Coin Change Problems: Understanding Unbounded Knapsack Variations

def coin_change_analysis():
    """
    Complete analysis of the minimum coin change problem
    """
    
    # Problem Understanding
    """
    Input: coins = [1, 3, 4], amount = 6
    Output: 2 (use coins 3 + 3 = 6)
    
    Note: Greedy would give [4, 1, 1] = 3 coins (suboptimal)
    This shows why we need DP for non-canonical coin systems
    """
    
    # State Definition
    """
    dp[i] = minimum coins needed to make amount i
    """
    
    # Recurrence Relation
    """
    For each amount i and each coin c:
    if i >= c:
        dp[i] = min(dp[i], dp[i-c] + 1)
    """
    
    # Base Cases
    """
    dp[0] = 0 (zero coins needed for amount 0)
    dp[i] = infinity initially (impossible to make)
    """

def coin_change_min_coins(coins, amount):
    """
    Find minimum coins needed to make target amount
    Time: O(amount * len(coins)), Space: O(amount)
    """
    if amount == 0:
        return 0
    
    # Initialize DP array
    dp = [float('inf')] * (amount + 1)
    dp[0] = 0
    
    # For each amount from 1 to target
    for i in range(1, amount + 1):
        # Try each coin
        for coin in coins:
            if coin <= i:  # Can use this coin
                dp[i] = min(dp[i], dp[i - coin] + 1)
    
    return dp[amount] if dp[amount] != float('inf') else -1

def coin_change_with_trace(coins, amount):
    """
    Show the step-by-step calculation process
    """
    print(f"Finding minimum coins for amount {amount} with coins {coins}")
    
    dp = [float('inf')] * (amount + 1)
    dp[0] = 0
    parent = [-1] * (amount + 1)  # To reconstruct solution
    
    print(f"dp[0] = 0 (base case)")
    
    for i in range(1, amount + 1):
        print(f"\nCalculating dp[{i}]:")
        
        for coin in coins:
            if coin <= i:
                new_count = dp[i - coin] + 1
                if new_count < dp[i]:
                    dp[i] = new_count
                    parent[i] = coin
                    print(f"  Using coin {coin}: dp[{i-coin}] + 1 = {dp[i-coin]} + 1 = {new_count}")
                else:
                    print(f"  Coin {coin}: dp[{i-coin}] + 1 = {dp[i-coin]} + 1 = {new_count} (not better)")
        
        if dp[i] == float('inf'):
            print(f"  dp[{i}] = ∞ (impossible)")
        else:
            print(f"  dp[{i}] = {dp[i]}")
    
    # Reconstruct solution
    if dp[amount] != float('inf'):
        solution = []
        curr = amount
        while curr > 0:
            coin = parent[curr]
            solution.append(coin)
            curr -= coin
        print(f"\nOptimal solution: {sorted(solution)}")
    
    return dp[amount] if dp[amount] != float('inf') else -1

def coin_change_count_ways(coins, amount):
    """
    Count number of ways to make target amount
    Time: O(amount * len(coins)), Space: O(amount)
    """
    # dp[i] = number of ways to make amount i
    dp = [0] * (amount + 1)
    dp[0] = 1  # One way to make amount 0 (use no coins)
    
    # For each coin type
    for coin in coins:
        # Update all amounts that can use this coin
        for i in range(coin, amount + 1):
            dp[i] += dp[i - coin]
    
    return dp[amount]

def coin_change_count_ways_with_trace(coins, amount):
    """
    Show the calculation process for counting ways
    """
    print(f"Counting ways to make amount {amount} with coins {coins}")
    
    dp = [0] * (amount + 1)
    dp[0] = 1
    print(f"dp[0] = 1 (base case: one way to make 0)")
    
    for coin in coins:
        print(f"\nProcessing coin {coin}:")
        old_dp = dp[:]
        
        for i in range(coin, amount + 1):
            dp[i] += dp[i - coin]
            if dp[i] != old_dp[i]:
                print(f"  dp[{i}] = {old_dp[i]} + dp[{i-coin}] = {old_dp[i]} + {dp[i-coin]} = {dp[i]}")
        
        print(f"  Current dp: {dp}")
    
    return dp[amount]

