Introduction

In this blog post, we will explore how to solve the LeetCode problem "2502" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

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: Allocate a block of size consecutive free memory units and assign it the id mID. 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.

Explanation

Here's a breakdown of the solution:

• Core Idea: Maintain a simple array representing the memory. allocate scans for a contiguous block of free memory (represented by 0) of the required size and assigns the given mID. freeMemory iterates through the array and sets all cells with the given mID back to 0.
• Optimization: While more sophisticated data structures could be used for larger n, the problem constraints (n <= 1000, max 1000 calls) allow for a straightforward linear scan approach, keeping the code simple and efficient enough.

• Implementation Details: The solution provides direct implementation of requested operations.

• Complexity: Time Complexity: O(n) for both allocate and freeMemory due to the linear scans. Space Complexity: O(n) to store the memory array.

Code

    class Allocator:

    def __init__(self, n: int):
        self.memory = [0] * n
        self.size = n

    def allocate(self, size: int, mID: int) -> int:
        count = 0
        start_index = -1
        for i in range(self.size):
            if self.memory[i] == 0:
                if count == 0:
                    start_index = i
                count += 1
                if count == size:
                    for j in range(start_index, start_index + size):
                        self.memory[j] = mID
                    return start_index
            else:
                count = 0
                start_index = -1
        return -1

    def freeMemory(self, mID: int) -> int:
        freed_units = 0
        for i in range(self.size):
            if self.memory[i] == mID:
                self.memory[i] = 0
                freed_units += 1
        return freed_units