Ace Google/Meta Coding Interviews Using AI-Inspired Patterns

def pattern_state_space(problem):
    """
    Think: What's my current state? What moves can I make?
    Classic: BFS, DFS, backtracking, dynamic programming
    """
    # Template structure
    visited = set()
    queue = [initial_state]
    
    while queue:
        current = queue.pop(0)
        
        # Goal check first (AI prunes early)
        if is_goal(current):
            return current
            
        # Generate next states (this is your "move generator")
        for next_state in get_neighbors(current):
            if next_state not in visited:
                visited.add(next_state)
                queue.append(next_state)
    
    return None

# Example: Word Ladder (Google L5 real question)
def ladderLength(beginWord, endWord, wordList):
    """
    State = current word
    Moves = change one letter to another word in list
    Goal = reach endWord
    """
    if endWord not in wordList:
        return 0
    
    wordList = set(wordList)
    queue = [(beginWord, 1)]  # (word, steps)
    
    while queue:
        word, steps = queue.pop(0)
        
        if word == endWord:
            return steps
        
        # Generate neighbors: try changing each position
        for i in range(len(word)):
            for c in 'abcdefghijklmnopqrstuvwxyz':
                next_word = word[:i] + c + word[i+1:]
                if next_word in wordList:
                    wordList.remove(next_word)  # AI pruning: don't revisit
                    queue.append((next_word, steps + 1))
    
    return 0

def pattern_constraint_satisfaction(problem):
    """
    Think: What are my constraints? Can I place this value?
    Classic: Backtracking, greedy with validation
    """
    def is_valid(state, new_element):
        # Check all constraints
        return all_constraints_satisfied(state, new_element)
    
    def solve(state):
        if is_complete(state):
            return state
        
        for candidate in get_candidates():
            if is_valid(state, candidate):
                state.add(candidate)
                
                result = solve(state)
                if result:
                    return result
                
                state.remove(candidate)  # Backtrack
        
        return None

# Example: N-Queens (Meta E5 real question)
def solveNQueens(n):
    """
    Constraint: No two queens attack each other
    States: Board configurations
    Validation: Check row/col/diagonal constraints
    """
    def is_safe(board, row, col):
        # Check column
        for i in range(row):
            if board[i] == col:
                return False
        
        # Check diagonals (AI optimization: only check upward)
        for i in range(row):
            if abs(board[i] - col) == abs(i - row):
                return False
        
        return True
    
    def solve(board, row):
        if row == n:
            return [["." * i + "Q" + "." * (n-i-1) for i in board]]
        
        solutions = []
        for col in range(n):
            if is_safe(board, row, col):
                board[row] = col
                solutions.extend(solve(board, row + 1))
                board[row] = -1  # Backtrack
        
        return solutions
    
    return solve([-1] * n, 0)

def pattern_transform_reduce(problem):
    """
    Think: Can I convert this to a problem I already know?
    Classic: Sorting + two pointers, hashing, math transformations
    """
    # Step 1: Transform to canonical form
    transformed = transform(problem)
    
    # Step 2: Solve the simpler problem
    result = solve_simple(transformed)
    
    # Step 3: Map back to original problem
    return inverse_transform(result)

# Example: Two Sum (but thinking like AI)
def twoSum(nums, target):
    """
    Transform: Array search → Hash lookup
    Reduce: O(n²) → O(n)
    """
    # Transform: Build inverse mapping (value → index)
    seen = {}
    
    for i, num in enumerate(nums):
        complement = target - num
        
        # Reduced problem: "Does complement exist in seen?"
        if complement in seen:
            return [seen[complement], i]
        
        seen[num] = i
    
    return []

# Example: Meeting Rooms II (Google real question)
def minMeetingRooms(intervals):
    """
    Transform: Intervals → Events (start/end times)
    Reduce: Overlap problem → Count simultaneous events
    """
    events = []
    for start, end in intervals:
        events.append((start, 1))   # Meeting starts
        events.append((end, -1))    # Meeting ends
    
    # Sort events (AI loves sorted data)
    events.sort()
    
    # Reduce: Count maximum simultaneous meetings
    rooms = max_rooms = 0
    for time, delta in events:
        rooms += delta
        max_rooms = max(max_rooms, rooms)
    
    return max_rooms

def pattern_incremental_refinement(problem):
    """
    Think: What do I know after processing first K elements?
    Classic: Sliding window, running calculations, monotonic structures
    """
    state = initialize()
    
    for element in data_stream:
        # Update state incrementally
        state = update(state, element)
        
        # Maintain invariant (AI keeps properties true)
        state = enforce_invariant(state)
        
        # Emit result if needed
        if should_output():
            yield get_result(state)

# Example: LRU Cache (Meta E5 favorite)
class LRUCache:
    """
    Invariant: Most recently used items at front
    Incremental: Update on each get/put
    """
    def __init__(self, capacity):
        self.capacity = capacity
        self.cache = {}  # key → node
        self.head = Node(0, 0)  # Dummy head
        self.tail = Node(0, 0)  # Dummy tail
        self.head.next = self.tail
        self.tail.prev = self.head
    
    def get(self, key):
        if key not in self.cache:
            return -1
        
        node = self.cache[key]
        # Incremental update: Move to front (most recent)
        self._remove(node)
        self._add_to_front(node)
        return node.value
    
    def put(self, key, value):
        if key in self.cache:
            self._remove(self.cache[key])
        
        node = Node(key, value)
        self._add_to_front(node)
        self.cache[key] = node
        
