CF 102697143 - Zoom Meeting
The task is to summarize the countries represented in a Zoom meeting. There are (n) attendees, and each attendee contributes one country name. We need to count how many attendees belong to each distinct country, then order the countries by decreasing attendee count.
Rating: -
Tags: -
Solve time: 1m 3s
Verified: yes
Solution
Problem Understanding
The task is to summarize the countries represented in a Zoom meeting. There are (n) attendees, and each attendee contributes one country name. We need to count how many attendees belong to each distinct country, then order the countries by decreasing attendee count. When two countries have the same count, their names are compared alphabetically.
Only the first five countries in this ordering are printed. If fewer than five distinct countries exist, all distinct countries are printed. Each output line contains the country name followed by its frequency.
The official statement does not give a numerical upper bound for (n), but the time limit is only 1 second and the memory limit is 256 MB. That makes a solution with quadratic work unsafe for large inputs. We should aim for a linear pass over the attendees followed by sorting the distinct countries. If there are (k) distinct countries, that gives (O(n + k\log k)), and (k\le n), so the worst case is (O(n\log n)).
There are two subtle cases that can easily cause incorrect output. First, fewer than five distinct countries can occur even when (n\ge5). For example,
9
USA
Germany
Brazil
China
USA
China
China
USA
USA
has only four distinct countries, so the correct output is
USA 4
China 3
Brazil 1
Germany 1
A careless implementation that always prints exactly five lines would attempt to print a nonexistent fifth country.
Second, equal frequencies must be resolved alphabetically. For example,
4
USA
USA
Mexico
Mexico
produces
Mexico 2
USA 2
even though both countries occur twice. Sorting only by frequency would leave their relative order unspecified and could produce the wrong answer.
Approaches
The direct brute-force approach is to examine every country and count how many times it appears in the complete attendee list. If we do this independently for every attendee, the worst case is when all (n) country names are different. The first country requires (n) comparisons, the second requires another (n), and so on, giving (n^2) comparisons in the worst case. More precisely, if we avoid recounting countries already processed, the number of comparisons is still on the order of (n^2), with roughly (n(n+1)/2) checks. That is unnecessary work because the same names are repeatedly compared.
The key observation is that every attendee contributes to exactly one country count. A dictionary can store the current frequency for each country, so each input name updates its count in expected (O(1)) time. After this pass, the dictionary contains exactly the information needed for the output.
We then sort the distinct countries using a key consisting of negative frequency followed by the country name. The negative value makes larger frequencies come first, while the name naturally supplies the required alphabetical tie-breaker.
The brute-force works because repeatedly scanning the original list eventually discovers the exact frequency of every country, but it fails because it recomputes information that has already been discovered. The dictionary changes the process from repeatedly asking "how many times does this name occur?" to maintaining the answer as each attendee is read. Once the counts are known, sorting only the distinct countries finishes the task.
| Approach | Time Complexity | Space Complexity | Verdict |
|---|---|---|---|
| Brute Force | (O(n^2)) | (O(n)) | Too slow for large (n) |
| Dictionary + Sorting | (O(n+k\log k)), where (k) is the number of distinct countries | (O(k)) | Accepted |
Algorithm Walkthrough
- Read (n), then process the next (n) country names one at a time. Store each name in a dictionary and increment its associated frequency. A country that has not appeared before starts with frequency zero, so the update naturally creates its first entry.
- Convert the dictionary entries into pairs containing the country name and its frequency. At this point there is exactly one pair for every distinct country, so no attendee information needs to be scanned again.
- Sort these pairs by decreasing frequency and, for equal frequencies, increasing country name. Using the key
(-count, country)expresses both ordering rules directly. - Take at most the first five sorted pairs. If there are fewer than five distinct countries, slicing automatically returns every available country instead of inventing extra output lines.
- Print each selected country followed by its count. The sorted order already matches the required ranking, so no additional processing is necessary.
The invariant is that after processing any prefix of the attendee list, the dictionary contains the exact number of occurrences of every country seen in that prefix. When the entire input has been processed, every country's frequency is exact. Sorting those exact frequencies with the specified tie-breaker produces precisely the required ranking, and taking the first five gives the required output.
