CF 1225B1 - TV Subscriptions (Easy Version)

CF 1225B1 - TV Subscriptions (Easy Version)

Rating: 1000
Tags: implementation
Solve time: 2m 41s
Verified: yes

Solution

Problem Understanding

We are given a schedule of TV shows over a sequence of days. Each day broadcasts exactly one show, and if we buy a subscription to a show, we gain access to all of its episodes for the entire timeline.

The task is not to cover all days, but to choose a set of shows such that there exists at least one contiguous block of exactly d days where every show appearing in that block is included in our purchased set. We want to minimize how many different shows we subscribe to in order to make such a block possible.

Another way to view it is that we are allowed to pick a window of length d, and for that window we must ensure that all distinct values inside it are “covered” by our chosen set. We are free to choose the best window after choosing the set, so the optimal strategy depends on finding a window whose distinct count is minimal.

The constraints are small: n ≤ 100 per test case and total n across tests also ≤ 100. This means an O(n^2) or even O(n^3) solution is acceptable, and we can safely examine every window explicitly without worrying about performance.

A subtle edge case appears when all elements are distinct in every window. In that case, the answer becomes exactly d, since any window of size d forces us to subscribe to all d shows. Another edge case is when a show repeats heavily, so that a window contains only one distinct show. Then the answer becomes 1, because buying that single show suffices for that window.

A naive mistake would be to think we must globally minimize distinct shows in the entire array. That is incorrect because the requirement applies only to one chosen window, not the whole schedule.

Approaches

A direct brute-force approach is to consider every contiguous segment of length d. For each segment, we compute how many distinct shows appear in it. If we buy subscriptions for exactly those shows, we can watch that segment, so the number of subscriptions needed for that segment is its distinct count. We take the minimum across all segments.

This works because the chosen segment determines exactly which shows we must subscribe to. However, recomputing distinct counts from scratch for each segment leads to inefficiency if done carelessly, since each window could cost O(d) and there are O(n) windows.

The key observation is that we do not need any global structure or advanced data structures. Since n is small, even recomputing a frequency map for each window is fast enough. Alternatively, we can maintain a sliding window frequency table and update it in O(1) per step, but it is not necessary.

The structure that makes this problem simple is that the decision is fully local to a window, and there is no interaction between different windows except taking a minimum.

Approach	Time Complexity	Space Complexity	Verdict
Brute Force (recompute per window)	O(n · d)	O(k)	Accepted
Sliding Window Frequency	O(n)	O(k)	Accepted

Algorithm Walkthrough

We iterate over all possible starting positions of a length d segment and compute the number of distinct elements inside it.

1. For each test case, read n, k, and d, along with the array of shows.
2. For every index i from 0 to n - d, consider the segment [i, i + d - 1]. This represents a possible block of consecutive days we might want to “support” with subscriptions.
3. For each segment, count how many distinct values appear inside it using a set or frequency array. This count represents the number of shows we must subscribe to if we choose this segment.
4. Track the minimum such distinct count over all segments.
5. Output this minimum value.

The key idea is that once we fix a segment, the minimal subscription set is forced: it is exactly the set of distinct shows in that segment. There is no benefit in subscribing to any show outside the segment, because it does not help satisfy the condition for that specific block.

Why it works

For any chosen segment of length d, we must be able to watch all its days. Since a subscription includes all episodes of a show, the only requirement is that every show appearing in that segment is included in our purchased set. Therefore, the optimal set for a fixed segment is exactly its distinct elements. Since we are allowed to choose the best segment, the final answer is the minimum distinct count over all segments. This ensures we are not overcounting or considering irrelevant parts of the array.

Python Solution

import sys
input = sys.stdin.readline

t = int(input())
for _ in range(t):
    n, k, d = map(int, input().split())
    a = list(map(int, input().split()))

    ans = float('inf')

    for i in range(n - d + 1):
        seen = set()
        for j in range(i, i + d):
            seen.add(a[j])
        ans = min(ans, len(seen))

    print(ans)

The core loop checks every possible window of length d. Inside each window, we build a set of encountered shows. The size of this set is exactly the number of subscriptions needed if we choose that window.

The initialization of ans as infinity ensures that any valid window improves it. Since constraints are small, rebuilding the set each time is sufficient and avoids complexity from maintaining incremental state.

Worked Examples

Example 1

Input:

5 2 2
1 2 1 2 1

We examine all windows of size 2:

Window start	Window	Distinct shows	Best so far
0	[1,2]	2	2
1	[2,1]	2	2
2	[1,2]	2	2
3	[2,1]	2	2

Every window has two distinct shows, so the answer is 2.

This shows the case where no overlap reduces complexity inside any window.

Example 2

Input:

9 3 3
3 3 3 2 2 2 1 1 1

Window start	Window	Distinct shows	Best so far
0	[3,3,3]	1	1
1	[3,3,2]	2	1
2	[3,2,2]	2	1
3	[2,2,2]	1	1
4	[2,2,1]	2	1
5	[2,1,1]	2	1
6	[1,1,1]	1	1

The minimum distinct count is 1 because there exist pure segments of identical shows.

This demonstrates that the optimal window is not necessarily unique and can appear anywhere in the array.

Complexity Analysis

Measure	Complexity	Explanation
Time	O(n · d)	We scan each window and recompute a set of size at most d
Space	O(k)	Frequency tracking or set size is bounded by number of shows

Given that n ≤ 100 and total n across tests is small, this computation is comfortably within limits even in Python.

Test Cases

import sys, io

def run(inp: str) -> str:
    sys.stdin = io.StringIO(inp)
    import sys
    input = sys.stdin.readline

    t = int(input())
    out = []
    for _ in range(t):
        n, k, d = map(int, input().split())
        a = list(map(int, input().split()))

        ans = float('inf')
        for i in range(n - d + 1):
            seen = set()
            for j in range(i, i + d):
                seen.add(a[j])
            ans = min(ans, len(seen))

        out.append(str(ans))
    return "\n".join(out)

# provided samples
assert run("""4
5 2 2
1 2 1 2 1
9 3 3
3 3 3 2 2 2 1 1 1
4 10 4
10 8 6 4
16 9 8
3 1 4 1 5 9 2 6 5 3 5 8 9 7 9 3
""") == """2
1
4
5"""

# all equal inside window
assert run("""1
5 5 3
7 7 7 7 7
""") == "1"

# all distinct
assert run("""1
5 5 3
1 2 3 4 5
""") == "3"

# tight window overlap
assert run("""1
6 10 2
1 2 3 1 2 3
""") == "2"

Test input	Expected output	What it validates
all equal	1	minimum distinct case
all distinct	d	worst-case distinct window
repeating pattern	2	overlapping structure correctness

Edge Cases

A fully uniform array shows the simplest structure. Every window contains a single value, so the algorithm consistently returns 1, matching the fact that subscribing to one show suffices.

A fully distinct array forces every window of length d to have exactly d unique shows. The algorithm correctly reflects this since the set size equals d for every segment.

Repetitive patterns like alternating or periodic sequences confirm that overlapping windows do not affect correctness, since each window is evaluated independently and only its internal distinct set matters.