CF 102697140 - Continental Breakfast

Rating: -
Tags: -
Solve time: 1m 32s
Verified: yes

Solution

Problem Understanding

We have two hotel breakfast menus. Each menu is represented by a list of food item names. The task is to find the food items that appear on both menus, then print how many such items there are followed by the common items in alphabetical order. The official problem specifies that the common items are printed once each, even though the input itself is presented as a list of menu entries.

For example, if the first menu contains Pastries Waffles Eggs and the second contains Waffles FrenchToast Eggs Cereal, the intersection is {Eggs, Waffles}. Alphabetical ordering gives Eggs before Waffles, so the output contains 2, followed by those two names.

The problem uses a one-second limit and 256 MB of memory. The statement does not expose explicit upper bounds for the menu sizes, so the natural target is linear or near-linear work in the total number of input items. A quadratic comparison of every item from the first menu against every item from the second can become expensive as the menus grow. A hash set gives expected linear membership checks, after which sorting only the common names determines the required output order.

There are several small cases that can cause an otherwise correct idea to fail in implementation. If the menus have no common item, for example

3Eggs Waffles Pastries2FrenchToast Cereal

the correct output is

There must not be an extra line containing a food name.

If the common items are not already alphabetically ordered, for example

3Waffles Eggs Pastries2Waffles Eggs

the correct output is

2EggsWaffles

A careless solution that preserves input order would incorrectly print Waffles first.

A duplicate name also should not make the intersection contain the same food multiple times. For example,

3Eggs Eggs Waffles2Eggs Waffles

has two distinct common food items, so the result is

2EggsWaffles

Using sets naturally handles this case.

Approaches

The direct brute-force solution stores the two menus as lists and compares every item in the first menu with every item in the second. Whenever two names match, the item is a common food. This works because an item belongs to both menus exactly when some occurrence in the first list equals some occurrence in the second list.

If the first menu has n 1 entries and the second has n 2 , this performs n 1 n 2 comparisons in the worst case. With both menus containing 10 5 entries, that is 10 10 comparisons, far beyond what a one-second program can handle. Handling duplicates correctly would also require additional bookkeeping.

The key observation is that the question asked for each item in the second menu is simply whether that string exists in the first menu. A hash set is designed for exactly this operation. Put every item from the first menu into a set, then scan the second menu and test membership in expected O(1) time per item. The result is a set as well, so a food appearing multiple times does not get printed multiple times.

The brute-force works because equality testing tells us exactly whether two entries represent the same food, but it repeats essentially the same search many times. The observation that menu membership can be represented by a hash set removes those repeated searches, reducing the intersection phase to expected linear time. The final alphabetical requirement is handled by sorting the resulting common names.

Approach	Time Complexity	Space Complexity	Verdict
Brute Force	O(n 1 n 2 ) plus sorting	O(k)	Too slow for large menus
Optimal	Expected O(n 1 +n 2 +klogk)	O(n 1 +k)	Accepted

Here k is the number of distinct food items shared by the two menus.

Algorithm Walkthrough

Read the number of items in the first menu and read all of its item names. Store them in a set because the only information needed later is whether a particular food exists in this menu.
Read the number of items in the second menu and scan its item names. For every item, check whether it is present in the first-menu set.
If the item exists in the first menu, insert it into a second set containing the common food items. Using a set prevents duplicate input entries from producing duplicate output lines.
Convert the common-item set to a list and sort it alphabetically. The problem requires alphabetical output regardless of the order in which the menus were given.
Print the number of common items, followed by one common food name per line.

Why it works

After processing the second menu, the common-item set contains exactly the food names that occur in both menus. Every inserted name came from the second menu and was verified to exist in the first-menu set, so nothing outside the intersection can enter it. Conversely, every food appearing in both menus is encountered while scanning the second menu and passes the membership test, so it is inserted. Since the result is a set, each distinct common food appears exactly once. Sorting then produces precisely the required alphabetical order.

Python Solution

Pythonimport sysinput = sys.stdin.readline
n1 = int(input())menu1 = set(input().split())
n2 = int(input())menu2 = input().split()
common = {item for item in menu2 if item in menu1}answer = sorted(common)
print(len(answer))for item in answer:    print(item)

The first two input lines give the size and contents of the first menu. The size is read separately because the menu itself is represented by one space-separated line. Converting the names directly to a set gives expected constant-time membership checks.

The second menu is read as a list because we need to inspect each of its entries. The set comprehension keeps only names that occur in the first menu and automatically removes duplicates.

Sorting is performed only after the intersection has been constructed. Sorting both entire menus would do unnecessary work, while sorting the common set requires only klogk comparisons.

