Count House Segments After Destruction

You are monitoring the building density in a district of houses. The district is represented as a number line, where each house is located at an integer position. Some of the houses are gradually destroyed over time.
You are given houses, an array of integers representing the initial locations of all houses in the district. You are also given queries, an array of integers representing the locations of houses that will be destroyed, in the order in which they are destroyed.
After each house is destroyed, find the number of house segments remaining in the district. A house segment is a maximal group of one or more houses at consecutive integer positions.
Return an array of integers representing the number of house segments after each respective house from queries is destroyed.

Function
solution(houses: int[], queries: int[]) → int[]

Examples
Example 1
houses = [1, 2, 3, 6, 7, 9]
queries = [6, 3, 7, 2, 9, 1]
return = [3, 3, 2, 2, 1, 0]
The initial house segments are [1, 2, 3], [6, 7], and [9]. Destroying houses in the order [6, 3, 7, 2, 9, 1] leaves [3, 3, 2, 2, 1, 0] segments after the respective operations.

Constraints
All house locations in houses are distinct.
Every house location in queries is present in houses.
All house locations in queries are distinct.

Don't rebuild segments. Put all houses in a hash set and count the initial segments: a house starts a segment if house-1 isn't in the set. Then process each query x. Check whether x-1 and x+1 exist. Both present: removing x splits one segment into two, so count +1. Neither present: x was a lone segment, so count -1. Exactly one present: the segment just shrinks, count unchanged. Remove x from the set and push the count. That's O(n) total. The common pitfall is checking neighbors after deleting x, or forgetting the isolated house case. Another trap is handling the example's order wrong: the answer is after each destruction, not before. Walk Example 1 by hand once. Destroying 6 leaves 7 alone, count stays 3. Destroying 3 shrinks [1,2,3], still 3. If the neighbor cases slip your mind mid-assessment, StealthCoder can hand you the clean version while you keep typing normally.

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