Coin Change

The problem is asking for the minimum number of coins needed to make a particular amount. If it is not possible to make the amount with the given coins, the function should return -1. This is a classic dynamic programming problem, and we can solve it by using bottom-up dynamic programming.

Here’s the step-by-step explanation of the code:

Initialize an Array: Create an array dp of length amount + 1 and initialize it with a large value (e.g., float('inf')), except for dp[0], which should be 0. The array will be used to store the minimum number of coins needed for each amount from 0 to amount.
Iterate Through Coins and Amounts: For each coin, iterate through the amounts from the coin’s value up to the target amount. For each amount, update dp[amount] by comparing its current value with the value of dp[amount - coin] + 1. This checks if using the current coin would result in a smaller number of coins for the current amount.
Check the Result: After the iterations, check if dp[amount] is still float('inf'). If it is, that means it’s not possible to make the amount with the given coins, and the function should return -1. Otherwise, return dp[amount].

Here’s the code:

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class Solution:
    def coinChange(self, coins: List[int], amount: int) -> int:
        # Initialize dp with 'infinity' for all values except dp[0]
        dp = [float('inf')] * (amount + 1)
        dp[0] = 0

        # Iterate through the coins and amounts to fill dp
        for coin in coins:
            for a in range(coin, amount + 1):
                dp[a] = min(dp[a], dp[a - coin] + 1)

        # Return the result for the given amount or -1 if not possible
        return dp[amount] if dp[amount] != float('inf') else -1

The time complexity of this code is O(m * n), where m is the length of the coins array, and n is the value of amount. The space complexity is O(n), where n is the value of amount.

House Robber II

Circular structure, we can use mod operation to go around.
- We don’t have to use a circular data structure
Constraint is different than House Robber

Problem Statement

Look for clues in the problem space.
- Identify which clue is most significant Fewest number of coins => Optimization Problem
- Infinite number ????
The problem statement is given in the context of a real world scenario. We can abstract this into a mathematical problem statement.
[1,2,5], amount = 11 1,1,1,1….1 = 11 1,1, 1, 2 6 - 1s and 1 5
Combinations:
(1,1,1,1,1,1,5) (1,1,1,1,1,1,1,1,1,1,1,1) (5,5,1)
Approaches
- Brute Force Observe the overlapping subproblems T: O(n^2) O(k^n) where k is the coins array size O(3^size) width * depth TLE
- Fewest means we can use what other approach. Greedy
- Dynamic Programming
Classify the Problem
How to optimize. We can memoize. We need to decide the key and value to use for the memoize.
Cashier’s Algorithm Greedy Algorithm
coins = [1,3,4,5], amount = 7 (5, 1, 1) => 3 (3,4) => 2
coins = [1,2,5], amount = 11
Difference Top Down and Bottom Up
coin_change(11) = 1 + coin_change(10) coin_change(10) = 1 + coin_change(9) coin_change(9) = 1 + coin_change(8) coin_change(8) = 1 + coin_change(7) coin_change(7) = 1 + coin_change(6) coin_change(6) = 1 + coin_change(5) coin_change(5) = 1 + coin_change(4) coin_change(4) = 1 + coin_change(3) coin_change(3) = 1 + coin_change(2) coin_change(2) = 1 + coin_change(1) coin_change(1) = 1 + coin_change(0)
11 - 2 = 9 (amount - coin)
Coin 2
coin_change(11) = 1 + coin_change(9) …
Coin 5 coin_change(11) = 1 + coin_change(6) ….
f(amount) = min(1 + f(amount - 1), 1 + f(amount - 2), 1 + f(amount - 5))
Goal: Fewest number of coins to make the change for a given amount
Invariant number of coins or the change you provide - amount = 0
Identify the Subproblems Relate the Subproblems
Base Cases
Recurrence Relation

