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:

• Allocate a block of size consecutive free memory units and assign it the id mID.

• 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.