CF 2041A - The Bento Box Adventure
The problem presents a scenario where a person visits one restaurant each day from Monday to Thursday, choosing a different restaurant each day from a set of five possible restaurants. The input gives the sequence of four distinct restaurants visited, one per day.
CF 2041A - The Bento Box Adventure
Rating: 1300
Tags: implementation, sortings
Solve time: 1m 25s
Verified: yes
Solution
Problem Understanding
The problem presents a scenario where a person visits one restaurant each day from Monday to Thursday, choosing a different restaurant each day from a set of five possible restaurants. The input gives the sequence of four distinct restaurants visited, one per day. The task is to determine which restaurant has not yet been visited and should be chosen for Friday.
In formal terms, the input consists of four integers between 1 and 5, all distinct. The output is the single integer from 1 to 5 that is missing from this list. Given the small input size, efficiency is not a major concern, but correctness is paramount.
Non-obvious edge cases include situations where the missing restaurant is either the lowest number (1) or the highest number (5). A careless implementation might assume the missing number is always in the middle of the range, which would fail for inputs like 2 3 4 5 (missing 1) or 1 2 3 4 (missing 5).
Approaches
A straightforward brute-force approach is to iterate through the numbers 1 to 5 and check each against the visited list. If a number is not found, it is returned as the answer. This works because the list of visited restaurants is guaranteed to contain exactly four distinct numbers, so exactly one number will be missing. The brute-force iteration checks at most five numbers, which is trivial in terms of computation.
The optimal approach leverages the property that the numbers 1 through 5 form a complete set. By calculating the sum of the set {1, 2, 3, 4, 5} and subtracting the sum of the visited restaurants, we immediately obtain the missing number. This uses basic arithmetic instead of iteration and scales naturally if the range or number of restaurants changes, though in this problem the constant size makes both methods effectively instantaneous.
The key insight is recognizing that the sum of the first five positive integers is fixed (15), and subtracting the sum of the four visited restaurants isolates the unvisited one. This avoids unnecessary checks or complicated data structures.
| Approach | Time Complexity | Space Complexity | Verdict |
|---|---|---|---|
| Brute Force | O(5 * 4) = O(20) | O(1) | Accepted |
| Sum Calculation | O(1) | O(1) | Accepted |
Algorithm Walkthrough
- Read the four integers from input, representing the restaurants visited from Monday to Thursday. Store them in a list
visited. - Compute the sum of the first five positive integers, which is
15. - Compute the sum of the four visited restaurants.
- Subtract the sum of visited restaurants from
15. The result is the unvisited restaurant number. - Output the result.
This works because the sum of a complete set minus the sum of a subset leaves exactly the missing element. The algorithm does not depend on order or position of the input numbers, ensuring correctness for all valid inputs.
Python Solution
import sys
input = sys.stdin.readline
visited = list(map(int, input().split()))
all_sum = 15
visited_sum = sum(visited)
print(all_sum - visited_sum)
The first line ensures fast input reading. We map the four space-separated integers into a list. Calculating the sum of the four numbers and subtracting from 15 directly yields the missing restaurant. No loops or conditionals are needed beyond input parsing, minimizing room for off-by-one errors.
Worked Examples
Sample 1
Input: 1 3 2 5
| Step | Variable | Value |
|---|---|---|
| 1 | visited | [1, 3, 2, 5] |
| 2 | all_sum | 15 |
| 3 | visited_sum | 1 + 3 + 2 + 5 = 11 |
| 4 | result | 15 - 11 = 4 |
Output: 4
This demonstrates that subtracting the sum of visited restaurants isolates the missing one correctly.
Custom Example 2
Input: 2 3 4 5
| Step | Variable | Value |
|---|---|---|
| 1 | visited | [2, 3, 4, 5] |
| 2 | all_sum | 15 |
| 3 | visited_sum | 2 + 3 + 4 + 5 = 14 |
| 4 | result | 15 - 14 = 1 |
Output: 1
This confirms the algorithm correctly identifies the smallest missing number.
Complexity Analysis
| Measure | Complexity | Explanation |
|---|---|---|
| Time | O(1) | Only four numbers are summed and a subtraction is performed, independent of input order. |
| Space | O(1) | Stores only the list of four integers and a few scalar variables. |
Given the constraints, this solution executes in negligible time and memory, well within the 1-second, 1024 MB limit.
Test Cases
import sys, io
def run(inp: str) -> str:
sys.stdin = io.StringIO(inp)
visited = list(map(int, input().split()))
all_sum = 15
visited_sum = sum(visited)
return str(all_sum - visited_sum)
# Provided sample
assert run("1 3 2 5") == "4", "sample 1"
# Custom cases
assert run("2 3 4 5") == "1", "missing smallest"
assert run("1 2 3 4") == "5", "missing largest"
assert run("1 2 4 5") == "3", "missing middle"
assert run("1 3 4 5") == "2", "missing lower-middle"
| Test input | Expected output | What it validates |
|---|---|---|
| 1 3 2 5 | 4 | Standard case |
| 2 3 4 5 | 1 | Missing smallest number |
| 1 2 3 4 | 5 | Missing largest number |
| 1 2 4 5 | 3 | Missing middle number |
| 1 3 4 5 | 2 | Missing lower-middle |
Edge Cases
For input 2 3 4 5, the algorithm correctly computes 15 - (2+3+4+5) = 1. It does not assume missing numbers are in the middle. For input 1 2 3 4, it computes 15 - 10 = 5, showing it also handles the upper boundary. Any permutation of four distinct numbers between 1 and 5 will be handled identically, confirming robustness.