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public class Solution {
    public int coinChange(int[] coins, int amount) {
        if(amount==0)
            return 0;
        int n = amount+1;
        for(int coin : coins) {
            int curr = 0;
            if (amount >= coin) {
                int next = coinChange(coins, amount-coin);
                if(next >= 0)
                    curr = 1+next;
            }
            if(curr > 0)
                n = Math.min(n,curr);
        }
        int finalCount = (n==amount+1) ? -1 : n;
        return finalCount;
    }
}

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def changer(coins, amount, memo)
  if amount == 0
      return 0
  end
   if memo.key?(amount)
      return memo[amount] 
   end
    n = amount + 1

    coins.each do |coin|
       current = 0
        if amount >= coin
            result = changer(coins, amount-coin, memo)
            if result >= 0
                current = 1 + result
            end
        end
        if current > 0
            n = [n, current].min
        end
    end
    final = (n == amount + 1) ? -1 : n

    memo[amount] = final

    return final
end

# @param {Integer[]} coins
# @param {Integer} amount
# @return {Integer}
def coin_change(coins, amount)
    memo = {}
    changer(coins, amount, memo)
end

The problem “Coin Change” (LeetCode 322) is a classic dynamic programming problem. Before tackling it, solve some simpler problems that introduce you to the concepts of dynamic programming, recursion, and array manipulation.

Climbing Stairs (LeetCode 70): A simpler dynamic programming problem that requires finding the number of distinct ways to reach the top of the staircase.
Best Time to Buy and Sell Stock (LeetCode 121): This problem introduces the concept of finding maximum and minimum in an array, which is useful in dynamic programming problems.
Maximum Subarray (LeetCode 53): A simpler dynamic programming problem that asks for the maximum sum of a contiguous subarray.
House Robber (LeetCode 198): This problem is another introduction to dynamic programming, where you need to maximize the sum of numbers without picking two adjacent numbers.
Fibonacci Number (LeetCode 509): This is a straightforward problem to understand recursion and memoization, two fundamental concepts for dynamic programming.
Partition Equal Subset Sum (LeetCode 416): This problem requires similar dynamic programming skills as “Coin Change”, but is considered a bit simpler.
Unique Paths (LeetCode 62): A problem involving dynamic programming, it helps in understanding the concept of calculating the number of unique paths from top left to bottom right of a grid.
Longest Increasing Subsequence (LeetCode 300): This is a dynamic programming problem where the target is to find length of the longest increasing subsequence.
Palindrome Partitioning II (LeetCode 132): Though a hard problem, it sets a good stage for understanding the need of dynamic programming in certain scenarios.
Minimum Path Sum (LeetCode 64): It’s a problem to find a path from top left to bottom right which minimizes the sum of all numbers along its path.

It’s not necessary to solve these problems in order, but they’re intended to provide a gradient of difficulty leading up to “Coin Change”.

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class Solution:
    def coinChange(self, coins: List[int], amount: int) -> int:        
        dp=[math.inf] * (amount+1)
        dp[0]=0

        for coin in coins:
            for i in range(coin, amount+1):
                if i-coin>=0:
                    dp[i]=min(dp[i], dp[i-coin]+1)
        
        return -1 if dp[-1]==math.inf else dp[-1]

Problem Classification

This problem is a classic example of a problem in the field of Combinatorial Optimization. It can be categorized under Dynamic Programming, which is a method for solving a complex problem by breaking it down into a collection of simpler subproblems, solving each of those subproblems just once, and storing their solutions. The hope is that the solution to a given subproblem will help solve larger, more complex problems.

Problem Components:

Input:
- An array of integers representing coins of different denominations.
- An integer amount representing a total amount of money.
Output:
- Return the minimum number of coins needed to make the target amount.
- If the amount cannot be made by any combination of the coins, return -1.
Constraints:
- There is an infinite number of each kind of coin available.
- The length of the coins array is between 1 and 12, inclusive.
- Each coin’s value is a positive integer less than 2^31.
- The target amount is a non-negative integer less than or equal to 10,000.

