The four elimination probes

Exam questions present plausible near-misses. The skill is elimination by contract, not by vibes. The four elimination probes:

1. The empty probe. What happens with length 0 / "" / empty list? A solution that reads element 0 unguarded dies here.
2. The single probe. One element: does a seeded-max or adjacency scan handle it? Adjacency loops from 1 or to length − 1 do nothing — correct or not, depending on the contract.
3. The all-equal probe. Every value identical: tie-breaking (strict vs non-strict) decides.
4. The order probe. Run the trace on a sorted and a reverse input: order-dependent bugs (first vs last, min vs max) expose themselves in one of the two.

Apply to this candidate solution for "index of the smallest value, earliest on ties":

int best = 0;
for (int i = 1; i < arr.length; i++) {
    if (arr[i] <= arr[best]) { best = i; }   // probe 3: fails
}

The all-equal probe {7, 7}: i=1, 7 <= 7 true, best becomes 1 — the LAST tie. Contract says earliest. The <= must be <. One probe, one character, one eliminated answer.