The core idea isn't hard once you see it: precompute prefix sums of the distances, then for any pair the clockwise distance is just a prefix difference, and counterclockwise is total minus that.
First, clarify the problem constraints and edge cases, then propose an efficient solution using prefix sums to answer each query in O(1) time. Explain how to compute the clockwise and counterclockwise distances and take the minimum, summing across all queries.
Pro tip: Mention that the total sum of all distances is constant, so the counterclockwise distance is totalSum minus clockwise distance, avoiding a second traversal. Also, discuss handling large N and Q with 64-bit integers to prevent overflow.
Ask about constraints (N, Q, distance ranges), whether nodes are 0-indexed or 1-indexed, and if the array is given as distances between consecutive nodes. Confirm that queries are independent and we need the sum of shortest paths.
Build a prefix sum array where prefix[i] is the sum of distances from node 0 to node i clockwise. Compute totalSum as the sum of all distances. This allows O(1) clockwise distance between any two nodes.
For nodes u and v, compute clockwise distance as (prefix[v] - prefix[u] + totalSum) % totalSum. The counterclockwise distance is totalSum minus clockwise distance. The shortest path is the minimum of the two.
Accumulate the shortest distances for all queries into a result variable. Use a 64-bit integer to avoid overflow. Return the total sum.
Time: O(N + Q) for preprocessing and answering queries. Space: O(N) for prefix sums. This is optimal for the given constraints.
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