Problem Description

The Incremental Memory Leak problem is a simulation-based problem where you are given three integer values: memory1, memory2, and memory3. Each integer represents the amount of memory available in three memory banks.

The process works in seconds. On the ith second (starting from 1), you must allocate i units of memory from the bank with the largest available memory (if there is a tie, choose the bank with the smallest index). If the chosen bank does not have enough memory to allocate i units, the process stops.

The task is to return an array of length 4:

• The first element is the time (in seconds) when the process stops.
• The next three elements are the final amounts of memory left in memory1, memory2, and memory3, in that order.

• All input values are positive integers.
• At each step, only one bank can be selected for allocation.
• There is only one valid way to simulate the process given the rules.

Thought Process

When approaching this problem, the first thought is to simulate the process step by step, as described. At each second, we need to determine which memory bank has the largest available memory and perform the allocation if possible.

A brute-force way would be to simply repeat this process, updating the memory values and incrementing the time, until we can no longer allocate the required amount of memory. This is acceptable because the constraints are small and the process is deterministic.

An optimization would be to avoid unnecessary checks by always picking the maximum efficiently and stopping immediately when allocation is not possible. Since there are only three banks, comparing them is trivial and fast.

The main challenge is to carefully follow the tie-breaking rule and update the correct memory bank at each step.

Solution Approach

The solution follows a straightforward simulation, step by step:

1. Initialize a variable time to 1 (representing the first second).
2. At each step:
   • Determine which of the three memory banks has the largest available memory. If there is a tie, choose the bank with the smallest index.
   • Check if the chosen bank has at least time units of memory.
   • If yes, subtract time units from that bank, increment time by 1, and repeat.
   • If not, stop the process.
3. Return an array with the current time (when the process stopped) and the final values of memory1, memory2, and memory3.

We use a simple array or variables to keep track of the memory banks. Since there are only three, manual comparison is efficient and clear. There is no need for advanced data structures.

Example Walkthrough

Let's use the input memory1 = 2, memory2 = 2, memory3 = 3.

• Second 1: Need 1 unit. Largest is memory3 (3). Subtract 1 from memory3: [2, 2, 2]
• Second 2: Need 2 units. All banks have 2. Tie, choose the smallest index (memory1). Subtract 2 from memory1: [0, 2, 2]
• Second 3: Need 3 units. Largest is memory2 and memory3 (both 2), but 2 < 3, so can't allocate. Process stops.

Return: [3, 0, 2, 2]

This matches the expected output, and shows how at each step we pick the correct bank, update, and stop when allocation fails.

Time and Space Complexity

Brute-force Approach:

• Each step can require up to the total sum of memory across the three banks, as the allocation increases by one each time.
• Let N be the maximum sum of the three banks. The loop runs at most O(sqrt(N)) times, since the sum of the first k natural numbers is O(k^2).
• At each step, we do O(1) work (comparing three values).
• Total Time Complexity: O(sqrt(N))
• Space Complexity: O(1), as we only use a few variables.

• The approach is already optimal for the problem size, as we can't skip steps in the simulation.

Summary

The Incremental Memory Leak problem is a simulation that requires careful tracking of which bank to allocate from at each step, following the rules for tie-breaking. The solution is elegant in its simplicity: step through each second, always pick the largest bank, and stop as soon as you can't allocate. The process is efficient due to the small number of banks and the clear stopping condition.

Code Implementation

class Solution:
    def memLeak(self, memory1: int, memory2: int, memory3: int) -> list[int]:
        mem = [memory1, memory2, memory3]
        time = 1
        while True:
            # Find the index of the largest memory (with tie-breaker)
            max_idx = 0
            for i in range(1, 3):
                if mem[i] > mem[max_idx]:
                    max_idx = i
            # Check if we can allocate
            if mem[max_idx] < time:
                break
            mem[max_idx] -= time
            time += 1
        return [time, mem[0], mem[1], mem[2]]
      

class Solution {
public:
    vector<int> memLeak(int memory1, int memory2, int memory3) {
        vector<int> mem = {memory1, memory2, memory3};
        int time = 1;
        while (true) {
            int max_idx = 0;
            for (int i = 1; i < 3; ++i) {
                if (mem[i] > mem[max_idx]) max_idx = i;
            }
            if (mem[max_idx] < time) break;
            mem[max_idx] -= time;
            ++time;
        }
        return {time, mem[0], mem[1], mem[2]};
    }
};
      

class Solution {
    public int[] memLeak(int memory1, int memory2, int memory3) {
        int[] mem = new int[] {memory1, memory2, memory3};
        int time = 1;
        while (true) {
            int maxIdx = 0;
            for (int i = 1; i < 3; ++i) {
                if (mem[i] > mem[maxIdx]) maxIdx = i;
            }
            if (mem[maxIdx] < time) break;
            mem[maxIdx] -= time;
            time++;
        }
        return new int[] {time, mem[0], mem[1], mem[2]};
    }
}
      

var memLeak = function(memory1, memory2, memory3) {
    let mem = [memory1, memory2, memory3];
    let time = 1;
    while (true) {
        let maxIdx = 0;
        for (let i = 1; i < 3; ++i) {
            if (mem[i] > mem[maxIdx]) maxIdx = i;
        }
        if (mem[maxIdx] < time) break;
        mem[maxIdx] -= time;
        time++;
    }
    return [time, mem[0], mem[1], mem[2]];
};