CF 166A - Rank List

Rating: 1100
Tags: binary search, implementation, sortings
Solve time: 1m 31s
Verified: yes

Solution

Problem Understanding

We are given the final results of a programming contest. Each team has two values attached to it: how many problems it solved and its total penalty time.

The ranking rule is the standard ICPC-style ordering. A team ranks higher if it solved more problems. If two teams solved the same number of problems, the one with the smaller penalty ranks higher. Teams with exactly the same pair (solved, penalty) share the same positions in the standings.

The task is to determine how many teams share the k-th place.

The key detail is that the input is not guaranteed to already be sorted by rank. We must first reconstruct the standings according to the contest rules, then identify the team sitting at position k, and finally count how many teams have the same result.

The constraints are very small. There are at most 50 teams, so even quadratic algorithms are easily fast enough. A sort of 50 elements is essentially instantaneous. This means the problem is not about optimization pressure, it is about implementing the ranking rules correctly without subtle mistakes.

The most common mistake is misunderstanding what “sharing the k-th place” means. The problem is not asking for the team currently stored at index k. It asks for all teams whose result is identical to the team occupying the k-th position after sorting.

Consider this example:

After sorting, the standings become:

1st: (5,1)
2nd-4th: (4,10)
5th: (3,1)

The 4th place belongs to the (4,10) group, so the answer is 3.

A careless implementation might only check the exact index k-1 and output 1.

Another subtle case appears when all teams are identical:

Every team shares every place from 1 to 4. The correct answer is 4.

An incorrect implementation that tries to assign unique ranks manually may accidentally separate equal teams.

There is also a potential off-by-one mistake because Codeforces positions are 1-indexed while Python lists are 0-indexed. If k = 1, the relevant team is at index 0 after sorting.

Approaches

The brute-force idea is straightforward. We can compare every team against every other team and manually determine ranking groups. For each team, we could count how many teams are strictly better and then identify which teams belong to the same place group.

This works because the ranking rule is fully determined by pairwise comparisons. With n ≤ 50, even an O(n²) solution performs only 2500 comparisons in the worst case.

The weakness of this approach is not runtime, but complexity of implementation. Manually reconstructing ranks introduces unnecessary bookkeeping and makes edge cases around tied groups easier to mishandle.

The cleaner observation is that the contest ranking is exactly a sorting order. Teams are ordered by:

Larger solved count first.
Smaller penalty first.

Once the teams are sorted, the team at index k-1 defines the entire rank group for the k-th place. Every team with the same (solved, penalty) pair shares that place.

This turns the problem into two simple operations:

Sort the teams correctly.
Count how many teams equal the team at position k-1.

The structure of the ranking rule makes sorting the natural solution because equal results form contiguous blocks in sorted order.

Approach	Time Complexity	Space Complexity	Verdict
Brute Force	O(n²)	O(1)	Accepted
Optimal	O(n log n)	O(1) excluding sorting space	Accepted

Algorithm Walkthrough

Read the number of teams n and the target place k.
Store every team as a pair (problems_solved, penalty).
Sort the teams using the contest ranking rule.

Teams with more solved problems must come first, so we sort that value in descending order. For equal solved counts, smaller penalties must come first, so we sort penalties in ascending order. 4. Convert the contest position into a Python index.

Since contest places are 1-indexed, the team representing the k-th place is stored at index k-1. 5. Save the pair at index k-1.

This pair uniquely identifies the entire tied group sharing the k-th place. 6. Traverse the sorted list and count how many teams have exactly the same pair. 7. Print the count.

Why it works

After sorting, teams with identical results become adjacent because the ordering depends only on (solved, penalty). The team at position k-1 represents the exact performance associated with the k-th place. Every team with that same pair shares the place, and every team with a different pair belongs to another ranking group. Counting equal pairs gives the correct answer.

Python Solution

import sys
input = sys.stdin.readline

n, k = map(int, input().split())

teams = []

for _ in range(n):
    p, t = map(int, input().split())
    teams.append((p, t))

teams.sort(key=lambda x: (-x[0], x[1]))

target = teams[k - 1]

answer = 0

for team in teams:
    if team == target:
        answer += 1

print(answer)

The first section reads the input and stores each team as a tuple. Using tuples is convenient because equality checks become automatic and reliable.

The sorting line is the core of the solution:

teams.sort(key=lambda x: (-x[0], x[1]))

The negative sign on x[0] creates descending order for solved problems. The penalty remains positive because smaller penalties should rank higher.