Python Solution
import sys
input = sys.stdin.readline
def solve():
n = int(input())
count = {}
for _ in range(n):
country = input().strip()
count[country] = count.get(country, 0) + 1
ranking = sorted(count.items(), key=lambda item: (-item[1], item[0]))
for country, frequency in ranking[:5]:
print(country, frequency)
if __name__ == "__main__":
solve()
The dictionary count implements the first two algorithm steps. get(country, 0) handles both cases, whether the country has already appeared or is being encountered for the first time.
The sorting key is the central implementation detail. For an entry such as ("USA", 4), the key becomes (-4, "USA"). Python sorts tuples lexicographically, so a smaller first component means a larger original frequency comes first. If two frequencies are equal, their first components are equal and Python compares the country names alphabetically.
The slice ranking[:5] is also deliberate. It handles both the normal case with at least five distinct countries and the case where fewer than five distinct countries exist. In particular, it does not confuse the number of attendees with the number of distinct countries.
Python integers do not overflow, and the largest possible frequency is simply (n). The input is read with sys.stdin.readline, which is sufficient for the potentially large number of country lines.
Worked Examples
For the first sample, every country occurs exactly once. The dictionary therefore contains eleven entries, all with frequency one. Since the frequencies are tied, alphabetical order determines the ranking.
| Step | Country read | Updated count | Relevant ranking state |
|---|---|---|---|
| 1 | USA | USA = 1 | USA 1 |
| 2 | Sweden | Sweden = 1 | Sweden 1, USA 1 |
| 3 | Argentina | Argentina = 1 | Argentina 1, Sweden 1, USA 1 |
| 4 | Cameroon | Cameroon = 1 | Cameroon 1, Argentina 1, Sweden 1, USA 1 |
| 5 | Egypt | Egypt = 1 | Egypt 1, Cameroon 1, Argentina 1, Sweden 1, USA 1 |
| 6 | Australia | Australia = 1 | Australia 1, Argentina 1, Cameroon 1, Egypt 1, Sweden 1, ... |
| 7 | England | England = 1 | England 1, Australia 1, Argentina 1, ... |
| 8 | France | France = 1 | France 1, England 1, Australia 1, ... |
| 9 | Russia | Russia = 1 | Russia 1, France 1, England 1, ... |
| 10 | Mongolia | Mongolia = 1 | Mongolia 1, Russia 1, France 1, ... |
| 11 | India | India = 1 | India 1, Mongolia 1, Russia 1, ... |
After sorting, all counts are equal, so the first five names alphabetically are Argentina, Australia, Cameroon, Egypt, and England. The output is consequently:
Argentina 1
Australia 1
Cameroon 1
Egypt 1
England 1
This example confirms that the alphabetical tie-breaker is applied globally after frequencies have been determined.
For the second sample, there are only two distinct countries.
| Step | Country read | Updated count | Current counts |
|---|---|---|---|
| 1 | USA | USA = 1 | USA 1 |
| 2 | USA | USA = 2 | USA 2 |
| 3 | Mexico | Mexico = 1 | USA 2, Mexico 1 |
| 4 | Mexico | Mexico = 2 | USA 2, Mexico 2 |
The final frequencies are equal, so the country names decide the order. Mexico comes before USA, producing:
Mexico 2
USA 2
The example also exercises the distinction between the number of attendees and the number of distinct countries. Even though (n=4), the same logic works for any (n), and the output contains only as many lines as there are distinct countries, capped at five.
Complexity Analysis
| Measure | Complexity | Explanation |
|---|---|---|
| Time | (O(n+k\log k)) | Counting takes (O(n)), and sorting the (k) distinct countries takes (O(k\log k)). |
| Space | (O(k)) | The dictionary and sorted list store one entry per distinct country. |
Since (k\le n), the total worst-case time is (O(n\log n)), with (O(n)) auxiliary space. This comfortably avoids the quadratic behavior of repeatedly scanning the attendee list and fits the 1 second, 256 MB limits for the intended input sizes.
