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The LeetCode Framework for Rotting Brains

Start with just one solution—understand it fully instead of guessing

Chaïmae SritiOctober 2025

⚠️ Friendly Disclaimer: This framework is NOT for people who solve hards under 5 min. This is for us—the regular, normal people—who spent a lot of time brainrotting and now need 15+ minutes to write down a decent brute force solution. If you're some leetcode genius, this article will bore you.

1. The Problem with Zero Training Data

If your brain is rotting—from doom-scrolling, context-switching, or general mental entropy—attempting to solve LeetCode problems from scratch is training your brain on failure patterns, not solution patterns.

Core insight: Like a machine learning model needs training data to recognize patterns, your brain needs exposure to solutions before it can generate good ones. When you have zero training data, you're not problem-solving—you're speculating. And speculation with broken heuristics produces garbage.

The traditional advice is "struggle is good for learning." That's true when your cognitive machinery is functioning. But when your brain is mush, struggle reinforces the mush. You're practicing being bad at problem-solving.

2. Assess Your Brainrot Level

Use Binary Search on Your Capacity

Use binary search on your capacity to solve a medium problem in under 15 minutes. Based on your level of brainrot, here's how many solutions you should consume before attempting to solve problems independently:

Solutions perPatternBrainrot Level05101520LowMediumHighExtreme2-37-1015-2025-30+

Low Brainrot (Solve medium in <15 min)

Your problem-solving scaffolding is intact. Consume 2-3 solutions per pattern, then start solving.

Medium Brainrot (15-30 min for medium)

You need more training data. Consume 7-10 solutions per pattern with spaced repetition before attempting problems.

High Brainrot (30+ min or can't solve medium)

Significant rewiring needed. Consume 15-20 solutions per pattern with active reviews at Day 1, 2, 4, 7 before solving.

Extreme Brainrot (Can't conceptualize brute force)

Neural pathways need complete reconstruction. Consume 25-30+ solutions per pattern. Don't attempt solving for at least 3-4 weeks of active review.

How to assess yourself using binary search:

1. Pick 5 random LeetCode mediums you've never seen
2. Time yourself solving each (no looking at solutions)
3. Take the median time from the 3 middle results
4. If median < 15 min → Low brainrot
5. If median 15-30 min → Medium brainrot
6. If median > 30 min or couldn't solve → High brainrot
7. If you can't even write a brute force → Extreme brainrot

3. Case Study: Walls and Gates

Let's look at a concrete example: LeetCode 286 - Walls and Gates

Problem Statement

You are given an m×n grid where:

-1 represents a wall or obstacle
0 represents a gate
INF represents an empty room (use 2147483647)

Fill each empty room with the distance to its nearest gate. If impossible to reach a gate, leave it as INF.

Example Input/Output
Input:
INF  -1   0   INF
INF  INF  INF  -1
INF  -1   INF  -1
 0   -1   INF  INF

Output:
 3   -1   0   1
 2    2   1  -1
 1   -1   2  -1
 0   -1   3   4

Scenario A: Guessing (Zero Training Data)

You stare at this problem. Your brain, with zero training data on graph traversal, generates this thought process:

"Okay, so I need to find distances... maybe I can loop through each empty room, and for each room, search outward until I hit a gate?"

→ For each empty room, do BFS to find nearest gate
→ Need to track: visited cells (per BFS), current distance, queue of cells to explore
→ Where to store? New visited set each time, new queue each time
→ Time complexity: O(m²n²) — BFS from every single empty room

The Brute Force Code:

def wallsAndGates(rooms):
    if not rooms:
        return

    m, n = len(rooms), len(rooms[0])
    INF = 2147483647

    # For each empty room, find nearest gate
    for i in range(m):
        for j in range(n):
            if rooms[i][j] == INF:  # Empty room
                # BFS from this room to find nearest gate
                queue = [(i, j, 0)]  # (row, col, distance)
                visited = set()
                visited.add((i, j))
                min_dist = INF

                while queue:
                    x, y, dist = queue.pop(0)

                    # Found a gate
                    if rooms[x][y] == 0:
                        min_dist = min(min_dist, dist)
                        break  # BFS guarantees shortest path

                    # Explore neighbors
                    for dx, dy in [(0,1), (1,0), (0,-1), (-1,0)]:
                        nx, ny = x + dx, y + dy

                        if (0 <= nx < m and 0 <= ny < n and
                            (nx, ny) not in visited and
                            rooms[nx][ny] != -1):  # Not wall
                            visited.add((nx, ny))
                            queue.append((nx, ny, dist + 1))

                rooms[i][j] = min_dist

You came up with this yourself? You're not as brainrotted as you think. You have basic problem-solving scaffolding. Now go read the optimal solution and see the insight you missed.