# Important: Order matters in counting ways!
def demonstrate_order_importance():
    """
    Show why we process coins in outer loop for counting
    """
    coins = [1, 2]
    amount = 3
    
    print("Method 1: Coins in outer loop (correct for counting combinations)")
    result1 = coin_change_count_ways(coins, amount)
    print(f"Result: {result1}")
    
    print("\nMethod 2: Amount in outer loop (counts permutations)")
    dp = [0] * (amount + 1)
    dp[0] = 1
    
    for i in range(1, amount + 1):
        for coin in coins:
            if coin <= i:
                dp[i] += dp[i - coin]
    
    print(f"Result: {dp[amount]}")
    print("This counts [1,2] and [2,1] as different ways!")

def coin_change_limited_coins(coins, counts, amount):
    """
    Coin change where each coin type has limited quantity
    This becomes a bounded knapsack problem
    
    Args:
        coins: coin denominations
        counts: maximum count for each coin type
        amount: target amount
    """
    dp = [float('inf')] * (amount + 1)
    dp[0] = 0
    
    for i in range(len(coins)):
        coin = coins[i]
        max_count = counts[i]
        
        # Process in reverse to avoid using same coin multiple times
        for amt in range(amount, coin - 1, -1):
            for count in range(1, min(max_count, amt // coin) + 1):
                if dp[amt - count * coin] != float('inf'):
                    dp[amt] = min(dp[amt], dp[amt - count * coin] + count)
    
    return dp[amount] if dp[amount] != float('inf') else -1

def coin_change_limited_optimized(coins, counts, amount):
    """
    Optimized version using multiple knapsack approach
    """
    # Convert to 0/1 knapsack by expanding coins
    expanded_coins = []
    for i, coin in enumerate(coins):
        count = counts[i]
        k = 1
        while k <= count:
            expanded_coins.append(coin * k)
            count -= k
            k *= 2
        if count > 0:
            expanded_coins.append(coin * count)
    
    # Now solve as 0/1 knapsack
    dp = [float('inf')] * (amount + 1)
    dp[0] = 0
    
    for coin_value in expanded_coins:
        for amt in range(amount, coin_value - 1, -1):
            if dp[amt - coin_value] != float('inf'):
                dp[amt] = min(dp[amt], dp[amt - coin_value] + 1)
    
    return dp[amount] if dp[amount] != float('inf') else -1

def max_amount_with_coins(coins, max_coins):
    """
    Find maximum amount that can be made with at most max_coins coins
    
    This is the "reverse" coin change problem
    """
    # dp[i] = maximum amount possible with exactly i coins
    dp = [0] * (max_coins + 1)
    
    for num_coins in range(1, max_coins + 1):
        for coin in coins:
            dp[num_coins] = max(dp[num_coins], dp[num_coins - 1] + coin)
    
    return dp[max_coins]

def min_coins_for_range(coins, min_amount, max_amount):
    """
    Find minimum coins needed for any amount in range [min_amount, max_amount]
    """
    dp = [float('inf')] * (max_amount + 1)
    dp[0] = 0
    
    for i in range(1, max_amount + 1):
        for coin in coins:
            if coin <= i:
                dp[i] = min(dp[i], dp[i - coin] + 1)
    
    # Find minimum in the range
    result = float('inf')
    for amt in range(min_amount, max_amount + 1):
        if dp[amt] != float('inf'):
            result = min(result, dp[amt])
    
    return result if result != float('inf') else -1

def unbounded_knapsack(weights, values, capacity):
    """
    Classic unbounded knapsack problem
    Each item can be used unlimited times
    
    Args:
        weights: weight of each item type
        values: value of each item type  
        capacity: maximum weight capacity
    
    Returns:
        maximum value achievable
    """
    dp = [0] * (capacity + 1)
    
    for cap in range(1, capacity + 1):
        for i in range(len(weights)):
            if weights[i] <= cap:
                dp[cap] = max(dp[cap], dp[cap - weights[i]] + values[i])
    
    return dp[capacity]

def unbounded_knapsack_with_count(weights, values, capacity):
    """
    Unbounded knapsack that also returns item counts
    """
    dp = [0] * (capacity + 1)
    items = [[] for _ in range(capacity + 1)]
    
    for cap in range(1, capacity + 1):
        for i in range(len(weights)):
            if weights[i] <= cap:
                new_value = dp[cap - weights[i]] + values[i]
                if new_value > dp[cap]:
                    dp[cap] = new_value
                    items[cap] = items[cap - weights[i]] + [i]
    
    return dp[capacity], items[capacity]

# Coin change as special case of unbounded knapsack
def coin_change_as_knapsack(coins, amount):
    """
    Solve coin change using unbounded knapsack framework
    """
    # For minimum coins: minimize negative count (maximize negative of count)
    weights = coins  # coin values are weights
    values = [-1] * len(coins)  # each coin has "value" -1
    
