Package g2501_2600.s2502_design_memory_allocator

Class Allocator

java.lang.Object
g2501_2600.s2502_design_memory_allocator.Allocator
public class Allocator extends Object
2502 - Design Memory Allocator.

Medium

You are given an integer n representing the size of a 0-indexed memory array. All memory units are initially free.

You have a memory allocator with the following functionalities:

  1. Allocate a block of size consecutive free memory units and assign it the id mID.
  2. Free all memory units with the given id mID.

Note that:

  • Multiple blocks can be allocated to the same mID.
  • You should free all the memory units with mID, even if they were allocated in different blocks.

Implement the Allocator class:

  • Allocator(int n) Initializes an Allocator object with a memory array of size n.
  • int allocate(int size, int mID) Find the leftmost block of size consecutive free memory units and allocate it with the id mID. Return the block’s first index. If such a block does not exist, return -1.
  • int free(int mID) Free all memory units with the id mID. Return the number of memory units you have freed.

Example 1:

Input [“Allocator”, “allocate”, “allocate”, “allocate”, “free”, “allocate”, “allocate”, “allocate”, “free”, “allocate”, “free”] [[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.free(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.free(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.free(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 free.

  • Constructor Details

    • Allocator

      public Allocator(int n)

  • Method Details

    • allocate

      public int allocate(int size, int mID)

    • free

      public int free(int mID)