How should I practice Nickname-Constrained Grouping?

Start by reading the full statement, solve the visible examples, then compare the pattern against related Zoox onsite interview problems in FastPrep.

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Problem · Array

Nickname-Constrained Grouping

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ZooxFULLTIMEONSITE INTERVIEW

Problem statement

You are given parallel arrays names and nicknames. Person i has name names[i], nickname nicknames[i], and original index i. Partition every person into groups of exactly k people so that no group contains the same nickname twice.

The input is guaranteed to admit the following canonical construction:

Number the g = names.length / k groups from 0 through g - 1, and set a group cursor to 0.
Process distinct nicknames in lexicographically increasing order.
For one nickname, order its people by name and then by original index.
If this nickname has f people, assign person j in that order to group (cursor + j) % g, then advance the cursor to (cursor + f) % g.
Sort the original indices inside every group increasingly.

Return the groups in group-number order as an array of index arrays.

Function

groupPeople(names: String[], nicknames: String[], k: int) → int[][]

Examples

Example 1

names = ["ana","ben","cara","drew","eli","faye"]nicknames = ["red","blue","red","green","blue","green"]k = 2return = [[1,5],[0,4],[2,3]]

Nicknames are processed as blue, green, then red. Cyclic assignment gives group 0 indices [1,5], group 1 indices [0,4], and group 2 indices [2,3]. Every group has size 2 and distinct nicknames.

Example 2

names = ["zoe","amy","kai","bo"]nicknames = ["x","x","y","z"]k = 2return = [[1,2],[0,3]]

Within nickname x, index 1 comes before index 0 because amy is lexicographically smaller than zoe. The two occurrences go to different groups before nicknames y and z complete them.

Example 3

names = ["c","a","b"]nicknames = ["z","x","y"]k = 1return = [[1],[2],[0]]

Each group has one person. Processing nicknames x, y, and z assigns original indices 1, 2, and 0 to groups 0, 1, and 2.

Constraints

1 <= names.length = nicknames.length <= 2000
1 <= k <= names.length
names.length is divisible by k.
Every name and nickname contains between 1 and 20 lowercase English letters.
Each nickname occurs at most min(k, names.length / k) times.