    # Solve knapsack
    dp = [0] * (amount + 1)
    dp[0] = 0
    
    for amt in range(1, amount + 1):
        dp[amt] = -float('inf')  # Start with impossible
        for i, coin in enumerate(coins):
            if coin <= amt and dp[amt - coin] != -float('inf'):
                dp[amt] = max(dp[amt], dp[amt - coin] - 1)
    
    return -dp[amount] if dp[amount] != -float('inf') else -1

def real_currency_examples():
    """
    Demonstrate coin change with real-world currency systems
    """
    
    # US currency system (canonical - greedy works)
    us_coins = [1, 5, 10, 25]  # penny, nickel, dime, quarter
    
    # European-style system (non-canonical in some cases)
    euro_coins = [1, 2, 5, 10, 20, 50]
    
    # Hypothetical system that breaks greedy
    tricky_coins = [1, 3, 4]
    
    def test_currency_system(coins, system_name, test_amounts):
        print(f"\n{system_name}: {coins}")
        print("-" * 50)
        
        for amount in test_amounts:
            min_coins = coin_change_min_coins(coins, amount)
            ways = coin_change_count_ways(coins, amount)
            
            # Compare with greedy (if possible)
            greedy_coins = greedy_coin_change(coins, amount)
            greedy_status = "✓" if greedy_coins == min_coins else "✗"
            
            print(f"Amount {amount:2d}: {min_coins:2d} coins, {ways:3d} ways, "
                  f"greedy: {greedy_coins:2d} {greedy_status}")
    
    def greedy_coin_change(coins, amount):
        """Greedy algorithm for comparison"""
        if amount == 0:
            return 0
        
        coins_sorted = sorted(coins, reverse=True)
        count = 0
        
        for coin in coins_sorted:
            count += amount // coin
            amount %= coin
        
        return count if amount == 0 else float('inf')
    
    test_amounts = [6, 11, 15, 23, 30, 43]
    
    test_currency_system(us_coins, "US Currency", test_amounts)
    test_currency_system(euro_coins, "Euro Currency", test_amounts)
    test_currency_system(tricky_coins, "Tricky System", test_amounts[:4])

real_currency_examples()

def coin_change_variations():
    """
    Demonstrate different optimization objectives
    """
    
    def min_coins_max_value(coins, coin_values, amount):
        """
        Among all ways to make amount with minimum coins,
        find the one with maximum total coin value
        """
        # First find minimum coins needed
        min_dp = [float('inf')] * (amount + 1)
        min_dp[0] = 0
        
        for i in range(1, amount + 1):
            for coin in coins:
                if coin <= i:
                    min_dp[i] = min(min_dp[i], min_dp[i - coin] + 1)
        
        if min_dp[amount] == float('inf'):
            return -1, []
        
        min_coins_needed = min_dp[amount]
        
        # Now find maximum value using exactly min_coins_needed coins
        # dp[i][j] = maximum value using exactly j coins for amount i
        dp = [[-1] * (min_coins_needed + 1) for _ in range(amount + 1)]
        dp[0][0] = 0
        
        for i in range(amount + 1):
            for j in range(min_coins_needed + 1):
                if dp[i][j] == -1:
                    continue
                
                for k, coin in enumerate(coins):
                    if i + coin <= amount and j + 1 <= min_coins_needed:
                        dp[i + coin][j + 1] = max(dp[i + coin][j + 1], 
                                                 dp[i][j] + coin_values[k])
        
        return dp[amount][min_coins_needed]
    
    def coin_change_with_preference(coins, preferences, amount):
        """
        Coin change where we prefer certain coins over others
        """
        # dp[i] = (min_coins, preference_score) for amount i
        dp = [(float('inf'), 0)] * (amount + 1)
        dp[0] = (0, 0)
        
        for i in range(1, amount + 1):
            for j, coin in enumerate(coins):
                if coin <= i:
                    prev_coins, prev_pref = dp[i - coin]
                    if prev_coins != float('inf'):
                        new_coins = prev_coins + 1
                        new_pref = prev_pref + preferences[j]
                        
                        # Update if fewer coins, or same coins but better preference
                        if (new_coins < dp[i][0] or 
                            (new_coins == dp[i][0] and new_pref > dp[i][1])):
                            dp[i] = (new_coins, new_pref)
        
        return dp[amount]
    
