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Explain that the key is to maintain a running sum of the window and update it incrementally as the cursor moves: subtract the element leaving the window and add the new element entering, handling wraparound with modulo arithmetic. Then compute the fraction as sum/w in O(1) per step.
Pro tip: Mention that this sliding window technique is a classic pattern for streaming or circular data, and emphasize that it avoids recomputing the sum from scratch, which would be O(w) per step. Also, note that using modulo for wraparound keeps the code clean and efficient.
Compute the sum of the first window of length w (indices 0 to w-1) in O(w) time. This sets the initial sum for r(0).
Calculate r(0) as the initial sum divided by w. Report it.
When moving from index i to i+1, subtract the element at index i (leaving the window) and add the element at index (i+w) mod n (entering the window). This maintains the sum in O(1).
Divide the updated sum by w to get r(i+1) and report it. Repeat for each step.
Use modulo arithmetic for indices to ensure the window wraps around the circular array correctly.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
Use a sparse table (binary lifting) to precompute jumps of power-of-two sizes, then decompose the arbitrary jump k into its binary representation to answer each query in O(log k) time. To achieve O(1) per query, combine binary lifting with a precomputed table for small jumps (e.g., up to sqrt(n)) and handle circularity by duplicating the array or using modulo arithmetic. Prove that preprocessing takes O(n log n) time and space, and each query is O(1) by using a two-level decomposition: large jumps via binary lifting and small jumps via direct lookup.
Pro tip: Mention that for circular arrays, you can avoid modulo in the inner loop by duplicating the array to length 2n, which simplifies indexing and reduces constant factors. Also, note that if k is large, you can reduce it modulo n because jumping n times returns to the same position.
Restate that r(i) is the result after k jumps from index i in a circular array, and that k is arbitrary (k >= 0). Confirm that preprocessing is allowed and queries must be O(1).
Propose a sparse table (binary lifting) where up[i][j] stores the index after 2^j jumps from i. For O(1) queries, add a second level: precompute direct jumps for all k up to B = sqrt(n) (or a constant) to handle small k quickly.
Duplicate the array to length 2n to avoid modulo operations, or use modulo n when computing indices. Note that jumping n times returns to the same index, so k can be reduced modulo n.
For a query (i, k), if k <= B, return small[i][k] directly. Otherwise, decompose k into a large part and a small part: use binary lifting for the large part (e.g., k - (k mod B)) and then use small table for the remainder. This yields O(1) time.
Preprocessing: O(n log n) time and space for sparse table, O(n sqrt(n)) for small table (which is O(n sqrt(n)) space, but can be optimized to O(n) by storing only one row? Actually, small table is O(n * B) = O(n sqrt(n)) space, which may be too large. Instead, use a different approach: precompute for each index the result after B jumps, then use binary lifting for the rest. But to get O(1), we can use the method of 'jump pointers' with block decomposition. Alternatively, use the fact that k can be reduced modulo n, and if n is small, O(n) per query is not O(1). The standard solution for O(1) arbitrary jump on a functional graph is to use binary lifting with O(log n) per query, but to get O(1) we need a more advanced technique like the 'level ancestor' problem on trees, which can be solved in O(1) with O(n) preprocessing using ladder decomposition or Euler tour + RMQ. However, for a simple circular array, we can precompute the answer for all possible k? That would be O(n^2) space. So O(1) per query with O(n log n) preprocessing is not trivial. Actually, the problem likely expects O(log n) per query, but the question says O(1). So we need to think: if k is arbitrary, we can precompute a table of size n x n? No. But we can use the fact that the jump is uniform: r(i) = (i + k) mod n if the jump is just adding k? Wait, the problem says 'jump size k', but it's not clear if it's a fixed step or a jump to the k-th successor. Typically, in such problems, r(i) is the index after k steps in a linked list or array where each step moves to the next element. If the array is circular and each step moves to the next index, then r(i) = (i + k) mod n, which is O(1) trivially. But the problem likely implies a more complex structure, like a functional graph where each node has a pointer to another node, and we need to find the k-th successor. That is the level ancestor problem. For a functional graph (each node out-degree 1), the k-th successor can be found in O(1) after O(n) preprocessing using the 'binary lifting' with O(log n) per query, but O(1) is possible with the 'jump pointer' technique combined with 'ladder decomposition' for trees. However, for a general functional graph, it's more complex. Given the context, the candidate should propose a solution that achieves O(1) per query by using a combination of binary lifting and a small table, but acknowledge that O(1) might require O(n log n) preprocessing and O(n log n) space, and that the small table can be of size O(n) if we choose B = log n? Actually, if we choose B = log n, then small table is O(n log n) space, and for large k we use binary lifting which takes O(log n) steps, so not O(1). To get O(1), we need to precompute for all k up to n, which is O(n^2) space. So maybe the intended solution is to use the fact that k can be reduced modulo n, and if n is small, we can precompute all answers in O(n^2) time and space, but that's not efficient. Alternatively, the problem might be about a specific data structure like a 'jump table' where each node has a pointer to the next, and we want to answer queries of the form 'what is the k-th successor?' in O(1) after O(n log n) preprocessing. This is possible using the 'level ancestor' technique with Euler tour and RMQ, but that gives O(1) query after O(n) preprocessing for trees. For a functional graph, it's a set of trees with cycles. We can handle cycles by contracting them and then using the tree technique. So the candidate should describe that: decompose the functional graph into cycles and trees attached to cycles, then for each tree, use level ancestor with O(1) query, and for cycles, use modulo arithmetic. This yields O(1) per query after O(n) preprocessing. That is a strong answer.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.