We can classify this problem as a Minimum Coin Change problem. It’s a well-known problem in the Computer Science domain that falls under the category of Combinatorial Optimization problems. It has practical applications in many areas, including operations research, algorithmic trading, complexity theory, and cryptography.

The problem is to find the least number of coins (of certain denominations) that add up to a given amount of money. It’s a variation of the Knapsack problem, and is a typical example of a problem that can be solved efficiently using Dynamic Programming techniques.

Given the constraints of the problem, it’s essential to come up with a solution that doesn’t just solve the problem, but does so efficiently, and Dynamic Programming allows us to do exactly that. This way, we avoid solving the same subproblems over and over again, which results in a significant reduction in the time complexity of the solution.

Clarification Questions

What are the clarification questions we can ask about this problem?

Identifying Problem Isomorphism

“Coin Change” can be mapped to “Minimum Path Sum”.

The reasoning is that both problems share the concept of finding an optimal solution given a set of decisions. In “Coin Change”, we’re trying to find the fewest number of coins that sum up to a given amount. This is akin to the problem in “Minimum Path Sum” where we need to find the minimum sum of a path from the top left to bottom right of a grid.

The isomorphism is in the dynamic programming approach required to solve these problems. For “Coin Change”, the state transitions are from the current amount to a smaller amount, while for “Minimum Path Sum”, the state transitions are from one grid cell to another. In both problems, we need to make an optimal choice at each step.

While the fundamental concepts are similar, the specific details of the problems are different. For example, in the “Coin Change” problem, you’re dealing with an array of coins and a target amount, whereas in “Minimum Path Sum”, you’re working with a 2D grid. Hence, this is an approximate isomorphism.

“Coin Change” is more complex due to the possibility of having different coin combinations leading to the same sum, unlike in “Minimum Path Sum”, where the path is more straightforwardly determined by the values in the grid.

Problem Analysis and Key Insights

Analyzing the problem statement for the Minimum Coin Change problem can provide us with several key insights that may guide our approach to a solution:

Optimization Problem: The problem is asking for the “fewest number of coins,” which indicates it is an optimization problem. Specifically, it is a minimization problem. In such problems, dynamic programming is often a useful approach.
Infinite Coins: The problem statement specifies that you have an infinite number of each kind of coin. This suggests that the problem is an unbounded knapsack problem (as opposed to a 0/1 knapsack problem where each item—or coin, in this case—can be used only once).
Possibility of No Solution: The problem allows for a scenario where no amount of coins can add up to the specified amount. In such a case, we are required to return -1. This should be an edge case that our solution must account for.
Order Doesn’t Matter: The problem does not indicate that the order of the coins matters. Choosing coin A before coin B is the same as choosing coin B before coin A. This helps us reduce the number of scenarios we need to consider.
DP State Transition: Each amount from 0 to the target can be reached from a smaller amount. The smaller amount is determined by subtracting one coin’s value. This insight can help form the state transition rule in our dynamic programming approach.
Choice of Coins: From the given set of coins, for each amount, we can choose to either include a coin or exclude it. This gives a hint towards a decision-making process at each step, a characteristic of dynamic programming problems.

Understanding these key insights will allow us to devise an algorithmic solution to the problem that is both correct and efficient.

Problem Boundary

The scope of the “Minimum Coin Change” problem is well-defined, with clear constraints and requirements.

The task is to design an algorithm that takes an array of different coin denominations and a target amount of money, and returns the fewest number of coins needed to make up that amount. The algorithm also needs to handle cases where no combination of coins can make up the target amount, returning -1 in such situations.

The problem is defined for:

The length of the coins array, which is between 1 and 12.
The denomination of each coin, which is a positive integer less than 2^31.
The target amount of money, which is a non-negative integer less than or equal to 10,000.

The constraints ensure that the problem can be solved within reasonable computational resources, while the large range of possible target amounts and coin denominations make it non-trivial and interesting from an algorithmic point of view.

