Editorial for K-Distinct Windows

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Approach

Maintain the frequency of each value inside the current window of length $x$ ~x~, together with the number of distinct values currently present.

Build the first window, check whether it contains at least $k$ ~k~ distinct values, then slide the window one position at a time:

remove the value leaving the window;
add the value entering the window;
update the distinct counter whenever a frequency changes between $0$ ~0~ and $1$ ~1~.

Each slide takes $O(1)$ ~O(1)~, so the full scan runs in $O(n)$ ~O(n)~ time.

Solution (Python)

from collections import defaultdict


n, x, k = map(int, input().split())
values = list(map(int, input().split()))

freq: dict[int, int] = defaultdict(int)
distinct = 0

for i in range(x):
    if freq[values[i]] == 0:
        distinct += 1
    freq[values[i]] += 1

answer = 1 if distinct >= k else 0

for right in range(x, n):
    left_value = values[right - x]
    freq[left_value] -= 1
    if freq[left_value] == 0:
        distinct -= 1

    if freq[values[right]] == 0:
        distinct += 1
    freq[values[right]] += 1

    if distinct >= k:
        answer += 1

print(answer)

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