3161. Block Placement Queries - Detailed Explanation

Problem Statement

Description:
You are given a sequence of queries, where each query is an integer that represents a desired block placement index on an infinite one-dimensional line (or array). Initially, all positions are free.

For each query, you need to “place” a block at the desired position or—if that position is already occupied—at the next available (free) position to the right. Once a block is placed at a position, that position becomes occupied, and subsequent queries that request that same position (or any position whose available spot was already taken) must move to the next free slot.

Your Task:
Return the result for each query as the final index where the block was placed.

Example Inputs, Outputs, and Explanations

1. Example 1:

   • Input: queries = [1, 2, 1, 2]
   • Output: [1, 2, 3, 4]
   • Explanation:
     • Query 1: Request index 1 → free, so place a block at 1.
     • Query 2: Request index 2 → free, so place a block at 2.
     • Query 3: Request index 1 → already occupied; next free index is 3 → place a block at 3.
     • Query 4: Request index 2 → already occupied; next free index is 4 → place a block at 4.

2. Example 2:

   • Input: queries = [5, 5, 5]
   • Output: [5, 6, 7]
   • Explanation:
     • Query 1: Request index 5 → free, so place at 5.
     • Query 2: Request index 5 → index 5 is occupied; next free is 6.
     • Query 3: Request index 5 → indices 5 and 6 are occupied; next free is 7.

3. Example 3 (Edge Case):

   • Input: queries = [10]
   • Output: [10]
   • Explanation:
     • A single query that requests index 10, which is free, so the block is placed at 10.

Constraints

• The number of queries can be large.
• The queries are non-negative integers.
• The positions extend to infinity (or a very large number), so the solution should not preallocate an array for all positions.
• A block, once placed at a position, makes that position unavailable for future queries.

Hints to Approach the Solution

1. Next Available Position Problem:

   • Think of each query as “find the next free slot at or after index x.”
   • A naïve approach might iterate from x upward until a free position is found, but that would be too slow for many queries.

2. Union-Find / Disjoint Set (DSU):

   • Notice that the problem can be modeled using union–find (or DSU) by “linking” an occupied position to the next candidate free position.
   • When you place a block at position x, update the structure so that the next query for position x (or any query landing on it) will immediately jump to the next free index (i.e. x + 1).

3. Path Compression:

   • Using union–find with path compression ensures that subsequent queries run in nearly constant time (amortized).

Approach 1: Naïve Linear Scan

Explanation

• Method: For each query, starting from the requested index, iterate to the right until you find a free position.
• Drawback: In the worst case, if many consecutive positions are occupied, the scan can be very slow (O(n) per query), leading to an overall inefficient solution.

Approach 2: Union-Find (Optimal)

Explanation

• Idea: Use a union–find data structure (also known as DSU) to track the next available free index.
• How It Works:
  • Initialization:
    • No positions are occupied initially.
  • Find Function:
    • Implement a find(x) function that returns the smallest available position at or after x. If x is free, then find(x) = x; if not, recursively find the next free position.
  • Placement (Union):
    • When a block is placed at position x, mark it as occupied by “unioning” it with x + 1. This means that for future queries, the next available position will be determined by find(x + 1).
  • Processing a Query:
    • For a query with desired index x, compute pos = find(x). Then place the block at pos and update the union–find structure by setting parent[pos] = pos + 1. Return pos as the answer for that query.

Pseudocode

initialize parent as an empty hash map

function find(x):
    if x is not in parent:
         return x
    parent[x] = find(parent[x])
    return parent[x]

function placeBlock(x):
    pos = find(x)
    parent[pos] = pos + 1
    return pos

for each query in queries:
    answer[i] = placeBlock(query)
return answer

Code Implementations

Python Code