How should I practice Precision of Stacked PII Detector Flags?

Start by reading the full statement, solve the visible examples, then compare the pattern against related Mercor phone screen problems in FastPrep.

Is Precision of Stacked PII Detector Flags associated with Mercor?

FastPrep tags this as a Mercor phone screen coding practice problem when the problem catalog has that company and stage context.

Precision of Stacked PII Detector Flags

Mercor● EasyFULLTIMEPHONE SCREEN

Problem statement

A pipeline anonymizes customer data by replacing real names with fake ones. First, a PII detector reads a document token by token and flags every token that looks like personal information (PII).

A document has piiTokens real PII tokens and otherTokens tokens that are not PII. Every detector has these properties:

It flags every real PII token.
It flags each non-PII token with probability falseFlagPercent / 100.

The pipeline runs detectors equally good detectors and replaces a token only when all of them flag it. The detectors make their mistakes independently.

Pick a token uniformly at random from the document. Given that every detector flagged it, what is the probability that it is real PII? Return the exact answer as a reduced fraction string "numerator/denominator" with no spaces. Write a probability of 1 as "1/1".

The reported question used 100 PII tokens, 100000 other tokens, and a 1% false-flag rate, first with one detector and then with two.

Function

flaggedPiiProbability(piiTokens: int, otherTokens: int, falseFlagPercent: int, detectors: int) → String

Examples

Example 1

piiTokens = 100otherTokens = 100000falseFlagPercent = 1detectors = 1return = "1/11"

The detector flags all 100 PII tokens. Each of the 100000 other tokens is flagged with probability 1/100, which adds 1000 expected false flags.

So the probability is 100 / (100 + 1000) = 1/11, about 9%.

Example 2

piiTokens = 100otherTokens = 100000falseFlagPercent = 1detectors = 2return = "10/11"

Both detectors still flag all 100 PII tokens. A non-PII token is flagged by both with probability (1/100)^2 = 1/10000, so there are 10 expected false flags.

The probability is 100 / (100 + 10) = 10/11.

Example 3

piiTokens = 3otherTokens = 7falseFlagPercent = 50detectors = 2return = "12/19"

A non-PII token is flagged by both detectors with probability (1/2)^2 = 1/4, so the expected false flags are 7/4.

The probability is 3 / (3 + 7/4) = 12/19.

Constraints

1 <= piiTokens <= 10^6
0 <= otherTokens <= 10^6
0 <= falseFlagPercent <= 100
1 <= detectors <= 3