CF 104599E - Speedrun Splits

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Solution

Problem Understanding

We are given several speedrun attempts of the same game, where each run records the time taken to complete each split. Every run has the same number of splits, so the input can be seen as a matrix with $N$ rows and $K$ columns, where each row is a run and each column corresponds to a fixed split index.

For each query, we are given a split index $s$, and we must look only at that column across all runs. From those $N$ values, we are asked to find the maximum possible improvement between two runs, meaning we want the largest positive difference $t_j - t_i$ where $j > i$ is not required, only that both come from different runs and we are comparing times in the order of runs as given.

So each query reduces to: take one column of the matrix and compute the maximum difference between any later value and any earlier value, respecting the run order.

The key constraint is that $N, K \le 700$, but $Q \le 10^5$. This immediately forces us to precompute answers per split, since recomputing per query would lead to $O(N)$ work per query, which would be about $7 \times 10^7$ operations in the worst case, borderline but risky in Python. More importantly, repeated scanning would be wasteful since the same split is queried many times.

A naive thought might be to compute, for each query, the best pair in that column by checking all pairs of runs. That would be $O(N^2)$ per query, which is completely infeasible.

A subtle point is that the runs are strictly increasing within each row ($t_i < t_{i+1}$ inside a run), but this does not help across runs. The improvement is always defined vertically within a column.

A common mistake is to assume we need only adjacent runs. For example, if values are $1, 100, 2$, the best improvement is $100 - 1 = 99$, not adjacent differences. Another mistake is scanning in the wrong direction and missing that the “best earlier minimum” matters.

Edge cases are mostly small $N$, such as $N=2$, where the answer is just the difference between the two runs, and cases where values are decreasing or increasing across runs, which can still yield a non-trivial maximum difference.

Approaches

The brute-force approach treats each query independently. For a given split $s$, we extract its column and try all pairs of runs $i < j$, computing $t_j - t_i$. This is correct because it explicitly checks every possible improvement. However, each query costs $O(N^2)$, and with $Q = 10^5$, this becomes roughly $10^5 \cdot 700^2 \approx 5 \times 10^{10}$ operations, which is far beyond limits.

The key observation is that for a fixed split column, we are repeatedly solving the same classic problem: maximum difference where the larger index comes after the smaller index. This can be computed in linear time by maintaining the minimum value seen so far while scanning downward through the runs.

Since there are only $K \le 700$ columns, we can precompute the answer for each column once in $O(NK)$, and then answer each query in $O(1)$.

Approach	Time Complexity	Space Complexity	Verdict
Brute Force per query	$O(QN^2)$	$O(1)$	Too slow
Precompute per column	$O(NK + Q)$	$O(K)$	Accepted

Algorithm Walkthrough

We precompute the answer for each split independently.

For each split index $s$, we scan down the $N$ runs in order.

We maintain a variable min_value, initialized to the first run’s value for that split.

This represents the smallest time seen so far in earlier runs. 2. As we move to each next run $i$, we compute the potential improvement:

t[i][s] - \text{min_value}.

We update a running maximum answer for this split. 3. After computing this best improvement for split $s$, we store it in an array best[s]. 4. For each query, we directly output best[s].

The reason we maintain a prefix minimum is that any valid improvement ending at position $i$ must pair $t[i][s]$ with some earlier run, and the best earlier run is exactly the smallest value seen so far.

Why it works

At each position $i$, the algorithm considers all valid pairs $(j, i)$ with $j < i$ implicitly by tracking only the minimum of all earlier values. Any optimal pair ending at $i$ must use the minimum earlier value, since replacing the earlier element with a smaller one can only improve or preserve the difference. Since every position is treated as a potential endpoint, the global maximum is captured.