What's wrong with this:

• You're doing BFS from every empty room
• Each BFS creates its own queue and visited set (memory waste)
• You're traversing the same cells over and over for different rooms
• Result: O(m²n²) time, massive redundant work
→ You submit. Time Limit Exceeded.

"Maybe I need to optimize... add memoization? Try different pruning?"

→ You thrash for 30 more minutes trying to cache BFS results
→ Nothing works elegantly
→ You feel stupid, give up, move on

What just happened:

• Your brain had zero training data on multi-source BFS
• It defaulted to the obvious approach: BFS from each target (room)
• You didn't think about inverting the problem: BFS from sources (gates) instead
• You reinforced a bad pattern: "when stuck, add random optimizations"
• You learned nothing transferable

🚨 Ultimate Brainrot Check:

If you can't even conceptualize the brute force approach above—if you can't think through "loop through empty rooms, BFS from each to find nearest gate"—that's ultimate brainrot. You're not even at the level of generating bad solutions. You're at zero training data, no problem-solving scaffolding. This is precisely when you should stop trying and consume solutions instead.

Scenario B: Reading the Solution (Training Data)

Instead of guessing, you go straight to a high-quality solution. Here's what you consume:

Key Insight: Multi-Source BFS

Instead of starting BFS from each empty room, start BFS from all gates simultaneously. This way, you only traverse the grid once.

Algorithm:

1. Initialize a queue with all gate positions (cells with value 0)
2. Perform BFS: for each position, explore its 4 neighbors
3. If a neighbor is an empty room (INF), update its distance and add to queue
4. Continue until queue is empty

Time Complexity: O(mn)

Each cell is visited at most once.

Code:

from collections import deque

def wallsAndGates(rooms):
    if not rooms:
        return

    m, n = len(rooms), len(rooms[0])
    queue = deque()

    # Step 1: Add all gates to queue
    for i in range(m):
        for j in range(n):
            if rooms[i][j] == 0:
                queue.append((i, j))

    # Step 2: Multi-source BFS
    directions = [(0,1), (1,0), (0,-1), (-1,0)]

    while queue:
        x, y = queue.popleft()  # O(1) instead of pop(0)

        for dx, dy in directions:
            nx, ny = x + dx, y + dy

            # Check bounds and if it's an empty room
            if 0 <= nx < m and 0 <= ny < n and rooms[nx][ny] == INF:
                rooms[nx][ny] = rooms[x][y] + 1
                queue.append((nx, ny))

What you just learned:

Pattern recognition: "Multiple sources, find shortest distance" → multi-source BFS
Mental model: BFS from all sources simultaneously is more efficient than BFS from each target
Implementation details: How to set up queue, how to track visited cells implicitly by updating values
Transfer learning: This pattern applies to other problems (e.g., "01 Matrix", "Rotting Oranges")

Questions You MUST Ask While Understanding This Solution:

1. What is BFS? (For the ultimate brainrotted)

BFS = Breadth-First Search: Explore all neighbors at distance 1, then all at distance 2, then distance 3, etc.
Think of it like ripples in water spreading outward from a stone.
Key property: BFS finds the shortest path in an unweighted graph/grid
How: Use a queue (FIFO). Add starting point(s) → while queue not empty, pop front, explore neighbors, add them to back

2. Why BFS and not DFS?

BFS: Explores level-by-level → guarantees shortest path
DFS: Explores depth-first → might find a longer path first
For this problem: We need shortest distance to nearest gate → BFS is correct
If you used DFS, you'd have to explore ALL paths and track minimum, which defeats the purpose

3. Why use a queue? Why not a deque?

Queue (list with pop(0)): Works, but pop(0) is O(n) in Python
Deque (collections.deque): pop from left is O(1) → more efficient
Better code would use:
from collections import deque
queue = deque()
queue.popleft() # O(1) instead of queue.pop(0)
For LeetCode, either works. For production, use deque.

