Jest Test Case Minimization

You've encountered a flaky test suite where certain complex test cases intermittently fail. Instead of debugging the entire large test case, you want to find the smallest subset of the input that still causes the test to fail. This process, known as test case minimization, is crucial for efficient debugging and improving test suite stability.

Problem Description

Your task is to implement a function that takes an array of test cases (represented as data structures) and a test execution function. The test execution function will return true if the test passes for a given test case, and false if it fails. Your goal is to find the smallest possible subset of the original input array of test cases that, when passed to the test execution function, still results in at least one failure.

Requirements:

• You will be given an array of individual test cases (e.g., Array<TestCaseData>).
• You will be given a runTest function that accepts a single test case and returns a boolean: true for pass, false for fail.
• Your function should return a new array containing the minimized subset of test cases that, when all passed to runTest (collectively), still produce at least one false (failure).
• The order of test cases in the minimized output should be preserved relative to their original order.
• If the original input array has no failing test cases, return an empty array.

Expected Behavior:

The minimized subset should be the smallest possible contiguous sub-array (or a non-contiguous subset if that's how the minimization algorithm is implemented, but prioritizing contiguous for simplicity in this challenge) that, when its elements are processed by runTest individually or in sequence, results in at least one failure. If multiple minimal subsets exist, any one of them is acceptable.

Edge Cases:

• An empty input array of test cases.
• An input array where all test cases pass.
• An input array where only one test case fails.
• An input array where multiple, non-contiguous test cases cause failures.

Examples

Example 1:

Input Test Cases: [A, B, C, D, E]
runTest(testCase):
  - Returns false for C
  - Returns true for all others

Output: [C]
Explanation: The smallest subset that causes a failure is just [C].

Example 2:

Input Test Cases: [A, B, C, D, E, F, G]
runTest(testCase):
  - Returns false for B and F
  - Returns true for all others

Output: [B, F]  (or potentially [A, B, C, D, E, F] if a simpler binary search like approach is used that doesn't isolate non-contiguous failures as well)
Explanation: If we consider a binary-search like approach, we might check [A, B, C], then [D, E, F, G]. If [A, B, C] fails, we check [A, B], then [C]. If [A, B] fails, we check [A], then [B]. If [B] fails, then [B] is part of the minimal set. If the initial [A, B, C] passed, we'd then focus on [D, E, F, G]. This example assumes a more sophisticated minimization that can find multiple distinct failing elements. For this challenge, a simpler approach that might return a larger but still valid failing subset is acceptable. The key is *minimality* relative to the approach. Let's refine this to assume a "divide and conquer" style minimization where we try to isolate failing groups.

Let's rephrase Example 2 for clarity with a "divide and conquer" approach in mind:

Input Test Cases: [T1, T2, T3, T4, T5, T6, T7]
runTest(testCase):
  - Returns false for T3
  - Returns false for T6
  - Returns true for all others

Initial check: runTest for all [T1...T7] returns false (meaning at least one fails).
Divide: Check [T1, T2, T3] and [T4, T5, T6, T7].
  - runTest for [T1, T2, T3] returns false.
  - runTest for [T4, T5, T6, T7] returns false.

Now focus on [T1, T2, T3]: Divide into [T1] and [T2, T3].
  - runTest for [T1] returns true.
  - runTest for [T2, T3] returns false.

Now focus on [T2, T3]: Divide into [T2] and [T3].
  - runTest for [T2] returns true.
  - runTest for [T3] returns false.
  So, T3 is a failing case.

Now focus on [T4, T5, T6, T7]: Divide into [T4, T5] and [T6, T7].
  - runTest for [T4, T5] returns true.
  - runTest for [T6, T7] returns false.

Now focus on [T6, T7]: Divide into [T6] and [T7].
  - runTest for [T6] returns false.
  - runTest for [T7] returns true.
  So, T6 is a failing case.

Output: [T3, T6]
Explanation: By repeatedly dividing the test cases and checking for failures, we isolate T3 and T6 as the individual failing test cases.

Example 3 (Edge Case: All Pass):

Input Test Cases: [X, Y, Z]
runTest(testCase):
  - Returns true for X, Y, and Z

Output: []
Explanation: Since no test cases fail, the minimized subset of failing test cases is empty.

Constraints

• The input testCases array can contain up to 1000 elements.
• Each testCase can be any JSON-serializable data structure.
• The runTest function's execution time for a single test case is assumed to be constant and reasonably fast for the purpose of this algorithm. However, the overall number of runTest calls should be minimized for efficiency.
• The total number of runTest calls made by your minimization function should ideally be logarithmic in the number of test cases (e.g., O(log N) or O(N) in the worst-case for a naive linear scan, but a more efficient divide-and-conquer approach is encouraged).

Notes

• Consider a divide-and-conquer strategy. You can test a subset of the input test cases. If that subset contains a failure, you recursively apply minimization to that subset. If the subset passes, you can discard it and focus on the remaining test cases.
• The runTest function simulates the actual Jest test execution for a given input. Your minimization function needs to strategically call this runTest to pinpoint the failing inputs.
• Think about how to combine the results from recursive calls when using a divide-and-conquer approach.
• The definition of "smallest" is critical. Aim for the absolute minimum number of test cases in the output array that collectively demonstrate a failure.