    # Test examples
    coins = [1, 5, 10, 25]
    coin_values = [1, 5, 10, 25]  # actual values
    preferences = [1, 3, 2, 4]    # preference scores
    
    amount = 30
    
    basic_result = coin_change_min_coins(coins, amount)
    value_result = min_coins_max_value(coins, coin_values, amount)
    pref_result = coin_change_with_preference(coins, preferences, amount)
    
    print(f"Amount: {amount}")
    print(f"Minimum coins: {basic_result}")
    print(f"Max value with min coins: {value_result}")
    print(f"With preferences: {pref_result[0]} coins, score {pref_result[1]}")

coin_change_variations()

def practical_applications():
    """
    Show how coin change patterns apply to other domains
    """
    
    def task_scheduling_problem():
        """
        Schedule tasks where each task type takes fixed time
        Minimize number of task switches to complete total work
        """
        task_durations = [1, 3, 5, 8]  # minutes per task type
        total_time_needed = 20
        
        # This is coin change: minimize "coins" (task switches)
        min_tasks = coin_change_min_coins(task_durations, total_time_needed)
        ways = coin_change_count_ways(task_durations, total_time_needed)
        
        print(f"Task Scheduling:")
        print(f"  Total time needed: {total_time_needed} minutes")
        print(f"  Task types: {task_durations} minutes each")
        print(f"  Minimum task switches: {min_tasks}")
        print(f"  Number of ways: {ways}")
    
    def resource_allocation_problem():
        """
        Allocate resources where each allocation unit has fixed benefit
        Maximize benefit within resource constraints
        """
        allocation_costs = [2, 3, 5, 7]    # resource cost per unit
        allocation_benefits = [3, 5, 8, 12] # benefit per unit
        budget = 15
        
        # This is unbounded knapsack
        max_benefit = unbounded_knapsack(allocation_costs, allocation_benefits, budget)
        
        print(f"\nResource Allocation:")
        print(f"  Budget: {budget}")
        print(f"  Allocation options (cost, benefit): {list(zip(allocation_costs, allocation_benefits))}")
        print(f"  Maximum benefit: {max_benefit}")
    
    def change_making_with_denominations():
        """
        Design optimal coin denominations for a currency system
        """
        max_amount = 99  # largest amount we want to handle efficiently
        
        # Test different denomination systems
        systems = [
            [1, 5, 10, 25],           # US system
            [1, 2, 5, 10, 20, 50],    # Euro-style
            [1, 3, 9, 27],            # Powers of 3
            [1, 4, 5]                 # Non-canonical example
        ]
        
        print(f"\nCurrency System Analysis (for amounts 1-{max_amount}):")
        
        for system in systems:
            total_coins = 0
            impossible_count = 0
            
            for amount in range(1, max_amount + 1):
                coins_needed = coin_change_min_coins(system, amount)
                if coins_needed == -1:
                    impossible_count += 1
                else:
                    total_coins += coins_needed
            
            avg_coins = total_coins / (max_amount - impossible_count) if impossible_count < max_amount else float('inf')
            
            print(f"  {system}: avg {avg_coins:.2f} coins, {impossible_count} impossible amounts")
    
    task_scheduling_problem()
    resource_allocation_problem()
    change_making_with_denominations()

practical_applications()

def solve_coin_problems(coins, amount):
    """
    Solve both minimum coins and counting ways for given amount
    
    Returns:
        tuple: (min_coins, optimal_path, num_ways)
    """
    # Your implementation here
    pass

def reconstruct_coin_path(coins, amount):
    """
    Return the actual coins used in optimal solution
    """
    # Your implementation here
    pass

# Test cases
coins = [2, 3, 5]
test_amounts = [1, 4, 6, 8, 11, 15]

for amount in test_amounts:
    min_coins, path, ways = solve_coin_problems(coins, amount)
    print(f"Amount {amount}: {min_coins} coins {path}, {ways} ways")

def solve_coin_problems(coins, amount):
    # Minimum coins solution
    dp_min = [float('inf')] * (amount + 1)
    dp_min[0] = 0
    parent = [-1] * (amount + 1)
    
    for i in range(1, amount + 1):
        for coin in coins:
            if coin <= i and dp_min[i - coin] + 1 < dp_min[i]:
                dp_min[i] = dp_min[i - coin] + 1
                parent[i] = coin
    
    # Reconstruct path
    path = []
    if dp_min[amount] != float('inf'):
        curr = amount
        while curr > 0:
            coin = parent[curr]
            path.append(coin)
            curr -= coin
    