4. What data do I need to persist as I'm solving?

Option 1 (This solution): No separate visited set
→ Visited tracking is implicit: we modify rooms[nx][ny] from INF to a number
→ When we check "if rooms[nx][ny] == INF", we're checking if unvisited
→ Data persisted: just the rooms grid itself
Option 2 (Alternative): Explicit visited set
→ Keep visited = set() to track (i, j) tuples we've seen
→ Clearer, but uses extra O(mn) space
→ Use this approach when you CAN'T modify the input
Option 3 (Brute force used): New visited set per BFS
→ This is why brute force is slow: recreating data structures wastefully
Key insight: In this solution, we only need to persist:
• The queue (current frontier of BFS)
• The rooms grid (which we modify in-place for both output AND visited tracking)

5. Why start from gates instead of empty rooms?

Key insight: "Shortest distance from room to nearest gate" = "Shortest distance from nearest gate to room"
It's the same distance, just traversed in opposite direction!
Why this matters:
• Starting from gates: One BFS that processes all rooms
• Starting from rooms: N BFS operations where N = number of empty rooms
• Same result, vastly different performance
This is the "inversion" insight: When you have multiple targets and one source, flip it to multiple sources and check targets

6. How does the distance calculation work?

When we process cell (x, y) and find an empty neighbor (nx, ny):
rooms[nx][ny] = rooms[x][y] + 1
Why this works:
• Gates start with distance 0
• Neighbors of gates get distance 0 + 1 = 1
• Neighbors of those get 1 + 1 = 2
• BFS guarantees we process closer cells first
• So the first time we reach a cell, that's the shortest distance

Next Steps:

1. Answer all the questions above without looking. If you can't, you don't understand the solution yet.
2. Type it out: Implement the solution from memory (peek if you forget)
3. Internalize the pattern: Write down "multi-source BFS = start from all sources at once"
4. Find similar problems: Do "01 Matrix" and "Rotting Oranges" next—same pattern

The Difference:

Scenario A: You spent 45 minutes reinforcing bad problem-solving habits. You learned nothing about BFS, nothing about multi-source search, nothing transferable.

Scenario B: You spent 15 minutes absorbing a high-quality solution. You now have training data. Next time you see "multiple sources, shortest path," your brain will fire: "multi-source BFS." You didn't guess—you learned.

4. The Framework

Here's the process:

Step 1: Read the Problem (Don't Attempt More than 5 Minutes ^^ )

Understand what it's asking. Notice the constraints, the input/output format. Don't try to solve it.

Step 2: Go Straight to the Solution

Find a well-explained solution (NeetCode, LeetCode Discuss, etc.). Read it fully.

Step 3: Understand It Deeply

Ask yourself:

• What pattern is being used?
• Why does this pattern work for this problem?
• What's the time/space complexity?
• What are the edge cases?
• Where have I seen this pattern before? Where might I see it again?

Step 4: Implement It Yourself

Type it out from memory. If you forget, look back. Repeat until you can implement it without looking.

Step 5: Build the Pattern Library

Keep a note:

Pattern: Multi-source BFS
When to use: Multiple starting points, need shortest distance to any of them
Examples: Walls and Gates, 01 Matrix, Rotting Oranges

Step 6: Consume More of the Same Pattern

Don't jump to a different pattern yet. Find 2-3 more problems with the same pattern. Consume their solutions. Reinforce the training data.

Step 7: Spaced Repetition with Active Review (CRITICAL)

This is where most people fail. Consuming solutions once is not enough. Your brain will forget. You need active review with spaced repetition.

