2587. Design Memory Allocator¶

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2587. Design Memory Allocator

Medium

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¶

class Allocator {
    private int memory[];
    public Allocator(int n) {
        memory = new int[n];
    }
    public int allocate(int size, int mID) {
        int startIdx = -1;
        int cons = 0;
        int count = 0;
        boolean flag = false;
        for (int i = 0; i < memory.length; i++) {
            if (memory[i] == 0) {
                count++;
                if (startIdx == -1)
                    startIdx = i;
                if (count == size) {
                    flag = true;
                    break;
                }
            } else {
                count = 0;
                startIdx = -1;
            }
        }
        if (flag == false)
            return -1;

        for (int i = startIdx; i < startIdx + size; i++)
            memory[i] = mID;
        return startIdx;
    }
    public int freeMemory(int mID) {
        int count = 0;
        for (int i = 0; i < memory.length; i++) {
            if (memory[i] == mID) {
                memory[i] = 0;
                count++;
            }
        }
        return count;
    }
}

/**
    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.freeMemory(mID);
*/

Complexity Analysis¶

• Time Complexity: O(?)
• Space Complexity: O(?)

Approach¶

Detailed explanation of the approach will be added here