After sorting, teams[k - 1] identifies the performance corresponding to the k-th place. The subtraction by one is necessary because Python uses zero-based indexing.

The final loop counts every occurrence of that exact tuple. Since tied teams are defined by equality of both values, tuple comparison perfectly matches the contest rules.

A common implementation mistake is sorting penalties in descending order as well. That would incorrectly rank larger penalties ahead of smaller ones.

Another frequent mistake is counting only consecutive equal teams around index k-1. While that also works because equal teams are contiguous after sorting, a full traversal is simpler and less error-prone.

Worked Examples

Sample 1

Input:

After sorting:

Index	Team `(p,t)`	Same as target?	Count
0	(4,10)	Yes	1
1	(4,10)	Yes	2
2	(4,10)	Yes	3
3	(3,20)	No	3
4	(2,1)	No	3
5	(2,1)	No	3
6	(1,10)	No	3

The target position is k = 2, so the target team is at index 1, which is (4,10).

There are three teams with this exact result, so the answer is 3.

This example demonstrates that a tied group can span multiple ranking positions simultaneously.

Sample 2

Input:

After sorting:

Index	Team `(p,t)`	Same as target?	Count
0	(5,3)	No	0
1	(3,1)	Yes	1
2	(3,1)	Yes	2
3	(3,1)	Yes	3
4	(3,1)	Yes	4

The target position is k = 4, so the target team is (3,1).

Four teams share that result, so four teams share the 4th place.

This example confirms that the requested place may lie inside a large tied block rather than at its beginning.

Complexity Analysis

Measure	Complexity	Explanation
Time	O(n log n)	Sorting dominates the runtime
Space	O(1) excluding sorting space	Only a few extra variables are used

With at most 50 teams, the runtime is tiny. Even much slower approaches would fit comfortably within the limits. The sorting-based solution is both efficient and very simple to reason about.

Test Cases

# helper: run solution on input string, return output string
import sys
import io

def solve():
    input = sys.stdin.readline

    n, k = map(int, input().split())

    teams = []

    for _ in range(n):
        p, t = map(int, input().split())
        teams.append((p, t))

    teams.sort(key=lambda x: (-x[0], x[1]))

    target = teams[k - 1]

    print(sum(1 for team in teams if team == target))

def run(inp: str) -> str:
    backup_stdin = sys.stdin
    backup_stdout = sys.stdout

    sys.stdin = io.StringIO(inp)
    sys.stdout = io.StringIO()

    solve()

    output = sys.stdout.getvalue()

    sys.stdin = backup_stdin
    sys.stdout = backup_stdout

    return output

# provided samples
assert run(
"""7 2
4 10
4 10
4 10
3 20
2 1
2 1
1 10
"""
) == "3\n", "sample 1"

assert run(
"""5 4
5 3
3 1
3 1
3 1
3 1
"""
) == "4\n", "sample 2"

# minimum size
assert run(
"""1 1
5 10
"""
) == "1\n", "single team"

# all equal
assert run(
"""4 2
3 20
3 20
3 20
3 20
"""
) == "4\n", "all teams tied"

# boundary around tied block
assert run(
"""5 3
5 1
4 10
4 10
4 10
3 5
"""
) == "3\n", "k inside tied group"

# different penalties with same solved count
assert run(
"""5 2
4 20
4 10
4 30
3 1
2 1
"""
) == "1\n", "penalty ordering"

Test input	Expected output	What it validates
Single team	1	Smallest possible input
All equal teams	4	Entire ranking is one tied block
`k` inside tied block	3	Shared positions handled correctly
Same solved count, different penalties	1	Secondary sorting key works correctly

Edge Cases

One tricky situation is when the requested place lies in the middle of a tied group.

Input:

After sorting:

(5,1)
(4,10)
(4,10)
(4,10)
(3,1)

The 4th place corresponds to index 3, which still belongs to (4,10). The algorithm stores that tuple as the target and counts all identical tuples, producing 3.

Another subtle case is when every team is identical.

Input:

Sorting changes nothing. The target tuple is (3,20), and every team matches it. The algorithm correctly returns 4.

An off-by-one boundary case appears when k = 1.

Input:

After sorting, the target is at index 0, namely (5,10). Only one team matches it, so the answer is 1.

Using teams[k] instead of teams[k - 1] would incorrectly select the second-ranked team.