README.md

2502. Design Memory Allocator

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.

Solutions (Rust)

1. Solution

struct Allocator {
    memory: Vec<i32>,
}

/**
 * `&self` means the method takes an immutable reference.
 * If you need a mutable reference, change it to `&mut self` instead.
 */
impl Allocator {
    fn new(n: i32) -> Self {
        Self {
            memory: vec![0; n as usize],
        }
    }

    fn allocate(&mut self, size: i32, m_id: i32) -> i32 {
        let size = size as usize;
        let mut count = self.memory.iter().take(size).filter(|&&x| x == 0).count();

        for i in 0..=self.memory.len().saturating_sub(size) {
            if count == size {
                for j in 0..size {
                    self.memory[i + j] = m_id;
                }

                return i as i32;
            }

            count -= (self.memory[i] == 0) as usize;
            count += (*self.memory.get(i + size).unwrap_or(&1) == 0) as usize;
        }

        -1
    }

    fn free(&mut self, m_id: i32) -> i32 {
        let mut count = 0;

        for i in 0..self.memory.len() {
            if self.memory[i] == m_id {
                self.memory[i] = 0;
                count += 1;
            }
        }

        count
    }
}

/**
 * Your Allocator object will be instantiated and called as such:
 * let obj = Allocator::new(n);
 * let ret_1: i32 = obj.allocate(size, mID);
 * let ret_2: i32 = obj.free(mID);
 */