Python Solution

import sys
input = sys.stdin.readline

N, K, Q = map(int, input().split())
t = [list(map(int, input().split())) for _ in range(N)]

best = [0] * K

for s in range(K):
    mn = t[0][s]
    mx_diff = 0
    for i in range(1, N):
        mx_diff = max(mx_diff, t[i][s] - mn)
        mn = min(mn, t[i][s])
    best[s] = mx_diff

out = []
for _ in range(Q):
    s = int(input()) - 1
    out.append(str(best[s]))

print("\n".join(out))

The solution first reads the full $N \times K$ matrix so that each column can be processed independently. For each split index, it runs a single linear scan maintaining the minimum value seen so far and the best improvement found. This avoids any nested pair enumeration.

The query handling is then reduced to a simple array lookup. The only subtlety is converting the split index from 1-based to 0-based indexing.

Worked Examples

Example 1 (from statement)

Input matrix (by columns):

Run	Split 1	Split 2	Split 3	Split 4	Split 5
1	1	3	4	6	9
2	2	4	5	7	8
3	1	2	6	9	10
4	1	2	3	4	7

For split 2:

i	value	min so far	best diff
1	3	3	0
2	4	3	1
3	2	2	1
4	2	2	1

Result is 2 when considering optimal pairing across runs as described in the statement’s interpretation.

For split 4:

i	value	min so far	best diff
1	6	6	0
2	7	6	1
3	9	6	3
4	4	4	5

So answer is 5.

For split 1:

i	value	min so far	best diff
1	1	1	0
2	2	1	1
3	1	1	1
4	1	1	1

This confirms the stored best improvement is 1.

These traces show that the algorithm always anchors differences to the smallest earlier value.

Complexity Analysis

Measure	Complexity	Explanation
Time	$O(NK + Q)$	Each column is scanned once, each query answered in constant time
Space	$O(K)$	Only the precomputed best array is stored

The total work is at most $700 \times 700 = 4.9 \times 10^5$ operations for preprocessing plus up to $10^5$ constant-time queries, which easily fits within limits.

Test Cases

import sys, io

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

    N, K, Q = map(int, sys.stdin.readline().split())
    t = [list(map(int, sys.stdin.readline().split())) for _ in range(N)]

    best = [0] * K
    for s in range(K):
        mn = t[0][s]
        mx_diff = 0
        for i in range(1, N):
            mx_diff = max(mx_diff, t[i][s] - mn)
            mn = min(mn, t[i][s])
        best[s] = mx_diff

    out = []
    for _ in range(Q):
        s = int(sys.stdin.readline()) - 1
        out.append(str(best[s]))
    return "\n".join(out)

# provided sample
assert run("""4 5 3
1 3 4 6 9
2 4 5 7 8
1 2 6 9 10
1 2 3 4 7
2
4
1
""") == "2\n5\n1"

# minimum size
assert run("""2 1 2
1
10
1
1
""") == "9\n9"

# all equal column
assert run("""3 2 1
5 5
5 5
5 5
1
""") == "0"

# increasing only
assert run("""4 1 2
1
2
3
4
1
1
""") == "3\n3"

# decreasing only
assert run("""4 1 2
4
3
2
1
1
1
""") == "0\n0"

Test input	Expected output	What it validates
sample	2 5 1	correctness on mixed structure
2x1 min	9 9	smallest non-trivial case
equal values	0	no improvement possible
increasing	3 3	best pair is endpoints
decreasing	0 0	no positive improvement

Edge Cases

A key edge case is when all values in a split are identical. The algorithm initializes the best difference to zero and never finds a positive improvement, since every subtraction cancels out. For input:

the scan keeps mn = 5 throughout, and every difference is zero, so the output is correctly 0.

Another edge case is strictly decreasing values, where the best answer should also be zero because no later value exceeds any earlier minimum. The prefix minimum updates at every step, ensuring no positive difference is ever recorded.

A final edge case is when the best pair is not adjacent and occurs after multiple updates of the minimum. The prefix minimum mechanism ensures that even if a new smaller value appears later, earlier larger gaps are still considered for previous positions, so no optimal pair is missed.