How should I practice Top-Funded Villages and Projects?

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Top-Funded Villages and Projects

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Problem statement

You are given three finite relations encoded as string matrices:

Each villages row is [villageId, villageName].
Each projects row is [projectId, villageId, projectName].
Each funds row is [projectId, date, amount], where date uses YYYY-MM-DD format and amount is a nonnegative decimal integer.

Only funding rows whose date is in calendar year 2024 contribute. A project's funding is the sum of its contributing rows, and a project with no contributing row has funding 0. A village's total funding is the sum of the funding for all of its listed projects.

Consider every village that has at least one listed project. Keep every village tied for the maximum total funding. Within each selected village, keep every project tied for that village's maximum project funding.

Return one row [villageName, projectName, projectFunding] for every retained project. Encode projectFunding as a base-10 string. Sort the rows by village name, then project name, then village ID, then project ID, all in ascending lexicographic order. Return an empty matrix when there are no villages or no projects.

Function

topFundedVillageProjects(villages: String[][], projects: String[][], funds: String[][]) → String[][]

Examples

Example 1

villages = [["v1","Maple"],["v2","River"],["v3","Hill"]]projects = [["p1","v1","Clinic"],["p2","v1","Library"],["p3","v2","Bridge"],["p4","v2","School"],["p5","v3","Well"]]funds = [["p1","2024-01-10","50"],["p2","2024-05-20","50"],["p3","2024-02-01","60"],["p3","2024-12-15","40"],["p4","2023-08-01","200"],["p5","2024-04-12","80"]]return = [["Maple","Clinic","50"],["Maple","Library","50"],["River","Bridge","100"]]

Maple and River each receive a village total of 100, so both survive the first tie. Maple's two projects tie at 50; River's Bridge project has 100 while School has 0. Hill's total of 80 is not maximal.

Example 2

villages = [["v2","Beta"],["v1","Alpha"]]projects = [["p2","v2","Orchard"],["p1","v1","Garden"]]funds = [["p1","2023-12-31","90"],["p2","2025-01-01","90"]]return = [["Alpha","Garden","0"],["Beta","Orchard","0"]]

Neither funding row belongs to 2024. Both village totals and both project totals are therefore 0, so both tie levels retain every listed project. Village-name ordering places Alpha first.

Example 3

villages = []projects = []funds = []return = []

There is no village-project pair to rank, so the result is empty.

Constraints

0 <= villages.length, projects.length, funds.length <= 100000.
Village IDs are unique, project IDs are unique, and every project and funding row references an existing ID.
IDs and names contain between 1 and 100 printable ASCII characters.
Every date is a valid Gregorian date in YYYY-MM-DD format.
0 <= amount <= 10^9, and every project and village sum fits in a signed 64-bit integer.