As with many algorithmic problems, the scope doesn’t include building a user interface, connecting to a database, or any other application-specific requirements. The focus is solely on the algorithm that solves the problem.

It’s also worth mentioning that the problem is an instance of a classic problem in computer science and operations research, known as the change-making problem. This problem has many real-world applications, such as in vending machines or in financial domains, making this problem quite relevant and practical.

Establishing the boundaries of a problem means identifying its constraints and limitations, defining what inputs are valid, and how the outputs should be interpreted. For the “Minimum Coin Change” problem, these boundaries can be defined as follows:

Inputs:

An array of integers (coins) representing coins of different denominations. The length of this array is between 1 and 12, inclusive.
An integer (amount) representing a total amount of money. This amount is a non-negative integer less than or equal to 10,000.

Output:

The function should return an integer that represents the fewest number of coins needed to make up the amount. If no combination of coins can make up the amount, the function should return -1.

Constraints:

Each coin’s denomination is a positive integer less than 2^31.
There is an infinite number of each kind of coin available.

Assumptions:

The order of coins does not matter. For example, using a 1 cent coin followed by a 2 cent coin is considered the same as using a 2 cent coin followed by a 1 cent coin.
It’s possible that there is no combination of coins that can add up to the amount.

These boundaries define the scope of the problem, and any solution must respect these inputs, outputs, and constraints. The goal is to develop an efficient algorithm that correctly applies these rules in order to solve the problem.

Distilling the Problem to Its Core Elements

Fundamental Concept:

The fundamental concept that this problem is based on is Dynamic Programming (DP). Specifically, it uses the approach of solving overlapping subproblems and building up the solution iteratively. The nature of the problem is such that a larger problem (making up a higher amount) can be solved using the solutions of smaller problems (making up smaller amounts).

Simplest Description:

You have an infinite supply of coins in certain denominations (like 1 cent, 5 cents, 10 cents, etc.). You need to make up a certain amount of money. The problem is to find out the smallest number of coins you need to do this. If it’s impossible to make up that amount with your coins, you should say so.

Core Problem:

The core problem we are trying to solve is: Given unlimited quantities of coins with certain denominations, what is the minimum number of coins needed to sum up to a specific target amount?

Simplified Problem Statement:

Find the smallest number of coins needed to sum to a target amount. If it’s not possible, return -1.

Key Components:

Different coin denominations
A specific target amount
Calculation of minimum number of coins
Handling of cases where no combination of coins can reach the target

Minimal Set of Operations:

Initialize a large enough array to keep track of the minimum number of coins needed to make every amount up to the target.
Iterate through each coin denomination.
For each coin, update the array values for amounts that can be made up by including this coin. If including this coin leads to fewer coins than previously noted, update the value in the array.
At the end, check the array value for the target amount. If it is still at the initialized value, return -1 as no combination of coins can make up the target. Otherwise, return the value as the minimum number of coins needed.

Visual Model of the Problem

Visualizing the “Minimum Coin Change” problem can be accomplished using the concept of a table or a grid which is a common approach in Dynamic Programming problems. This tabulation method helps to break down the problem into sub-problems and visualize how their solutions contribute to the overall problem’s solution.

Let’s take the example where coins = [1,2,5] and amount = 11.

We can start by creating a table with columns ranging from 0 to 11 (the target amount), and rows for each coin type including a row for ‘0’ coin. Each cell [i][j] in the table will represent the minimum number of coins (of types 0 through i) needed to form the amount represented by column j.

    0 1 2 3 4 5 6 7 8 9 10 11 (amounts)
------------------------------------------
0 | 0 ∞ ∞ ∞ ∞ ∞ ∞ ∞ ∞ ∞ ∞ ∞
1 | 0 1 2 3 4 5 6 7 8 9 10 11
2 | 0 1 1 2 2 3 3 4 4 5 5 6
5 | 0 1 1 2 2 1 2 2 3 3 2 3
------------------------------------------
  (coin types)

Here, ‘∞’ represents an impossible amount (initially, no amount can be formed without using any coin).

