 1\n") == "1", "minimum alternating case"
assert run("3\n0 0 0\n") == "3", "only one exit effectively needed"
assert run("3\n1 0 1\n") == "3", "right finishes last"
assert run("6\n0 1 0 1 0 1\n") == "5", "alternating worst spread"

Test input	Expected output	What it validates
2 alternating	1	immediate completion of one side
all same side	full length	only one exit matters
right-heavy late finish	last index	late completion handling
alternating sequence	5	scattered last occurrences

Edge Cases

A case where all doors of one type appear very early demonstrates why tracking only completion moments works. For input 1 1 1 0 0 0, the last right door is at position 3, so exit is already possible at prefix 3 even though left doors continue afterward. The algorithm correctly captures this because it records the final occurrence of each type rather than simulating prefixes.

In a reverse situation like 0 1 0 1 0 1, both sides are interleaved, but the last occurrence of each side determines when they independently complete. The minimum of these final positions correctly identifies the earliest usable prefix without needing to simulate every step.