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:
size
consecutive free memory units and assign it the id mID
.mID
.Note that:
mID
.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.For example:
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.
How would you implement the Allocator
class to efficiently manage memory allocation and deallocation?
When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:
The brute force strategy for memory allocation involves checking every possible place to put a new block of data. Think of it like trying to fit a new box into a shelf by trying every available space, one by one. We will keep trying until we find an available space that works.
Here's how the algorithm would work step-by-step:
class MemoryAllocator:
def __init__(self, memory_size):
self.memory_size = memory_size
self.memory = [0] * memory_size # 0 means free, 1 means allocated
def allocate(self, size):
# Iterate through memory to find a suitable block
for start_index in range(self.memory_size - size + 1):
is_available = True
for index in range(start_index, start_index + size):
if self.memory[index] == 1:
is_available = False
break
if is_available:
# Mark the block as allocated, since it's available
for index in range(start_index, start_index + size):
self.memory[index] = 1
return start_index
return -1 # Indicate allocation failure
def free(self, start_index, size):
# Ensure the start index and size are within bounds
if 0 <= start_index < self.memory_size and start_index + size <= self.memory_size:
# Mark the specified memory region as free
for index in range(start_index, start_index + size):
self.memory[index] = 0
The best way to manage memory efficiently is to keep track of what parts are free and what parts are used. We'll use a way to quickly find a free chunk big enough when someone asks for memory, and then update our records to show that chunk is now used.
Here's how the algorithm would work step-by-step:
class MemoryAllocator:
def __init__(self, memory_size):
self.memory_size = memory_size
self.memory = [False] * memory_size
self.free_blocks = [(0, memory_size)]
def allocate(self, size):
for i, (start, block_size) in enumerate(self.free_blocks):
if block_size >= size:
# Found a block large enough
allocated_start = start
if block_size > size:
# Split the block if it's much larger than requested
self.free_blocks[i] = (start + size, block_size - size)
else:
del self.free_blocks[i]
for j in range(allocated_start, allocated_start + size):
self.memory[j] = True
return allocated_start
return None
def deallocate(self, start, size):
for i in range(start, start + size):
self.memory[i] = False
# Inserting the freed block
self.free_blocks.append((start, size))
self.free_blocks.sort()
self.merge_free_blocks()
def merge_free_blocks(self):
# Iterate through the free blocks and merge contiguous blocks
i = 0
while i < len(self.free_blocks) - 1:
start1, size1 = self.free_blocks[i]
start2, size2 = self.free_blocks[i+1]
if start1 + size1 == start2:
# Merge adjacent free blocks
self.free_blocks[i] = (start1, size1 + size2)
del self.free_blocks[i+1]
else:
i += 1
#Merging is complete, so we can exit the function
return
Case | How to Handle |
---|---|
Memory size n is 0 | Allocate should always return -1 and Free should return 0 since no memory is available. |
Allocate size is 0 | Allocate should return a valid index if any is available, and assign the id to a block of size 0; otherwise return -1. |
Allocate size is greater than n | Allocate should always return -1 because requested size exceeds total memory. |
Allocate id already exists | The problem statement does not specify behavior; either reject the allocation, free the old allocation and proceed, or overwrite the old allocation directly, each should be clearly defined. |
Free id does not exist | Free should return 0 indicating that no blocks were freed. |
Multiple free blocks exist for a single ID (from potential multiple allocations). | Free should free *all* blocks allocated with the specified id and return the total number of freed blocks. |
Memory is completely full and allocate is called | Allocate should return -1 as there is no space available to allocate. |
Large n leading to potential memory constraints for internal data structures. | Choose efficient data structures like bit arrays or linked lists to represent memory allocation to avoid excessive memory usage. |