Leetcode 1860: Incremental Memory Leak

Input: The input consists of two integers, memory1 and memory2, representing the available memory on the two sticks.

Example: memory1 = 5, memory2 = 6

Constraints:

• 0 <= memory1, memory2 <= 2^31 - 1

Output: The output is an array with three elements: the crash time (the second at which the program crashes), the remaining memory on the first stick, and the remaining memory on the second stick.

Example: [4, 0, 2]

Constraints:

• The program crashes when one of the memory sticks cannot allocate the required memory.

Goal: Simulate the memory allocation process until the program crashes. At each second, allocate memory to the stick with more available memory, or to the first stick if both sticks have the same available memory. Stop when neither stick has enough memory to allocate.

Steps:

• At each second, compare the available memory on both sticks.

• Allocate i bits to the stick with more available memory.

• If both sticks have the same available memory, allocate to the first stick.

• If neither stick has enough memory to allocate i bits, return the crash time and the remaining memory on both sticks.

Goal: The input consists of two integers representing the available memory, and the output should indicate when the program crashes.

Steps:

• 0 <= memory1, memory2 <= 2^31 - 1

• The program will crash within a reasonable number of seconds, given the constraints.

Assumptions:

• Both memory sticks have a non-negative amount of memory.

• Input: Input: memory1 = 5, memory2 = 6

• Explanation: The memory is allocated as follows: At the 1st second, 1 bit is allocated to stick 2 (memory2 = 5). At the 2nd second, 2 bits are allocated to stick 2 (memory2 = 3). At the 3rd second, 3 bits are allocated to stick 1 (memory1 = 2). At the 4th second, 4 bits are allocated to stick 2 (memory2 = -1). The program crashes here.

Approach: The solution involves simulating the memory allocation process, allocating memory in increasing amounts, and stopping when the program crashes due to insufficient memory.

Observations:

• The program allocates memory to the stick with more available memory.

• By simulating the memory allocation process and checking if both sticks have enough memory, we can determine when the program crashes.

Steps:

• Initialize variables to keep track of the available memory on both sticks.

• Loop through each second and allocate memory to the appropriate stick.

• Check if either stick has insufficient memory to allocate the required amount, and if so, return the crash time and remaining memory.

Empty Inputs:

• There are no empty inputs to handle, as memory1 and memory2 will always be non-negative.

Large Inputs:

• If the values for memory1 or memory2 are extremely large, the loop may run for a large number of iterations.

Special Values:

• The program should handle cases where one memory stick has zero memory.

Constraints:

• The program should return the correct result even for edge cases.

vector<int> memLeak(int mem1, int mem2) {
    int i = 1;
    while(i <= mem1 || i <= mem2) {
        if(mem1 >= mem2) {
            mem1 -= i;
        } else mem2 -= i;
        i++;
    }
    return vector<int>{i, mem1, mem2};
}

1 : Function Definition

vector<int> memLeak(int mem1, int mem2) {

Define the function `memLeak`, which takes two integer inputs `mem1` and `mem2`, representing the memory blocks.

2 : Variable Initialization

    int i = 1;

Initialize the variable `i` to 1, which represents the amount of memory to be subtracted in each iteration.

3 : While Loop Condition

    while(i <= mem1 || i <= mem2) {

Start a `while` loop that continues as long as either memory block has sufficient memory to subtract (i.e., `i` is less than or equal to `mem1` or `mem2`).

4 : If Condition

        if(mem1 >= mem2) {

Check if `mem1` is greater than or equal to `mem2`. If true, reduce `mem1` by `i`.

5 : Memory Reduction (mem1)

            mem1 -= i;

Subtract the value of `i` from `mem1`.

6 : Else Condition

        } else mem2 -= i;

If the `if` condition is false, subtract the value of `i` from `mem2`.

7 : Increment i

        i++;

Increment the value of `i` to gradually increase the memory reduction in each iteration.

8 : Return Statement

    return vector<int>{i, mem1, mem2};

Return a vector containing the iteration count `i`, and the remaining memory in both `mem1` and `mem2`.

Best Case: O(1)

Average Case: O(n)

Worst Case: O(n)

Description: In the worst case, the program runs for n seconds, where n is the time before the program crashes.

Best Case: O(1)

Worst Case: O(1)

Description: The space complexity is constant as we only store the memory values and the current second.

Leetcode 1860: Incremental Memory Leak

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