- Difficulty: Medium
- Tags: LeetCode, Medium, Design, Array, Hash Table, Simulation, leetcode-2502, O(n), Sorted List
Problem
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 idmID
. - 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 anAllocator
object with a memory array of sizen
.int allocate(int size, int mID)
Find the leftmost block ofsize
consecutive free memory units and allocate it with the idmID
. 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 idmID
. 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 toallocate
andfree
.