Design Memory Allocator

The problem asks to simulate a memory allocator with a fixed-size array. You need to manage memory blocks, handle allocations, and free up memory based on requests. The solution involves efficient memory scanning with the use of hash tables for quick lookups.

Problem Statement

You are given an integer n representing the size of a 0-indexed memory array. All memory units are initially free. The task is to implement an allocator with three key operations: allocate(size, mID), freeMemory(mID), and freeMemoryAll(mID). The allocator must allocate consecutive free memory blocks of a given size and deallocate memory based on a specific memory identifier mID.

The allocator should handle allocation requests by finding the smallest consecutive memory block of sufficient size, then filling it with the requested memory identifier. When memory is freed, all units with the given mID should be released. If no available memory block meets the requested size, return -1. Additionally, a call to freeMemory(mID) should return the count of freed blocks, while freeMemoryAll(mID) must return 0 if no blocks are found.

Examples

Example 1

Input: See original problem statement.

Output: See original problem statement.

Input ["Allocator", "allocate", "allocate", "allocate", "freeMemory", "allocate", "allocate", "allocate", "freeMemory", "allocate", "freeMemory"] [[10], [1, 1], [1, 2], [1, 3], [2], [3, 4], [1, 1], [1, 1], [1], [10, 2], [7]] Output [null, 0, 1, 2, 1, 3, 1, 6, 3, -1, 0]

Explanation Allocator loc = new Allocator(10); // Initialize a memory array of size 10. All memory units are initially free. loc.allocate(1, 1); // The leftmost block's first index is 0. The memory array becomes [1,, ,, ,, ,, ,]. We return 0. loc.allocate(1, 2); // The leftmost block's first index is 1. The memory array becomes [1,2, ,, ,, ,, ,]. We return 1. loc.allocate(1, 3); // The leftmost block's first index is 2. The memory array becomes [1,2,3, ,, ,, ,,]. We return 2. loc.freeMemory(2); // Free all memory units with mID 2. The memory array becomes [1,, 3, ,, ,, ,,]. We return 1 since there is only 1 unit with mID 2. loc.allocate(3, 4); // The leftmost block's first index is 3. The memory array becomes [1,,3,4,4,4, ,, ,]. We return 3. loc.allocate(1, 1); // The leftmost block's first index is 1. The memory array becomes [1,1,3,4,4,4, ,, ,]. We return 1. loc.allocate(1, 1); // The leftmost block's first index is 6. The memory array becomes [1,1,3,4,4,4,1, ,,]. We return 6. loc.freeMemory(1); // Free all memory units with mID 1. The memory array becomes [, ,3,4,4,4,, ,,]. We return 3 since there are 3 units with mID 1. loc.allocate(10, 2); // We can not find any free block with 10 consecutive free memory units, so we return -1. loc.freeMemory(7); // Free all memory units with mID 7. The memory array remains the same since there is no memory unit with mID 7. We return 0.

Constraints

• 1 <= n, size, mID <= 1000
• At most 1000 calls will be made to allocate and freeMemory.

Solution Approach

Efficient Allocation with Array Scanning

The allocator can iterate through the memory array to find the first consecutive free memory block that meets the requested size. This can be done by scanning the array and looking for free memory units, using array indexing to track the beginning and end of the block.

Hash Table for Memory ID Tracking

Using a hash table to store memory block identifiers allows for quick lookups when freeing memory. Each memory block is tracked by its unique mID, making it efficient to find and deallocate the correct memory when a freeMemory(mID) operation is requested.

Handle Edge Cases and Inefficient Allocations

Edge cases include when there is insufficient memory for an allocation, or when attempting to free a memory block that was never allocated. Additionally, the algorithm needs to handle failed allocations gracefully by returning -1 if no suitable memory block is found.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

Time complexity depends on the allocation and scanning approach. Array scanning has a worst-case time complexity of O(n) for each allocation, while hash table lookups for freeing memory are O(1). The space complexity is O(n) for storing the memory array and hash table entries for mID tracking.

What Interviewers Usually Probe

• Look for understanding of memory management in an array and hash-based tracking.
• Test if the candidate can optimize the allocation search with efficient techniques.
• Gauge ability to handle edge cases such as memory exhaustion or invalid memory deallocation.

Common Pitfalls or Variants

Common pitfalls

• Not handling cases where memory blocks cannot be allocated due to insufficient consecutive free memory.
• Failing to correctly manage and track mID values, leading to incorrect deallocations.
• Inefficient allocation logic causing performance issues with large memory sizes or frequent operations.

Follow-up variants

• Increase memory array size or handle dynamic resizing of the allocator.
• Introduce memory fragmentation, requiring more complex memory allocation strategies.
• Add additional operations such as querying memory usage or checking for the largest free block.

FAQ

How do I handle allocation failures in the Design Memory Allocator problem?

You need to return -1 if there isn't enough free memory for the requested size. Efficient scanning of the memory array helps ensure this condition is checked early in the process.

What is the time complexity of the allocate operation in this problem?

The time complexity of the allocate operation is O(n) in the worst case, as you may need to scan the entire memory array to find a suitable free block.

How do I efficiently free memory in the Design Memory Allocator?

Use a hash table to track the memory blocks by mID. When freeing memory, simply look up the mID in the hash table to quickly access and deallocate the corresponding memory units.

Can I use other data structures apart from arrays and hash tables for the allocator?

While arrays and hash tables are the most straightforward and efficient for this problem, alternative data structures like linked lists or segment trees could be used for more complex scenarios, such as handling dynamic memory resizing or fragmentation.

What edge cases should I consider when implementing the Design Memory Allocator?

Edge cases include insufficient memory for allocation requests, invalid mID values for freeMemory operations, and handling deallocation of memory blocks that were never allocated.