We start filling the table row by row. Each cell is filled with the minimum of two values:

The cell directly above it (i.e., we do not use the current coin)
1 plus the cell to the left, the number of columns away equal to the denomination of the current coin (i.e., we use the current coin)

For instance, when we’re filling for coin type ‘2’, for amount ‘3’, we can either not use ‘2’ (take the cell value from the cell directly above, which is ‘3’), or we can use ‘2’ (so we go ‘2’ columns back in the current row and add ‘1’ to it, getting ‘2’). We take the minimum of these two options, which is ‘2’.

This visualization helps understand the iterative process of breaking down the problem into smaller sub-problems, solving each sub-problem, and building up to the solution of the original problem. By the end, the bottom right cell will contain the minimum number of coins needed to make up the target amount, ‘11’ in this case, which is ‘3’. If the target amount is impossible to form with the given coins, the cell will contain ‘∞’.

Problem Restatement

The problem involves a collection of coins that you have in endless supply. Each coin has a particular value, and these values are unique and provided to you in a list. Your task is to figure out the smallest number of coins that you can use from this collection to add up to a given target amount of money.

There are some important conditions to consider. If you find that you can’t reach the target amount exactly using the coins you have, even if you use a large quantity of them, then you should indicate that it’s not possible by returning -1. Also, the target amount might be zero. In this case, you don’t need any coins at all, so you should return 0.

As far as the numbers are concerned, you can have between 1 and 12 different types of coins, and the value of each coin is a positive integer less than 2^31. The target amount is also a non-negative integer and could be as large as 10,000.

Abstract Representation of the Problem

In abstract terms, we can represent this problem as follows:

The problem revolves around a finite set S consisting of unique positive integer values. Each integer value in S represents a distinct element, and we have an infinite supply of each of these elements. The objective is to find the minimum number of elements from S that sum up to a given positive integer target T.

The conditions include:

If no combination of elements from S can sum up to T, we should return a distinct marker that denotes impossibility (in this problem, -1).
If T equals zero, it implies that no elements are needed, hence, we return 0.

The size of the set S can vary between 1 and 12, and each element within S is less than 2^31. The target T is a non-negative integer that can be as large as 10,000.

In summary, this problem is a form of the combinatorial optimization problem where we are tasked with finding the optimal subset from a larger set under given constraints. In this particular instance, we are looking for the smallest subset from S that adds up to T, under the constraints provided.

Terminology

There are several specialized terms and concepts that are key to understanding this problem and its solution:

Dynamic Programming (DP): This is a method for solving complex problems by breaking them down into simpler, overlapping subproblems that can be solved independently. The solutions to these subproblems are stored and reused to build up the solution to the larger problem. In the context of this problem, we use dynamic programming to keep track of the minimum number of coins required to reach each possible amount up to the target.
Tabulation: Tabulation is a common strategy used in dynamic programming where the problem is broken down into subproblems, and a table is used to store the results of these subproblems. This table aids in avoiding the computation of the same subproblem multiple times, thereby improving efficiency. In this problem, we use a table (array) where each index represents a sub-amount, and the value at that index is the minimum number of coins needed to make that sub-amount.
Combinatorial Optimization: This term refers to searching for an optimal solution in a problem where the solution space is combinatorial in nature (i.e., it consists of combinations of choices). In this problem, we are looking for the optimal (i.e., minimum) combination of coins that sums to the target amount.
Denomination: This is a term used to describe the value of a particular coin. In this problem, we have coins of different denominations, and we can use any number of coins of each denomination.

Understanding these terms and concepts is crucial to grasp the problem statement fully and to understand the approach towards the solution.

Problem Simplification and Explanation

Imagine you are packing for a vacation and your suitcase can only carry a certain weight limit (this is the “amount” in our problem). You have various types of items you can pack, like shirts, pants, shoes, etc., and each type has a different weight (these are the “coins” in our problem). You want to pack as efficiently as possible, so you want to reach the weight limit exactly without going over. What is the fewest number of items you can pack to reach that weight limit?

