0808MediumArrayDynamic Programming

Best Window Across Repeated Cycles

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Trains the technique from

LeetCode 1191K-Concatenation Maximum Sum

This is an original problem, written from a brief that listed the technique, the difficulty, the topics, the function shape and the input bounds — none of that problem's wording, examples, hints or editorials. The link is there so you can map your practice onto the standard set.

Same function shape, different story and different numbers.

A conveyor test rig runs the same programme during every shift. The programme is given as pattern, a list of the net balance changes it records in each of its slots, and the rig runs repeats identical shifts back to back. The full log is therefore pattern written out repeats times in order, so the log holds pattern.length * repeats entries.

A window is a contiguous run of entries of the full log, and its value is the sum of the entries it covers. The empty window is allowed and its value is 0.

Return the largest window value in the full log, taken modulo 1000000007. Compare windows by their exact sums and reduce only the number you return.

Examples

Example 1

Input: pattern = [4, -3, 6], repeats = 3
Output: 21

The full log is 4, -3, 6, 4, -3, 6, 4, -3, 6. The window that covers all nine entries has value 21, and 21 modulo 1000000007 is 21.

Example 2

Input: pattern = [-7, -2], repeats = 4
Output: 0

Every entry of the log is negative. The empty window is allowed and its value is 0, so 0 is reported.

Example 3

Input: pattern = [5, -11, 5], repeats = 3
Output: 10

The full log is 5, -11, 5, 5, -11, 5, 5, -11, 5. The window covering the third and fourth entries has value 10.

Constraints

1 <= pattern.length <= 10^5
1 <= repeats <= 10^5
-10^4 <= pattern[i] <= 10^4

The signature

The editor is preloaded with this. It matches the parent problem's shape, so a solution that works here transfers to a judge unchanged.

Python

def best_repeated_window(pattern: list[int], repeats: int) -> int:

Java

public int bestRepeatedWindow(int[] pattern, int repeats)

The intended approach

Time O(n) for a cycle of n slots, independent of repeats.Space O(1) beyond the input.

The full log is never materialised. Three walks over one cycle give the best window confined to a single cycle, the best window that starts at a cycle's first slot, the best window that ends at a cycle's last slot, and the cycle total; each of those bests is floored at zero because the empty window is legal. A window that spans more than one cycle is a suffix of some cycle, then a whole number of complete cycles, then a prefix of a later cycle, and the complete cycles in the middle are worth taking only when a cycle total is positive, in which case all `repeats - 2` of them are taken. The answer is the larger of the single-cycle best and that spanning value, reduced at the very end.

Pythonentry point: best_repeated_window

MOD = 1000000007


def best_repeated_window(pattern, repeats):
    # One walk over a single cycle collects the four facts the answer needs:
    # the best block that stays inside one cycle, the best block that ends at a
    # cycle boundary, the best block that starts at one, and the cycle total.
    # Empty blocks are allowed, so the running bests never fall below zero.
    best_inside = 0
    run = 0
    for v in pattern:
        run = v if run + v < v else run + v
        if run > best_inside:
            best_inside = run

    total = 0
    best_prefix = 0
    for v in pattern:
        total += v
        if total > best_prefix:
            best_prefix = total

    best_suffix = 0
    acc = 0
    for i in range(len(pattern) - 1, -1, -1):
        acc += pattern[i]
        if acc > best_suffix:
            best_suffix = acc

    best = best_inside
    if repeats >= 2:
        joined = best_suffix + best_prefix
        if total > 0:
            joined += (repeats - 2) * total
        if joined > best:
            best = joined
    # the maximum is taken over exact sums; only the reported value is reduced
    return best % MOD

The obvious version

Materialise the whole log by repeating the cycle, then add up every window by walking each start forward over every end, keeping the largest sum seen and zero for the empty window. Quadratic in the log length, so it is only usable on tiny logs, but it assumes nothing about cycles.

This is not filler. It was written independently of the fast solution and the two were run against thousands of random inputs to confirm they agree — which is how the answer above is known to be right. Read it when the intended approach feels like a rabbit pulled out of a hat.

Python

def bf_best_repeated_window(pattern, repeats):
    """Write the whole log out and add up every block, empty block included."""
    full = list(pattern) * repeats
    best = 0
    for i in range(len(full)):
        acc = 0
        for j in range(i, len(full)):
            acc += full[j]
            if acc > best:
                best = acc
    return best % 1000000007

How this problem was checked

Reference solution executed against 15 test cases — 15 passed
3 worked examples confirmed by running the code
0 eight-word sequences shared with the parent problem's text
Agrees with an independent brute-force implementation on random inputs
Test set proven to catch deliberately broken solutions
Reviewed by a separate quality pass
Function arity matches the parent problem's

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