        # Enforce invariant: Maintain capacity
        if len(self.cache) > self.capacity:
            lru = self.tail.prev
            self._remove(lru)
            del self.cache[lru.key]
    
    def _remove(self, node):
        # Constant-time removal
        node.prev.next = node.next
        node.next.prev = node.prev
    
    def _add_to_front(self, node):
        # Constant-time insertion
        node.next = self.head.next
        node.prev = self.head
        self.head.next.prev = node
        self.head.next = node

class Node:
    def __init__(self, key, value):
        self.key = key
        self.value = value
        self.prev = None
        self.next = None

def pattern_divide_conquer_memo(problem):
    """
    Think: Can I break this into smaller identical problems?
    Classic: Dynamic programming, recursive with cache
    """
    memo = {}
    
    def solve(subproblem):
        # Base case
        if is_trivial(subproblem):
            return base_solution(subproblem)
        
        # Check cache (AI doesn't recompute)
        if subproblem in memo:
            return memo[subproblem]
        
        # Divide: Break into smaller problems
        subproblems = divide(subproblem)
        
        # Conquer: Solve recursively
        results = [solve(sub) for sub in subproblems]
        
        # Combine: Merge solutions
        solution = combine(results)
        
        memo[subproblem] = solution
        return solution

# Example: Longest Common Subsequence (Google real question)
def longestCommonSubsequence(text1, text2):
    """
    Subproblem: LCS(i, j) = LCS of text1[0:i] and text2[0:j]
    Recurrence: If match → 1 + LCS(i-1, j-1)
                Else → max(LCS(i-1, j), LCS(i, j-1))
    """
    memo = {}
    
    def dp(i, j):
        if i == 0 or j == 0:
            return 0
        
        if (i, j) in memo:
            return memo[(i, j)]
        
        if text1[i-1] == text2[j-1]:
            # Match: Include this character
            result = 1 + dp(i-1, j-1)
        else:
            # No match: Try both options
            result = max(dp(i-1, j), dp(i, j-1))
        
        memo[(i, j)] = result
        return result
    
    return dp(len(text1), len(text2))

# Bottom-up version (what you'd actually write in interview)
def longestCommonSubsequence(text1, text2):
    m, n = len(text1), len(text2)
    # dp[i][j] = LCS length of text1[0:i] and text2[0:j]
    dp = [[0] * (n + 1) for _ in range(m + 1)]
    
    for i in range(1, m + 1):
        for j in range(1, n + 1):
            if text1[i-1] == text2[j-1]:
                dp[i][j] = 1 + dp[i-1][j-1]
            else:
                dp[i][j] = max(dp[i-1][j], dp[i][j-1])
    
    return dp[m][n]

# Always start with the pattern template
def solution(input):
    # 1. State your pattern
    # "Using state space search with BFS"
    
    # 2. Initialize pattern-specific structures
    visited = set()
    queue = [initial_state]
    
    # 3. Core pattern loop
    while queue:
        current = queue.pop(0)
        
        # 4. Pattern-specific logic
        if is_goal(current):
            return current
        
        for next_state in neighbors(current):
            if valid(next_state):
                queue.append(next_state)
    
    return result

# AI-inspired test cases (from AlphaCode paper)

# 1. Edge case: Empty input
assert solution([]) == expected_empty

# 2. Base case: Single element
assert solution([x]) == expected_single

# 3. Adversarial: Breaks naive solutions
assert solution(worst_case_input) == expected_hard

# 4. Random: Your original example
assert solution(example_input) == expected_output

# Immediate recognition
"This is Transform & Reduce. I'll transform intervals into 
 sorted form, then reduce by merging overlaps linearly."

def merge(intervals):
    if not intervals:
        return []
    
    # Transform: Sort by start time
    # Why: AI loves sorted data—makes comparisons O(1)
    intervals.sort(key=lambda x: x[0])
    
    # Reduce: Merge overlapping intervals
    merged = [intervals[0]]
    
    for current in intervals[1:]:
        last = merged[-1]
        
        # Incremental refinement: Update last interval or add new
        if current[0] <= last[1]:
            # Overlap: Extend last interval
            last[1] = max(last[1], current[1])
        else:
            # No overlap: Start new interval
            merged.append(current)
    
    return merged

# Test cases (said out loud)
# Edge: [] → []
# Base: [[1,3]] → [[1,3]]
# Adversarial: [[1,4],[2,3]] → [[1,4]] (nested interval)
# Example: [[1,3],[2,6],[8,10],[15,18]] → [[1,6],[8,10],[15,18]]