Priority, Expiration, and LRU Eviction

A cache currently contains n items. Item i has key keys[i], priority priorities[i], expiration time expirations[i], and last-used time lastUsed[i].
Select the key of the one item to evict at time now using this precedence:
If at least one item is expired, only expired items are eligible. An item is expired when expirations[i] <= now. Return the smallest eligible key.
Otherwise, keep only items with the smallest priority value.
If several items have that priority, choose the least recently used one: the item with the smallest lastUsed value.
If a tie still remains, return the smallest key.

Function
selectEvictionKey(keys: int[], priorities: int[], expirations: int[], lastUsed: int[], now: int) → int

Examples
Example 1
keys = [10,20,30]
priorities = [5,1,1]
expirations = [100,40,40]
lastUsed = [9,8,7]
now = 50
return = 20
Keys 20 and 30 are expired. Expiration takes precedence over priority and recency, and the smaller eligible key is 20.
Example 2
keys = [1,2,3,4]
priorities = [2,1,1,3]
expirations = [100,100,100,100]
lastUsed = [5,8,3,1]
now = 50
return = 3
No item is expired. Keys 2 and 3 have the minimum priority, and key 3 has the earlier last-used time.
Example 3
keys = [9,4]
priorities = [1,1]
expirations = [100,100]
lastUsed = [7,7]
now = 50
return = 4
The items tie on expiration status, priority, and recency, so the smaller key is returned.

Constraints
1 <= n <= 10^5.
All four arrays have length n.
All keys are distinct.
0 <= keys[i], priorities[i], expirations[i], lastUsed[i], now <= 10^9.
Smaller lastUsed values represent less recent access.

The trick is that the rules are a lexicographic ordering with one twist. Expiration isn't a sort key, it's a filter. So first scan for any item where expirations[i] <= now. If at least one exists, restrict to those. Then pick the minimum by the tuple (priority, lastUsed, key) over the eligible set. For expired items, the spec says return the smallest key, so don't apply priority there. That's the common pitfall: Example 1 has key 30 with lower priority tie but key 20 wins by key only. Keep it O(n) with a single scan, no sorting needed, though sorting works at 10^5. Watch the <= on expiration, since equal to now counts as expired. If you freeze on the live OA, StealthCoder can hand you the filter-then-min structure so you just verify against the three examples.

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