Server Selection

You are given an array server, where server[i] is the request-handling capacity of one server. Every capacity is a power of 2. You are also given a target load expected_load.
Return the minimum number of servers whose capacities sum to exactly expected_load. Return -1 if no such selection exists.
For this exercise, assume the following selection rules:
Each array element represents a separate server that may be selected at most once. Servers with equal capacities remain separate choices.
A selected server contributes its entire capacity; capacities cannot be split or partially used. A total greater than expected_load is not valid.
If expected_load is 0, the empty selection is valid and the answer is 0.
The callable minimumServers(server, expected_load) returns only the minimum count, not the selected indices. The target uses a long integer in Java and a long long integer in C++.

Function
minimumServers(server: int[], expected_load: long) → int

Examples
Example 1
server = [1,2,8,4,2]
expected_load = 13
return = 3
Select capacities 8, 4, and 1, totaling 13. The two largest servers total only 12, so fewer than three cannot work. This is an authored practice example.
Example 2
server = [2,2,8]
expected_load = 7
return = -1
Every possible total is even, so no selection totals 7. The server with capacity 8 cannot be used partially. This is an authored practice example.
Example 3
server = [16,8,4]
expected_load = 0
return = 0
Select no servers to meet the zero load. This is an authored practice example.

Constraints
1 <= server.length <= 100000.
For this exercise, assume server[i] = 2^b for an integer b with 0 <= b <= 30.
For this exercise, assume 0 <= expected_load <= 10^15.

The trick: count how many servers exist at each power, cnt[b] for b from 0 to 30. Then walk the bits of expected_load from the lowest. If bit b is set, you need one server of 2^b. If cnt[b] is 0, you can't use a smaller leftover to cover it, so you must break a bigger one, which doesn't work because you can't split. Better approach: go from the highest power down. Keep a remaining target, and at each b take min(cnt[b], remaining / 2^b) servers. Since capacities are powers of 2, taking the biggest fit is optimal. If remaining ends at 0, return the count, else -1. Pitfalls: use long for the target, since it goes up to 10^15, and the loads above 2^30 times 100000 still fit in long. Handle expected_load 0 first and return 0. StealthCoder is your hedge if the greedy proof slips your mind live.

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