My first instinct was to just compute the clockwise sum for each query by looping through the relevant segment, then subtract from total to get the counterclockwise distance.
Precompute prefix sums of the clockwise distances to enable O(1) distance calculations between any two stops. For each query, compute the clockwise distance and subtract from total to get the counterclockwise distance, then take the minimum. Sum these minimums across all queries.
Pro tip: Mention that you can handle large inputs by using 64-bit integers to avoid overflow and by processing queries in O(1) time after O(n) preprocessing. Also, clarify edge cases like when stops are the same or when the route has only one stop.
Clarify that the circular route has n stops, distances are given clockwise, and queries ask for shortest distance between two stops. Note that you can travel either clockwise or counterclockwise.
Compute an array where prefix[i] is the sum of distances from stop 0 to stop i clockwise. Also compute total circumference. This allows O(1) clockwise distance between any two stops.
For a query (u, v), compute clockwise distance as (prefix[v] - prefix[u] + total) % total. The shortest distance is min(clockwise, total - clockwise).
Accumulate the minimum distances for all queries. Ensure to handle cases where u == v (distance 0) and use 64-bit integers for sums to prevent overflow.
State that preprocessing is O(n) and each query is O(1), so total time is O(n + q). Mention that this is optimal for large inputs.
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