Forward Block Placement Queries
Reported by candidates from Capital One's online assessment. Pattern, common pitfall, and the honest play if you blank under the timer.
The edge case that kills the naive solution here is scale. Capital One reported this OA in October 2026, and the positions go up to 10^9 with lengths up to 10^9, so you can't mark cells in an array or scan the interval. You get obstacle placements and range-empty queries on a huge line, and 10^5 operations to process. The pattern is an ordered set with a successor lookup. If you blank on the data structure under the clock, StealthCoder is the safety net that runs invisibly during the live OA and hands you the approach.
The problem
Process operations on a nonnegative integer line. Initially, no position contains an obstacle. [1, x] places an obstacle at position x. Placing an obstacle at an occupied position has no additional effect. [2, start, length] asks whether every integer position in the inclusive interval [start, start + length - 1] is empty. Append 1 when the block fits and 0 otherwise. A query does not place a block. Return the appended bits as one string in query order. Function forwardBlockPlacementQueries(operations: int[][]) → String Examples Example 1 operations = [[1,2],[1,5],[2,3,2],[2,3,3],[2,1,1],[2,1,2]] return = "1010" Positions 3 through 4 are empty, but 3 through 5 contains an obstacle. Position 1 is empty, while 1 through 2 contains an obstacle. Example 2 operations = [[2,0,1],[1,0],[2,0,1],[1,0],[2,1,3]] return = "101" The first query succeeds. After position 0 is occupied, the second fails; repeating the build is idempotent, and positions 1 through 3 remain empty. Constraints 1 <= operations.length <= 10^5. Each operation is either [1, x] or [2, start, length]. 0 <= x, start <= 10^9. 1 <= length <= 10^9. Interval endpoints fit in signed 64-bit integers.
Reported by candidates. Source: FastPrep
Pattern and pitfall
The trick: a query [start, start+length-1] is empty exactly when no obstacle lies in that range. So find the smallest obstacle at or after start. If it exists and is at most start+length-1, answer 0. Otherwise answer 1. That's a successor query on a sorted set, which is O(log n) per operation. The pitfall is iterating the interval, which is up to 10^9 steps. Another pitfall is using a plain hash set, which can't give you the next obstacle. Duplicate placements are harmless in a set, so don't count them. Also watch the inclusive end: it's start+length-1, not start+length. In Python there's no built-in tree set, so either use sorted list with bisect (insertion cost is acceptable here) or process offline with coordinate compression and a segment tree. If the successor idea won't come, StealthCoder is the hedge on the live OA.
Drill it cold or hedge it with StealthCoder. Either way, don't walk into the OA hoping you remember the trick.
You can drill Forward Block Placement Queries cold, or you can hedge it. StealthCoder runs invisibly during screen share and surfaces a working solution in under 2 seconds. The proctor sees the IDE. They don't see what's behind it. Made for the candidate who got the OA invite this morning and has 72 hours, not six months.
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Forward Block Placement Queries FAQ
What's the trick in Forward Block Placement Queries?+
Don't check every cell. Keep obstacles in a sorted structure and find the first obstacle at or after start. If it's within start+length-1, the block doesn't fit. Otherwise it does. One binary search per query.
Why can't I use a boolean array?+
Positions go up to 10^9 and lengths up to 10^9. An array that size won't fit in memory, and scanning an interval per query would blow past 10^5 operations. You need a sparse structure with logarithmic lookup.
Do duplicate obstacle placements matter?+
No. Placing an obstacle on an occupied position has no effect, as Example 2 shows. A set handles this naturally. With a sorted list, check whether the value is already there before inserting.
How do I get the output string right?+
Only type 2 operations append a bit. Type 1 appends nothing. Collect 1 or 0 for each query in order and join them at the end. Example 1 has four queries and returns 1010.
How do I prepare for this in 48 hours?+
Practice successor and predecessor lookups with bisect or a TreeSet, plus inclusive-interval arithmetic. Then write this one from scratch once. The Capital One October 2026 report suggests the difficulty is the scale, not the logic.