Aggressive Cows

Given distinct integer stall positions stalls and an integer cows, place exactly cows cows in different stalls.
Return the largest possible value of the minimum distance between every pair of placed cows.

Function
aggressiveCows(stalls: int[], cows: int) → int

Examples
Example 1
stalls = [1,2,4,8,9]
cows = 3
return = 3
Placing cows at 1, 4, and 8 gives a minimum distance of 3, which cannot be improved.
Example 2
stalls = [10,1,2,7,5]
cows = 3
return = 4
After sorting, positions 1, 5, and 10 achieve minimum distance 4.
Example 3
stalls = [0,5]
cows = 2
return = 5
Both stalls must be used, so their distance is the answer.

Constraints
2 <= stalls.length <= 10^5.
2 <= cows <= stalls.length.
0 <= stalls[i] <= 10^9.
All stall positions are distinct.

The trick: don't search placements, search the distance. Sort the stalls. Binary search d between 1 and max minus min. For each d, run a greedy check: put the first cow in the first stall, then drop the next cow at the first stall at least d away from the last one. If you place at least cows cows, d is feasible, so try bigger. Otherwise go smaller. Feasibility is monotonic, which is why binary search works. Cost is O(n log n) for the sort plus O(n log 10^9) for the search, fine for n up to 10^5. Common pitfalls: forgetting to sort, using a check that counts cows wrong, off-by-one in the loop bounds, and returning the wrong boundary. Track the last feasible d. If your head goes blank on the live Amazon OA, StealthCoder is the hedge that hands you the structure fast.

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