The Schedule:

Day 1: Consume the solution (Walls and Gates)
→ Read, understand deeply, implement from memory
Day 2: Active review
→ Close the solution. Try to implement from scratch. If you can't, peek at the pattern notes, try again
Day 4: Active review again
→ Implement from scratch. Explain the approach out loud as if teaching someone
Day 7: Final review
→ Implement from scratch. Should be automatic by now

What is Active Review?

✓ Active Review (Good):

• Close the solution and try to implement from memory
• Explain the approach out loud without looking
• Draw the algorithm flow on paper
• Identify where you got stuck and why
• Implement variations (what if we had 3 sources instead of gates?)

✗ Passive Review (Useless):

• Re-reading the solution
• Looking at the code and thinking "yeah, I remember this"
• Clicking through your notes without implementing
• Watching someone else solve it again

Why This Works:

Retrieval practice: Forcing recall strengthens memory far more than re-reading
Spacing effect: Reviews at increasing intervals (1 day, 3 days, 7 days) lock in long-term memory
Failure points: When you get stuck during review, you identify exactly what you don't understand
Pattern recognition: Multiple exposures train your brain to recognize the pattern signature instantly

⚠️ Without Spaced Active Review:

You'll consume 50 solutions, feel like you're learning, and then forget everything within 2 weeks. You'll see the same problem type in an interview and draw a blank. "I know I've seen this before, but..." → That's passive consumption without active review. Your brain never actually learned the pattern—it just recognized it temporarily.

Practical Implementation:

• Use a spreadsheet: Problem name | Pattern | Last reviewed | Next review date
• Set calendar reminders for review days
• Keep a "review queue" - 5 problems to actively review this week
• If you fail a review (can't implement), reset the spacing: review again tomorrow

When to Start Solving:

After you've consumed 5-10 solutions of the same pattern and done spaced active reviews on the first 3-5, then try solving a new problem with that pattern. Your brain now has training data that's actually consolidated in long-term memory. You'll recognize: "Oh, this is multi-source BFS" and implement it. Before, you were guessing. Now, you're applying learned patterns that you've actively rehearsed.

5. Warning: Corrupted Training Data

⚠️ When Solution Consumption Goes Wrong

Consuming solutions is powerful, but it has a dangerous failure mode: confusing similar-looking solutions when you don't actually understand the underlying problem structure.

Example: Walls and Gates vs Longest Increasing Path in a Matrix

Both are grid problems. Both use DFS/BFS. Both involve exploring neighbors. But they solve fundamentally different problems:

Walls and Gates (Multi-Source BFS)

Problem type: Find shortest distance from each room to ANY gate
Key insight: Multiple sources (gates), one traversal
Algorithm: BFS starting from all gates simultaneously
Why BFS? Need shortest path, BFS guarantees first visit = shortest distance
Data structure: Queue (deque)
Visited tracking: Modify grid in-place (INF → distance)

Longest Increasing Path in a Matrix (DFS + Memoization)

Problem type: Find longest path where values are strictly increasing
Key insight: Explore ALL paths, find maximum, avoid recomputation
Algorithm: DFS from each cell + memoization
Why DFS? Need to explore all possible paths to find longest
Data structure: Recursion stack + memo dictionary
Visited tracking: Memo cache stores max path length from each cell

What Goes Wrong Without Understanding:

You see "grid problem" and think "I'll use BFS like Walls and Gates"
You copy the multi-source BFS pattern without realizing this problem needs to explore ALL paths, not find shortest path
You spend 2 hours debugging why BFS doesn't work, get frustrated, give up
You've now reinforced a BAD pattern: "BFS works on all grid problems"

How to Consume Solutions Correctly:

Ask: What is the problem actually asking? (Shortest path vs longest path)
Ask: Why does this approach work for this specific problem? (BFS for shortest, DFS+memo for longest)
Ask: What constraints make this approach necessary? (All gates at once vs explore all paths)
Ask: How is this different from similar problems I've seen? (Compare side-by-side)
Write it down: "Walls and Gates = multi-source BFS for shortest paths. Longest Increasing Path = DFS+memo for exploring ALL paths."