Test Cases
The original statement does not specify a numeric maximum for (n), so the maximum-size test below uses 100,000 distinct countries as a practical stress case rather than claiming that 100,000 is the official maximum.
import sys
import io
def solve():
input = sys.stdin.readline
n = int(input())
count = {}
for _ in range(n):
country = input().strip()
count[country] = count.get(country, 0) + 1
ranking = sorted(count.items(), key=lambda item: (-item[1], item[0]))
return "\n".join(
f"{country} {frequency}"
for country, frequency in ranking[:5]
)
def run(inp: str) -> str:
old_stdin = sys.stdin
sys.stdin = io.StringIO(inp)
try:
return solve() + "\n"
finally:
sys.stdin = old_stdin
# Sample 1
assert run("""11
USA
Sweden
Argentina
Cameroon
Egypt
Australia
England
France
Russia
Mongolia
India
""") == """Argentina 1
Australia 1
Cameroon 1
Egypt 1
England 1
""", "sample 1"
# Sample 2
assert run("""4
USA
USA
Mexico
Mexico
""") == """Mexico 2
USA 2
""", "sample 2"
# Sample 3
assert run("""9
USA
Germany
Brazil
China
USA
China
China
USA
USA
""") == """USA 4
China 3
Brazil 1
Germany 1
""", "sample 3"
# Minimum-size input
assert run("""1
Canada
""") == """Canada 1
""", "single attendee"
# All attendees from one country
assert run("""5
Japan
Japan
Japan
Japan
Japan
""") == """Japan 5
""", "only one distinct country"
# Exactly five distinct countries, all tied
assert run("""5
Z
Y
X
W
V
""") == """V 1
W 1
X 1
Y 1
Z 1
""", "five-way alphabetical tie"
# More than five distinct countries with frequency ties
assert run("""8
Zulu
Alpha
Bravo
Charlie
Delta
Echo
Alpha
Bravo
""") == """Alpha 2
Bravo 2
Charlie 1
Delta 1
Echo 1
""", "top five with tie-breaking"
# Large stress test with 100000 distinct countries
n = 100000
inp = str(n) + "\n" + "\n".join(f"C{i:06d}" for i in range(n)) + "\n"
expected = "\n".join(
f"C{i:06d} 1" for i in range(5)
) + "\n"
assert run(inp) == expected, "large distinct-country stress test"
| Test input | Expected output | What it validates |
|---|---|---|
1 / Canada |
Canada 1 |
Minimum-size input |
Five Japan entries |
Japan 5 |
All-equal values and fewer than five distinct countries |
Z, Y, X, W, V |
V 1 through Z 1 |
Exactly five distinct countries and alphabetical tie-breaking |
Zulu, Alpha, Bravo, Charlie, Delta, Echo, Alpha, Bravo |
Alpha 2, Bravo 2, Charlie 1, Delta 1, Echo 1 |
Frequency ordering followed by alphabetical ordering and the five-country cutoff |
| 100,000 distinct generated names | First five names alphabetically | Large input and sorting performance |
Edge Cases
A single attendee is the smallest meaningful input. For
1
Canada
the dictionary becomes {"Canada": 1}. Sorting leaves the only entry unchanged, and slicing to five entries still returns one entry. The output is
Canada 1
so the implementation does not assume that five countries always exist.
When every attendee has the same country, there is only one distinct entry regardless of how large (n) becomes. For
5
Japan
Japan
Japan
Japan
Japan
the five dictionary updates leave Japan at frequency 5. The sorted list has length one, so [:5] returns only that entry and the output is Japan 5. This catches implementations that incorrectly print five rows whenever (n\ge5).
Equal frequencies require the secondary alphabetical ordering. For
4
USA
USA
Mexico
Mexico
both dictionary entries finish with frequency 2. Their sorting keys are (-2, "USA") and (-2, "Mexico"). Since "Mexico" is alphabetically smaller, it comes first, giving Mexico 2 followed by USA 2.
The most revealing boundary case is when there are more than five distinct countries but fewer than five should be printed because the ranking is capped. Consider
8
Zulu
Alpha
Bravo
Charlie
Delta
Echo
Alpha
Bravo
The final counts are Alpha 2, Bravo 2, and four other countries with frequency 1. Sorting gives Alpha, Bravo, Charlie, Delta, Echo, Zulu, and slicing to five entries removes only Zulu. The output is
Alpha 2
Bravo 2
Charlie 1
Delta 1
Echo 1
This catches both common mistakes at once: sorting equal-frequency countries incorrectly and printing more than the required five ranked countries.