2502. Design Memory Allocator

Problem Link

Description

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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 freeMemory(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", "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

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Python3

class Allocator:
 
    def __init__(self, n: int):
        self.N = n
        self.mp = defaultdict(list)
        self.A = [0] * self.N
 
    def allocate(self, size: int, mID: int) -> int:
        curr = 0
        
        for i in range(self.N):
            if self.A[i] != 0:
                curr = 0
            else:
                curr += 1
                
                if curr == size:
                    start = i - size + 1
 
                    for k in range(start, start + size):
                        self.mp[mID].append(k)
                        self.A[k] = 1
 
                    return start
        
        return -1
 
    def free(self, mID: int) -> int:
        count = 0
        
        for index in self.mp[mID]:
            count += 1
            self.A[index] = 0
        
        self.mp[mID].clear()
        
        return count
 
 
# Your Allocator object will be instantiated and called as such:
# obj = Allocator(n)
# param_1 = obj.allocate(size,mID)
# param_2 = obj.free(mID)