CF 1644A - Doors and Keys

The problem describes a narrow hallway with three doors and three keys, each uniquely colored red, green, or blue. The knight starts at the left end of the hallway, and the princess waits at the far right.

codeforcescompetitive-programmingimplementation

CF 1644A - Doors and Keys

Rating: 800
Tags: implementation
Solve time: 1m 14s
Verified: yes

Solution

Problem Understanding

The problem describes a narrow hallway with three doors and three keys, each uniquely colored red, green, or blue. The knight starts at the left end of the hallway, and the princess waits at the far right. The hallway is represented as a six-character string: uppercase letters indicate doors (R, G, B) and lowercase letters indicate keys (r, g, b). Each color appears exactly once as a door and once as a key.

The goal is to determine if the knight can traverse the hallway from left to right, opening each door in sequence. The knight can pick up keys as he moves and use them immediately when he encounters the corresponding door. The output is "YES" if all doors can be opened, otherwise "NO".

Given the constraints, the hallway is always six characters long. The number of test cases is up to 720, which is small enough that any linear-time approach per test case is efficient. Since there are only three doors and three keys, a brute-force check that simulates picking up keys and opening doors in order is feasible, but careful attention is required because a door appearing before its key blocks progress.

Edge cases involve a door immediately appearing before its key. For instance, in the map RgbrBG, the red door R is first, and the red key r appears later. The knight cannot open the first door, so the output must be "NO". A naive approach that just counts keys and doors without considering order would incorrectly say "YES".

Approaches

The brute-force approach iterates through the string from left to right, tracking the keys collected in a set. Each time the knight encounters a door, we check if the corresponding key has already been collected. If so, he opens the door; if not, the hallway cannot be traversed, and the answer is "NO". This approach works because the hallway length is fixed at six, making iteration trivial. The worst case involves examining all six characters per test case, resulting in 6 × 720 = 4320 operations, which is negligible.

The optimal approach follows the same linear scan. The key insight is that because each door has a unique key and both appear exactly once, the only situation that blocks progress is encountering a door before acquiring its key. Therefore, we only need to maintain a set of collected keys and check doors against it. There is no need for complex data structures or backtracking.

Approach Time Complexity Space Complexity Verdict
Brute Force Simulation O(6 * t) = O(t) O(3) = O(1) Accepted
Optimal Linear Scan O(t) O(1) Accepted

Algorithm Walkthrough

  1. Read the number of test cases t.
  2. For each test case, read the six-character string representing the hallway.
  3. Initialize an empty set to store the keys collected so far.
  4. Iterate through each character in the string from left to right.
  5. If the character is a key (r, g, b), add it to the set of collected keys.
  6. If the character is a door (R, G, B), check if its corresponding lowercase key is in the set. If not, print "NO" and stop processing this test case.
  7. If the end of the string is reached without encountering a blocked door, print "YES".

Why it works: at any moment, the set contains all keys the knight has passed. Doors can only be opened if their key is in the set. Since each key and door is unique and appears once, tracking collected keys guarantees correctness. There are no cycles or revisits, so a single left-to-right pass suffices.

Python Solution

import sys
input = sys.stdin.readline

t = int(input())
for _ in range(t):
    hallway = input().strip()
    keys_collected = set()
    possible = True
    for ch in hallway:
        if ch.islower():
            keys_collected.add(ch)
        else:  # door
            if ch.lower() not in keys_collected:
                possible = False
                break
    print("YES" if possible else "NO")

The code first reads the number of test cases and iterates through them. For each hallway string, it initializes a set to track keys. Each character is processed: keys are added to the set, doors are checked against the set. If a door cannot be opened, the loop breaks early. The result is printed immediately. Subtle points include converting doors to lowercase when checking keys and stripping the newline from input strings.

Worked Examples

Example 1: rgbBRG

Step Character Keys Collected Door Check Result
1 r {r} - continue
2 g {r, g} - continue
3 b {r, g, b} - continue
4 B {r, g, b} yes continue
5 R {r, g, b} yes continue
6 G {r, g, b} yes continue

The knight collects all keys first and can open all doors. Output: YES.

Example 2: RgbrBG

Step Character Keys Collected Door Check Result
1 R {} no break

The red door appears first, but the red key has not been collected. Output: NO.

These examples confirm that the left-to-right simulation captures the necessary order of keys relative to doors.

Complexity Analysis

Measure Complexity Explanation
Time O(t) Each test case processes 6 characters, so total operations scale linearly with t.
Space O(1) Only a set of up to 3 keys is stored per test case.

Given the small input size, this solution runs efficiently within time and memory limits.

Test Cases

import sys, io

def run(inp: str) -> str:
    sys.stdin = io.StringIO(inp)
    output = []
    t = int(input())
    for _ in range(t):
        hallway = input().strip()
        keys_collected = set()
        possible = True
        for ch in hallway:
            if ch.islower():
                keys_collected.add(ch)
            else:
                if ch.lower() not in keys_collected:
                    possible = False
                    break
        output.append("YES" if possible else "NO")
    return "\n".join(output)

# Provided samples
assert run("4\nrgbBRG\nRgbrBG\nbBrRgG\nrgRGBb\n") == "YES\nNO\nYES\nNO", "Sample 1"

# Custom test cases
assert run("1\nrbgRGB\n") == "YES", "all keys first"
assert run("1\nRGBrgb\n") == "NO", "all doors first"
assert run("1\nrGBgRb\n") == "NO", "blocked by first door"
assert run("1\nrgBGRb\n") == "YES", "keys before doors but mixed"
Test input Expected output What it validates
rbgRGB YES Collect all keys first
RGBrgb NO Doors appear before keys
rGBgRb NO First door blocks progress
rgBGRb YES Mixed order but all doors can be opened

Edge Cases

The critical edge case occurs when a door appears before its key. For RGBrgb, the knight encounters the red door immediately and has no key, so traversal is impossible. Our algorithm detects this on the first step and outputs "NO". Another subtle case is when all keys appear first, e.g., rbgRGB, allowing immediate opening of doors, which the algorithm correctly outputs as "YES". These tests confirm that order is the only factor affecting solvability, not the relative positions of non-blocking doors or keys.