Minimum Non-Adjacent Selection Capability

You are given an array of positive integers values. Choose at least count positions so that no two chosen positions are adjacent.
The capability of a selection is the maximum value among its chosen positions. Return the minimum capability achievable by any valid selection.

Function
minimumSelectionCapability(values: int[], count: int) → int

Examples
Example 1
values = [2,3,5,9]
count = 2
return = 5
Choosing values 2 and 5 from positions 0 and 2 gives capability 5. No valid pair has a smaller maximum.
Example 2
values = [2,7,9,3,1]
count = 2
return = 2
Positions 0 and 4 are non-adjacent and hold values 2 and 1, so the minimum capability is 2.
Example 3
values = [10,1,8,2,7,3]
count = 3
return = 3
Choosing positions 1, 3, and 5 gives values 1, 2, and 3.

Constraints
1 <= values.length <= 10^5.
1 <= values[i] <= 10^9.
1 <= count <= (values.length + 1) / 2.

Binary search on the capability. Pick a candidate cap X and ask: can I choose at least count non-adjacent positions, all with value <= X? Check it greedily. Scan left to right, and whenever values[i] <= X, take it and skip i+1. Greedy works because taking the earliest eligible position never hurts later choices. If the taken count reaches count, X is feasible. Feasibility is monotonic, since a bigger X only adds options. Search between the min and max of the array, or just 1 to 10^9. Total cost is O(n log(maxValue)). The common pitfall is reaching for DP with a max-of-selection state, which gets messy and slow. Another is forgetting to skip the neighbor after a pick, or searching over indexes instead of values. If the live OA rattles you, StealthCoder is the safety net that surfaces this binary search plus greedy check so you can still submit clean code.

Drill it cold or hedge it with StealthCoder. Either way, don't walk into the OA hoping you remember the trick.