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Formatted question description: https://leetcode.ca/all/2502.html

2502. Design Memory Allocator

Description

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

  • class Allocator {
    private TreeMap<Integer, Integer> tm = new TreeMap<>();
    private Map<Integer, List<Integer>> d = new HashMap<>();

    public Allocator(int n) {
        tm.put(-1, -1);
        tm.put(n, n);
    }

    public int allocate(int size, int mID) {
        int s = -1;
        for (var entry : tm.entrySet()) {
            int v = entry.getKey();
            if (s != -1) {
                int e = v - 1;
                if (e - s + 1 >= size) {
                    tm.put(s, s + size - 1);
                    d.computeIfAbsent(mID, k -> new ArrayList<>()).add(s);
                    return s;
                }
            }
            s = entry.getValue() + 1;
        }
        return -1;
    }

    public int free(int mID) {
        int ans = 0;
        for (int s : d.getOrDefault(mID, Collections.emptyList())) {
            int e = tm.remove(s);
            ans += e - s + 1;
        }
        d.remove(mID);
        return ans;
    }
}

/**
 * Your Allocator object will be instantiated and called as such:
 * Allocator obj = new Allocator(n);
 * int param_1 = obj.allocate(size,mID);
 * int param_2 = obj.free(mID);
 */

  • class Allocator {
public:
    Allocator(int n) {
        tm[-1] = -1;
        tm[n] = n;
    }

    int allocate(int size, int mID) {
        int s = -1;
        for (auto& [v, c] : tm) {
            if (s != -1) {
                int e = v - 1;
                if (e - s + 1 >= size) {
                    tm[s] = s + size - 1;
                    d[mID].emplace_back(s);
                    return s;
                }
            }
            s = c + 1;
        }
        return -1;
    }

    int free(int mID) {
        int ans = 0;
        for (int& s : d[mID]) {
            int e = tm[s];
            tm.erase(s);
            ans += e - s + 1;
        }
        d.erase(mID);
        return ans;
    }

private:
    map<int, int> tm;
    unordered_map<int, vector<int>> d;
};

/**
 * Your Allocator object will be instantiated and called as such:
 * Allocator* obj = new Allocator(n);
 * int param_1 = obj->allocate(size,mID);
 * int param_2 = obj->free(mID);
 */

  • from sortedcontainers import SortedList


class Allocator:
    def __init__(self, n: int):
        self.sl = SortedList([(-1, -1), (n, n)])
        self.d = defaultdict(list)

    def allocate(self, size: int, mID: int) -> int:
        for (_, s), (e, _) in pairwise(self.sl):
            s, e = s + 1, e - 1
            if e - s + 1 >= size:
                self.sl.add((s, s + size - 1))
                self.d[mID].append((s, s + size - 1))
                return s
        return -1

    def free(self, mID: int) -> int:
        ans = 0
        for block in self.d[mID]:
            self.sl.remove(block)
            ans += block[1] - block[0] + 1
        del self.d[mID]
        return ans


# Your Allocator object will be instantiated and called as such:
# obj = Allocator(n)
# param_1 = obj.allocate(size,mID)
# param_2 = obj.free(mID)


  • type Allocator struct {
	rbt *redblacktree.Tree
	d   map[int][]int
}

func Constructor(n int) Allocator {
	rbt := redblacktree.NewWithIntComparator()
	rbt.Put(-1, -1)
	rbt.Put(n, n)
	return Allocator{rbt, map[int][]int{} }
}

func (this *Allocator) Allocate(size int, mID int) int {
	s := -1
	it := this.rbt.Iterator()
	for it.Next() {
		v := it.Key().(int)
		if s != -1 {
			e := v - 1
			if e-s+1 >= size {
				this.rbt.Put(s, s+size-1)
				this.d[mID] = append(this.d[mID], s)
				return s
			}
		}
		s = it.Value().(int) + 1
	}
	return -1
}

func (this *Allocator) Free(mID int) int {
	ans := 0
	for _, s := range this.d[mID] {
		if e, ok := this.rbt.Get(s); ok {
			this.rbt.Remove(s)
			ans += e.(int) - s + 1
		}
	}
	this.d[mID] = []int{}
	return ans
}

/**
 * Your Allocator object will be instantiated and called as such:
 * obj := Constructor(n);
 * param_1 := obj.Allocate(size,mID);
 * param_2 := obj.Free(mID);
 */

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