Side-by-Side Code Comparison:

Walls and Gates (Multi-Source BFS)

def wallsAndGates(rooms):
    if not rooms:
        return

    m, n = len(rooms), len(rooms[0])
    queue = deque()

    # Initialize: Add ALL gates to queue
    for i in range(m):
        for j in range(n):
            if rooms[i][j] == 0:  # Gate
                queue.append((i, j))

    directions = [(0,1), (1,0), (0,-1), (-1,0)]

    # BFS from all gates simultaneously
    while queue:
        x, y = queue.popleft()

        for dx, dy in directions:
            nx, ny = x + dx, y + dy

            # Only visit empty rooms (INF)
            if 0 <= nx < m and 0 <= ny < n and rooms[nx][ny] == INF:
                rooms[nx][ny] = rooms[x][y] + 1  # Distance from gate
                queue.append((nx, ny))

# Key: Start from sources (gates), propagate outward once
# Result: Each room gets shortest distance to nearest gate

Longest Increasing Path (DFS + Memo)

def longestIncreasingPath(matrix):
    if not matrix:
        return 0

    m, n = len(matrix), len(matrix[0])
    memo = {}

    def dfs(x, y):
        # Return cached result if exists
        if (x, y) in memo:
            return memo[(x, y)]

        max_length = 1  # Current cell counts as length 1
        directions = [(0,1), (1,0), (0,-1), (-1,0)]

        # Try all 4 directions
        for dx, dy in directions:
            nx, ny = x + dx, y + dy

            # Only visit if in bounds AND strictly increasing
            if (0 <= nx < m and 0 <= ny < n and
                matrix[nx][ny] > matrix[x][y]):
                length = 1 + dfs(nx, ny)  # Recursively explore
                max_length = max(max_length, length)

        memo[(x, y)] = max_length  # Cache result
        return max_length

    # Must try DFS from EVERY cell to find global max
    result = 0
    for i in range(m):
        for j in range(n):
            result = max(result, dfs(i, j))

    return result

# Key: Start from every cell, explore all paths, cache results
# Result: Find the single longest increasing path in entire grid

Critical Differences:

AspectWalls and GatesLongest Increasing Path
Starting pointsAll gates at once (multi-source)Every cell individually
TraversalBFS (queue, iterative)DFS (recursion stack)
Visit conditionVisit unvisited rooms (INF)Visit if value is increasing
GoalFind shortest distance (minimize)Find longest path (maximize)
Data structureQueue (deque)Recursion + memo dict
Visits per cellOnce (BFS guarantees shortest)Multiple times until memoized
Result storageModify grid in-placeSeparate memo dictionary
Why this approach?BFS explores level-by-level = shortest path naturallyDFS explores all paths + memo avoids recomputation

Why You Can't Swap Them:

Using BFS for Longest Increasing Path:
→ BFS stops at first visit, but we need to explore ALL paths to find longest
→ BFS doesn't naturally handle "strictly increasing" constraint across multiple paths
→ You'd get wrong answer: some shorter path, not the longest
Using DFS for Walls and Gates:
→ DFS from each gate would require exploring from EVERY gate to EVERY room
→ Can't guarantee you found shortest path without trying all possibilities
→ Would need complex logic to track minimum distance = much slower than BFS

The Pattern Library Should Store DISTINCTIONS:

Pattern: Multi-source BFS
When: Multiple starting points, need SHORTEST distance to ANY of them
Examples: Walls and Gates, Rotting Oranges, 01 Matrix
NOT for: Problems needing to explore ALL paths or find LONGEST path
Pattern: DFS + Memoization on grid
When: Need to explore ALL paths with some constraint, overlapping subproblems
Examples: Longest Increasing Path, Unique Paths II
NOT for: Shortest path problems (BFS is better)

⚠️ The Real Danger:

If you consume solutions without understanding the problem structure and why this solution fits this problem, you'll build a library of superficially similar patterns that you'll misapply. You become the person who says "I know BFS!" but then tries to use it for every grid problem, even when DFS+memo or backtracking is correct. This is worse than having zero training data—you have corrupted training data.

Final Word:

Don't attempt to solve problems with zero training data. Your brain needs examples before it can generalize. Start with just one solution—understand it fully, deeply, completely. Then do another. And another. Build your pattern library. Then solve. This isn't cheating. This is how learning actually works.