The output uses len(answer) rather than the original menu sizes because the required number is the number of distinct food items in the intersection. The loop prints each result on its own line, including the case where the answer is empty, in which case only 0 is printed.

Worked Examples

For the first sample, the first menu is Pastries Waffles Eggs, while the second menu is Waffles FrenchToast Eggs Cereal.

Item from second menu	In first-menu set?	Common set
Waffles	Yes	`{Waffles}`
FrenchToast	No	`{Waffles}`
Eggs	Yes	`{Waffles, Eggs}`
Cereal	No	`{Waffles, Eggs}`

After sorting, the common names become Eggs, Waffles, giving:

2EggsWaffles

The trace demonstrates that membership testing is independent of the positions of the items in either menu. It also shows why sorting must happen after finding the intersection.

For the second sample, the first menu is Eggs Waffles Pastries and the second is FrenchToast Cereal.

Item from second menu	In first-menu set?	Common set
FrenchToast	No	`{}`
Cereal	No	`{}`

The resulting list is empty, so its length is zero and there are no following food names:

This exercises the empty-intersection case and confirms that the program does not attempt to print a nonexistent item.

Complexity Analysis

Measure	Complexity	Explanation
Time	Expected O(n 1 +n 2 +klogk)	Set construction and membership checks are expected linear, followed by sorting the k common names
Space	O(n 1 +k)	The first-menu set and common-item set are stored

The linear intersection phase is suitable for large menus because each food name is processed only a constant number of times on average. The only superlinear part is sorting the actual answer, which is necessary to produce alphabetical order. Since the statement gives a one-second limit without exposing a stricter menu-size bound, this is the standard scalable solution for the problem.

Test Cases

Pythonimport sysimport io
def solve():    input = sys.stdin.readline
    n1 = int(input())    menu1 = set(input().split())
    n2 = int(input())    menu2 = input().split()
    common = {item for item in menu2 if item in menu1}    answer = sorted(common)
    print(len(answer))    for item in answer:        print(item)

def run(inp: str) -> str:    old_stdin = sys.stdin    old_stdout = sys.stdout
    sys.stdin = io.StringIO(inp)    sys.stdout = io.StringIO()
    try:        solve()        return sys.stdout.getvalue()    finally:        sys.stdin = old_stdin        sys.stdout = old_stdout

# Provided sample 1assert run(    """3Pastries Waffles Eggs4Waffles FrenchToast Eggs Cereal""") == """2EggsWaffles""", "sample 1"
# Provided sample 2assert run(    """3Eggs Waffles Pastries2FrenchToast Cereal""") == """0""", "sample 2"
# Minimum-size menus with one common itemassert run(    """1Eggs1Eggs""") == """1Eggs""", "minimum-size common item"
# All items equal, checking duplicate handlingassert run(    """4Eggs Eggs Eggs Eggs3Eggs Eggs Eggs""") == """1Eggs""", "duplicates must produce one answer"
# Common items deliberately given in reverse alphabetical orderassert run(    """5Waffles Toast Eggs Bacon Coffee4Toast Coffee Eggs Bacon""") == """4BaconCoffeeEggsToast""", "alphabetical ordering"
# No common itemsassert run(    """3Apple Banana Cherry4Durian Elderberry Fig Grape""") == """0""", "empty intersection"

Test input	Expected output	What it validates
`1 / Eggs / 1 / Eggs`	`1` and `Eggs`	Minimum-size valid intersection
Four `Eggs` in the first menu and three in the second	`1` and `Eggs`	Duplicate names are printed once
`Waffles Toast Eggs Bacon Coffee` against `Toast Coffee Eggs Bacon`	`Bacon`, `Coffee`, `Eggs`, `Toast`	Alphabetical ordering and arbitrary input order
Disjoint menus	`0`	Empty intersection and output boundary

Edge Cases

For an empty intersection, consider

3Eggs Waffles Pastries2FrenchToast Cereal

The first set is {Eggs, Waffles, Pastries}. Neither FrenchToast nor Cereal belongs to it, so the common set remains empty. Sorting an empty set produces an empty list, and the program prints only 0. This avoids the common mistake of assuming that at least one matching item exists.

For an unordered intersection, consider

3Waffles Eggs Pastries2Waffles Eggs

Both names pass the membership test, giving the set {Waffles, Eggs}. Sets have no useful output order, so sorting produces [Eggs, Waffles]. The result is consequently

2EggsWaffles

For repeated food names, consider

3Eggs Eggs Waffles2Eggs Waffles

Both names are common, but Eggs is represented only once in the set. The algorithm produces

2EggsWaffles

rather than counting occurrences. This matches the problem's interpretation of common food items as distinct menu entries by name.