Welcome to Subscribe On Youtube

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 int[] m;

    public Allocator(int n) {
        m = new int[n];
    }

    public int allocate(int size, int mID) {
        int cnt = 0;
        for (int i = 0; i < m.length; ++i) {
            if (m[i] > 0) {
                cnt = 0;
            } else if (++cnt == size) {
                Arrays.fill(m, i - size + 1, i + 1, mID);
                return i - size + 1;
            }
        }
        return -1;
    }

    public int free(int mID) {
        int ans = 0;
        for (int i = 0; i < m.length; ++i) {
            if (m[i] == mID) {
                m[i] = 0;
                ++ans;
            }
        }
        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) {
        m = vector<int>(n);
    }

    int allocate(int size, int mID) {
        int cnt = 0;
        for (int i = 0; i < m.size(); ++i) {
            if (m[i]) {
                cnt = 0;
            } else if (++cnt == size) {
                fill(i - size + 1, i + 1, mID);
                return i - size + 1;
            }
        }
        return -1;
    }

    int free(int mID) {
        int ans = 0;
        for (int i = 0; i < m.size(); ++i) {
            if (m[i] == mID) {
                m[i] = 0;
                ++ans;
            }
        }
        return ans;
    }

private:
    vector<int> m;

    void fill(int from, int to, int val) {
        for (int i = from; i < to; ++i) {
            m[i] = val;
        }
    }
};

/**
 * 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:
    def __init__(self, n: int):
        self.m = [0] * n

    def allocate(self, size: int, mID: int) -> int:
        cnt = 0
        for i, v in enumerate(self.m):
            if v:
                cnt = 0
            else:
                cnt += 1
                if cnt == size:
                    self.m[i - size + 1 : i + 1] = [mID] * size
                    return i - size + 1
        return -1

    def free(self, mID: int) -> int:
        ans = 0
        for i, v in enumerate(self.m):
            if v == mID:
                self.m[i] = 0
                ans += 1
        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 {
	m []int
}

func Constructor(n int) Allocator {
	return Allocator{make([]int, n)}
}

func (this *Allocator) Allocate(size int, mID int) int {
	cnt := 0
	for i, v := range this.m {
		if v > 0 {
			cnt = 0
		} else {
			cnt++
			if cnt == size {
				for j := i - size + 1; j <= i; j++ {
					this.m[j] = mID
				}
				return i - size + 1
			}
		}
	}
	return -1
}

func (this *Allocator) Free(mID int) (ans int) {
	for i, v := range this.m {
		if v == mID {
			this.m[i] = 0
			ans++
		}
	}
	return
}

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

All Problems

All Solutions