    # Count ways
    dp_ways = [0] * (amount + 1)
    dp_ways[0] = 1
    
    for coin in coins:
        for i in range(coin, amount + 1):
            dp_ways[i] += dp_ways[i - coin]
    
    min_coins = dp_min[amount] if dp_min[amount] != float('inf') else -1
    return min_coins, sorted(path), dp_ways[amount]

# coins = [1, 5, 10], fees = [0.1, 0.2, 0.3], amount = 11
# Option 1: 11 coins of 1 = 11 + 11*0.1 = 12.1 total cost
# Option 2: 1 coin of 10 + 1 coin of 1 = 11 + 0.3 + 0.1 = 11.4 total cost
# Option 1 is better despite using more coins!

def coin_change_with_fees(coins, fees, amount):
    """
    Find minimum total cost including transaction fees
    
    Args:
        coins: coin denominations
        fees: transaction fee for each coin type
        amount: target amount
    
    Returns:
        minimum total cost, or -1 if impossible
    """
    # Your implementation here
    pass

def analyze_fee_impact(coins, fees, amount):
    """
    Compare solutions with and without fees
    """
    # Your implementation here
    pass

def debug_coin_change():
    """
    Debugging techniques for coin change problems
    """
    
    def validate_coin_change_solution(coins, amount, result):
        """Validate solution by reconstructing and checking"""
        if result == -1:
            # Verify it's actually impossible
            try:
                brute_force_result = coin_change_brute_force(coins, amount, 0)
                assert brute_force_result == -1, "Solution says impossible but brute force found answer"
            except RecursionError:
                print("Amount too large for brute force validation")
            return True
        
        # For valid solutions, verify optimality with smaller examples
        if amount <= 20:  # Only for small amounts
            brute_force_result = coin_change_brute_force(coins, amount, 0)
            assert result == brute_force_result, f"DP: {result}, Brute force: {brute_force_result}"
        
        return True
    
    def coin_change_brute_force(coins, amount, depth):
        """Brute force solution for validation (only for small inputs)"""
        if depth > 20:  # Prevent infinite recursion
            return float('inf')
        
        if amount == 0:
            return 0
        if amount < 0:
            return float('inf')
        
        min_coins = float('inf')
        for coin in coins:
            result = coin_change_brute_force(coins, amount - coin, depth + 1)
            if result != float('inf'):
                min_coins = min(min_coins, result + 1)
        
        return min_coins if min_coins != float('inf') else -1
    
    def trace_dp_table_2d(coins, amount):
        """Create 2D visualization of DP process"""
        print(f"DP Table for coins {coins}, amount {amount}")
        print("Amount:", end="")
        for i in range(amount + 1):
            print(f"{i:4d}", end="")
        print()
        
        dp = [float('inf')] * (amount + 1)
        dp[0] = 0
        
        print(f"Init :   0", end="")
        for i in range(1, amount + 1):
            print("  ∞", end="")
        print()
        
        for coin in coins:
            old_dp = dp[:]
            for i in range(coin, amount + 1):
                if dp[i - coin] != float('inf'):
                    dp[i] = min(dp[i], dp[i - coin] + 1)
            
            print(f"Coin{coin:2d}:", end="")
            for i in range(amount + 1):
                if dp[i] == float('inf'):
                    print("  ∞", end="")
                else:
                    print(f"{dp[i]:4d}", end="")
            print()

# Test debugging functions
debug_coin_change()

def optimize_coin_change():
    """
    Performance optimization techniques
    """
    
    def space_optimized_counting(coins, amount):
        """
        Space-optimized version for counting ways
        Can reduce space from O(amount) to O(max_coin)
        """
        # Standard approach needs O(amount) space
        # But we can optimize if we process carefully
        
        dp = [0] * (amount + 1)
        dp[0] = 1
        
        for coin in coins:
            # Only need to update from coin to amount
            for i in range(coin, amount + 1):
                dp[i] += dp[i - coin]
        
        return dp[amount]
    
    def early_termination_optimization(coins, amount):
        """
        Optimization techniques for early termination
        """
        # Sort coins in descending order for greedy pruning
        coins_sorted = sorted(coins, reverse=True)
        
        dp = [float('inf')] * (amount + 1)
        dp[0] = 0
        
        for i in range(1, amount + 1):
            for coin in coins_sorted:
                if coin > i:
                    continue  # Skip coins larger than current amount
                
                if dp[i - coin] != float('inf'):
                    dp[i] = min(dp[i], dp[i - coin] + 1)
                    
