Question
Implement GROUP BY with an aggregate. Given rows (list of dicts), a group_col name, an agg_col name, and an agg function one of 'sum','count','avg', group rows by row[group_col] and compute the aggregate of row[agg_col] within each group. 'count' ignores agg_col and counts rows; 'avg' is the floating-point mean. Return a list of [group_value, aggregate] pairs sorted ascending by group_value. avg values must equal the exact float mean.
group_aggregate(rows: list[dict], group_col: str, agg_col: str, agg: str) → list[list][[{"sal":100,"dept":"eng"},{"sal":200,"dept":"eng"},{"sal":50,"dept":"hr"}],"dept","sal","sum"]out[["eng",300],["hr",50]]State your approach and its time/space complexity out loud before you optimize. Handle the edge cases (empty input, duplicates, overflow), and say why you chose this over the brute force. Green tests are the floor, not the grade.
Vibe coding: describe the solution in plain language (or narrate it) and the coach grades your approach. Generating runnable code from your description is coming next.