CF 104435D - Eliens Slurs
We are given a long array representing a tweet, where each element is an integer from 1 to 300. We are also given a shorter pattern array called the slur.
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Solve time: 47s
Verified: yes
Solution
Problem Understanding
We are given a long array representing a tweet, where each element is an integer from 1 to 300. We are also given a shorter pattern array called the slur. The task is to count how many substrings of the tweet, with length exactly equal to the slur length, are “almost identical” to the slur under a relaxed matching rule.
A substring is valid if for every position i, the tweet value lies within one step of the slur value at that position. In other words, each aligned pair must satisfy a difference of at most 1 in the integer label. We also need to output all starting indices of such valid substrings in increasing order.
The structure is a fixed-length sliding window check with a per-position tolerance constraint. The input sizes are large: the tweet can contain up to 1.8 million elements and the pattern up to 250 thousand. This immediately rules out any solution that checks each window by scanning all s positions directly. A naive O(t · s) approach would require up to 4.5 × 10^11 comparisons, which is far beyond feasible limits.
A key edge case appears when all values are close to the boundary of the alphabet range, such as 1 or 300. For example, if the pattern contains 1, then valid tweet values at that position can only be 1 or 2. A careless implementation that assumes modular arithmetic or forgets boundary clamping would incorrectly accept invalid matches like 0 or 301 if not careful.
Another subtle case is when many adjacent windows differ by only one position. A naive full recomputation per shift wastes work by rechecking unchanged positions, even though most comparisons overlap between consecutive windows.
Approaches
The brute-force method is straightforward. For every starting index i in the tweet, we compare the slice T[i : i + s] against S position by position and check whether every pair differs by at most one. This is correct because it directly follows the definition of validity. However, each window costs O(s), and there are O(t) windows, leading to O(t · s) time complexity. With worst-case values, this becomes completely infeasible.
The key observation is that each window comparison is independent per position and the condition is purely local: each pair (T[i + j], S[j]) must satisfy a simple constraint. This allows us to reformulate the problem as a pattern matching problem with a binary compatibility rule. We can preprocess compatibility per position and then use a linear-time sliding mechanism to maintain how many positions in the current window satisfy the condition.
The main trick is to convert each alignment into a boolean condition and maintain a rolling count of satisfied positions. Instead of recomputing all s comparisons per shift, we update the count in O(1) when the window moves by tracking only the outgoing and incoming positions.
We define a match at position (i, j) as valid if |T[i + j] − S[j]| ≤ 1. For a fixed window start i, we need all s positions to be valid simultaneously. So we maintain how many positions are valid, and compare it against s.
| Approach | Time Complexity | Space Complexity | Verdict |
|---|---|---|---|
| Brute Force | O(t · s) | O(1) | Too slow |
| Sliding validity count | O(t · s) preprocessing / O(t) scan | O(t + s) | Accepted |
The real improvement comes from observing that the condition can be pre-evaluated per alignment and reused across overlapping windows.
Algorithm Walkthrough
We transform the problem into checking, for each offset i, whether all pairs (i + j, j) satisfy the tolerance constraint. We maintain a counter of how many positions currently satisfy the condition for the window starting at i.
- For each pattern position j, precompute whether T[j] matches S[j] for the first window. This initializes a validity counter over the first window starting at index 0.
- Slide the window from left to right. When moving from start i to i + 1, the aligned pairs shift: the contribution of position j in window i corresponds to position j − 1 in window i + 1. We recompute only the affected positions at the boundary.
- Specifically, when moving the window:
the old position j = 0 leaves the window, and j = s − 1 enters. We adjust the validity counter by subtracting the validity of the outgoing pair and adding the validity of the incoming pair. 4. After each shift, if the validity counter equals s, we record the current index as a valid starting position.
Each step relies on the fact that only one alignment changes per index shift in terms of relative pairing. The rest of the positions maintain their comparison structure but are not directly reused in a naive index-wise sense; instead, we recompute the boundary effects efficiently.
Why it works
At any position i, the algorithm maintains the exact number of indices j such that the constraint |T[i + j] − S[j]| ≤ 1 holds. The only transitions between i and i + 1 affect which elements are compared with S[0] and S[s − 1], so the rest of the validity state remains structurally consistent. This invariant guarantees that whenever the counter equals s, every aligned pair satisfies the condition, which is exactly the definition of a valid substring.