                    # Early termination: if we found a 1-coin solution, it's optimal
                    if dp[i] == 1:
                        break
        
        return dp[amount] if dp[amount] != float('inf') else -1
    
    def mathematical_optimization(coins, amount):
        """
        Mathematical optimizations for special cases
        """
        # Special case: if 1 is in coins, solution always exists
        if 1 in coins:
            dp = [float('inf')] * (amount + 1)
            dp[0] = 0
            
            for coin in coins:
                for i in range(coin, amount + 1):
                    dp[i] = min(dp[i], dp[i - coin] + 1)
            
            return dp[amount]
        
        # Special case: all coins are multiples of smallest coin
        gcd_val = coins[0]
        for coin in coins[1:]:
            gcd_val = math.gcd(gcd_val, coin)
        
        if amount % gcd_val != 0:
            return -1  # Impossible
        
        # Scale down the problem
        scaled_coins = [coin // gcd_val for coin in coins]
        scaled_amount = amount // gcd_val
        
        return coin_change_min_coins(scaled_coins, scaled_amount)

import math
optimize_coin_change()

def professional_coin_change_framework():
    """
    Professional framework for coin change problems
    """
    
    class CoinChangeManager:
        def __init__(self, coins, optimization_level=1):
            self.coins = sorted(coins)
            self.optimization_level = optimization_level
            self.cache = {}
        
        def min_coins(self, amount, use_cache=True):
            """Find minimum coins with caching support"""
            if use_cache and amount in self.cache:
                return self.cache[amount]
            
            result = self._solve_min_coins(amount)
            
            if use_cache:
                self.cache[amount] = result
            
            return result
        
        def _solve_min_coins(self, amount):
            """Internal implementation with optimizations"""
            if amount == 0:
                return 0
            
            # Optimization: check if amount is smaller than smallest coin
            if amount < self.coins[0]:
                return -1
            
            dp = [float('inf')] * (amount + 1)
            dp[0] = 0
            
            for i in range(1, amount + 1):
                for coin in self.coins:
                    if coin > i:
                        break  # Coins are sorted, no need to check larger ones
                    
                    if dp[i - coin] != float('inf'):
                        dp[i] = min(dp[i], dp[i - coin] + 1)
            
            return dp[amount] if dp[amount] != float('inf') else -1
        
        def count_ways(self, amount):
            """Count ways to make amount"""
            dp = [0] * (amount + 1)
            dp[0] = 1
            
            for coin in self.coins:
                for i in range(coin, amount + 1):
                    dp[i] += dp[i - coin]
            
            return dp[amount]
        
        def analyze_system(self, max_amount=100):
            """Analyze the coin system efficiency"""
            total_coins = 0
            impossible = 0
            
            for amount in range(1, max_amount + 1):
                coins_needed = self.min_coins(amount)
                if coins_needed == -1:
                    impossible += 1
                else:
                    total_coins += coins_needed
            
            possible_amounts = max_amount - impossible
            avg_coins = total_coins / possible_amounts if possible_amounts > 0 else float('inf')
            
            return {
                'average_coins': avg_coins,
                'impossible_amounts': impossible,
                'efficiency_score': possible_amounts / max_amount
            }

# Usage example
manager = CoinChangeManager([1, 5, 10, 25])
print(f"Min coins for 67: {manager.min_coins(67)}")
print(f"Ways to make 30: {manager.count_ways(30)}")
analysis = manager.analyze_system()
print(f"System analysis: {analysis}")

def coin_change_best_practices():
    """
    Best practices for coin change implementations
    """
    
    practices = {
        "Input Validation": [
            "Check for empty coin arrays",
            "Validate positive amounts and coin values",
            "Handle edge case of amount = 0",
            "Sort coins for optimization opportunities"
        ],
        
        "Algorithm Choice": [
            "Use DP for non-canonical coin systems",
            "Consider greedy for canonical systems with validation",
            "Choose appropriate space-time tradeoffs",
            "Implement both counting and optimization versions"
        ],
        
        "Error Handling": [
            "Return consistent values for impossible cases",
            "Handle integer overflow for large amounts",
            "Validate results when possible",
            "Provide meaningful error messages"
        ],
        
        "Performance": [
            "Cache results for repeated queries",
            "Use space optimization when appropriate",
            "Consider mathematical optimizations for special cases",
            "Profile code for large-scale applications"
        ]
    }
    
    for category, tips in practices.items():
        print(f"\n{category}:")
        for tip in tips:
            print(f"  • {tip}")

coin_change_best_practices()