The Deep Dive

An easy Python interview practice problem on DataDriven. Write the Python and run it against test cases, with instant feedback.

Domain
Python
Difficulty
easy
Seniority
L4
Asked in
Meta

Problem

You're triaging a metrics export that maps each host to its request count, and several hosts can land on the exact same count. Given counts and a 1-based n, return the n-th highest distinct count, collapsing repeated counts to a single rank; return None when there aren't n distinct counts.

A specific position in the unsorted pile.

Example 1

Input:

counts = {"web1":10,"web2":30,"web3":20,"web4":30}, n = 1

Output:

30

Example 2

Input:

counts = {"web1":10,"web2":30,"web3":20,"web4":30}, n = 2

Output:

20

Worked solution and explanation

What this really is

Strip the host-metrics costume and this is a rank-over-distinct-values problem. The maximum is easy, and most people get a clean answer on n=1. What actually separates candidates is recognizing that two hosts sharing a count must collapse to one rank. The trap is reaching for sorted(counts.values(), reverse=True)[n-1]: the moment any value repeats, that repeat eats a rank, and your n-th element is a value you already counted, sitting one or more positions too high. The fix is one word, set, applied before you sort.

Walking it through

Step 1: Collapse to distinct counts

Pull the counts out of the dict and drop them into a set. Keys are noise here, and two hosts on the same count must become one entry, otherwise the rank math is wrong before you even sort.

Step 2: Sort distinct counts high to low

The n-th highest is position n in a descending list, so sort the distinct counts high to low. Sorting a deduplicated set, not the raw values, is the whole game.

Step 3: Index by rank, guard the edge

1-based n maps to index n-1, so the 1st highest sits at index 0. Before indexing, guard the boundary: if n is past the number of distinct counts, there is no n-th value, so return None rather than letting an IndexError escape.

The solution

Distinct counts, then ranked selection
def the_deep_dive(counts: dict[str, int], n: int):
    distinct = sorted(set(counts.values()), reverse=True)
    if n > len(distinct):
        return None
    return distinct[n - 1]

set() folds the ties, sorted(..., reverse=True) ranks them, the guard handles n out of range.

Common follow-up questions