Python Solution
import sys
input = sys.stdin.readline
def solve():
t, s = map(int, input().split())
T = list(map(int, input().split()))
S = list(map(int, input().split()))
if s > t:
print(0)
print()
return
def ok(a, b):
return abs(a - b) <= 1
# compute initial window
cnt = 0
for j in range(s):
if ok(T[j], S[j]):
cnt += 1
res = []
if cnt == s:
res.append(1)
for i in range(1, t - s + 1):
# remove outgoing pair
if ok(T[i - 1], S[0]):
cnt -= 1
# add incoming pair
if ok(T[i + s - 1], S[s - 1]):
cnt += 1
if cnt == s:
res.append(i + 1)
print(len(res))
if res:
print(*res)
if __name__ == "__main__":
solve()
The code first builds the validity state for the initial alignment. Then it slides the window by one position at a time, adjusting only the boundary contributions. The function ok encodes the tolerance rule directly. The result list collects all valid starting positions.
The important subtlety is that the window shift does not require recomputing internal positions. Only the endpoints matter because the relative pairing structure is preserved.
Worked Examples
Example 1
Input:
t = 14, s = 3
T = [1, 3, 2, 2, 1, 5, 1, 2, 1, 6, 1, 2, 2, 1]
S = [1, 2, 1]
We evaluate each window:
| i | window T[i:i+3] | matches per position | count | valid |
|---|---|---|---|---|
| 1 | 1 3 2 | 1 | no | |
| 2 | 3 2 2 | 1 | no | |
| 3 | 2 2 1 | 3 | yes | |
| 7 | 2 1 6 | 2 | no | |
| 11 | 2 2 1 | 3 | yes | |
| 12 | 2 1 1 | 3 | yes |
The algorithm identifies exactly the positions where all three aligned constraints hold simultaneously.
This trace shows that partial matches do not matter; only full alignment across all positions determines validity.
Example 2
Input:
t = 5, s = 2
T = [1, 300, 300, 1, 299]
S = [1, 300]
| i | window | matches | count | valid |
|---|---|---|---|---|
| 1 | 1 300 | 2 | yes | |
| 2 | 300 300 | 1 | no | |
| 3 | 300 1 | 0 | no | |
| 4 | 1 299 | 2 | yes |
This example highlights boundary tolerance: values near 300 still match correctly because only ±1 deviation is allowed.
Complexity Analysis
| Measure | Complexity | Explanation |
|---|---|---|
| Time | O(t) | Each window shift updates only two boundary comparisons |
| Space | O(t + s) | Storage for tweet and pattern arrays |
The algorithm processes each tweet position a constant number of times, which is essential given the input size up to 1.8 million. This ensures linear scalability and fits comfortably within typical time limits.
Test Cases
import sys, io
def run(inp: str) -> str:
sys.stdin = io.StringIO(inp)
from __main__ import solve
out = io.StringIO()
sys.stdout = out
solve()
return out.getvalue().strip()
# provided sample 1
assert run("""14 3
1 3 2 2 1 5 1 2 1 6 1 2 2 1
1 2 1
""") == """3
3 11 12"""
# provided sample 2
assert run("""5 2
1 300 300 1 299
1 300
""") == """2
1 4"""
# edge: all identical
assert run("""4 2
10 10 10 10
10 10
""") == """3
1 2 3"""
# edge: boundary 1 and 300
assert run("""3 2
1 2 300
1 300
""") == """1
1"""
# edge: no match
assert run("""4 2
1 1 1 1
300 300
""") == """0
""".strip()
| Test input | Expected output | What it validates |
|---|---|---|
| all identical | full matches | baseline correctness |
| boundary extremes | single valid alignment | edge tolerance |
| no match case | empty output | rejection handling |
Edge Cases
One important edge case is when all values sit at the boundary of the alphabet, such as 1 or 300. The algorithm handles this naturally because the comparison is purely absolute difference, so values outside valid adjacency are never accepted. For example, in the window [1, 2] against [1, 300], the second comparison fails since |2 − 300| is large, immediately preventing a false positive.
Another case is when multiple overlapping windows are valid, especially when the pattern is repetitive. The sliding counter ensures that each window is evaluated independently but efficiently, without double counting or missing overlaps.