This is similar to our problem of finding the fewest number of coins to make up a certain amount. The coins are the different types of items we can pack, the amount is the weight limit of the suitcase, and we want to find the smallest number of coins (items) that adds up exactly to the amount (weight limit).

In the packing analogy, let’s say we have 3 types of items: shirts (weight 1), pants (weight 2), and shoes (weight 5). And our suitcase can carry up to 11 pounds. We can reach that exact weight by packing 1 pair of shoes (5 pounds), another pair of shoes (5 pounds), and a shirt (1 pound) - a total of 3 items. We can’t reach 11 pounds with fewer items.

If there’s no way to reach the exact weight limit (for instance, if all we have are pants of weight 2 and the weight limit is 3), that’s like saying it’s not possible to make the amount with the coins we have, and we return -1.

If the weight limit of the suitcase is 0, that means we don’t need to pack anything, so we return 0, which is analogous to a target amount of 0 in our problem.

The key concept here is to find the smallest combination of items (coins) that can exactly match the target weight limit (amount). The problem involves concepts from combinatorial optimization (finding the best combination from a set of options) and dynamic programming (breaking down a complex problem into simpler subproblems and combining their solutions).

Constraints

The characteristics and conditions that could be exploited to our advantage in this problem are:

Finite and small size of coin denominations: The problem states that we have between 1 and 12 different types of coins. This small, finite size can be leveraged in our solution since we only need to loop through these coin denominations to check for potential combinations.
Infinite supply of each coin denomination: This implies we can use each coin denomination as many times as we need, allowing us to form different combinations of the coins to reach the target amount.
The target amount is a positive integer: This allows us to use dynamic programming to build up to the target amount. We can start from 0 and incrementally compute the minimum number of coins required to reach every amount up to the target.
The range of the target amount: The problem mentions that the target amount is a non-negative integer up to 10,000. This provides a manageable range that we can work within and further enables the use of dynamic programming, as we can easily store solutions for all amounts up to the maximum of 10,000 in an array.
Each coin has a positive denomination: The fact that each coin has a positive value is important. If we could have coins with a denomination of 0, the problem would become trivial (we could reach any amount with an infinite number of 0-valued coins) or impossible (if only 0-valued coins were available, we could never reach a positive target amount). Additionally, having negative values would add considerable complexity, as we would have to take into account the possibility of reducing a total amount by adding a coin.
Array index as the subproblem (dynamic programming): The index of the dynamic programming array can represent the subproblem (i.e., the amount), which simplifies the task of building the solution incrementally. We can calculate the minimum number of coins for each sub-amount, leading to the solution of the original problem.
Use of a distinct marker for impossibility (-1): If no combination can sum up to the target, we return -1. This allows us to handle “impossible” cases explicitly and to differentiate them from cases where the target amount is simply 0.

These characteristics, coupled with a suitable dynamic programming strategy, can guide us towards an efficient solution for the problem.

Analyzing the constraints of the problem can provide key insights that can help narrow down the possible solution space and guide the algorithm design. Here are some insights gained from the constraints of this problem:

Limited Coin Denominations: The problem statement specifies that there are between 1 and 12 different coin denominations. This is a relatively small range, so algorithms that involve looping through all coin denominations should be feasible in terms of time complexity.
Positive Coin Values: The constraint that each coin value is positive (1 <= coins[i] <= 231 - 1) ensures that adding coins will always increase the total amount, and we cannot have a coin with a value of 0 which could complicate the problem.
Infinite Number of Each Coin: The problem states that we have an infinite number of each kind of coin. This condition suggests that for each coin, we should consider not only the possibility of using it but also the possibility of using it multiple times.
Range of Amount: The total amount that we are trying to sum up to is a non-negative integer less than or equal to 10,000. The fact that this value is non-negative and of a manageable magnitude suggests that we could use a dynamic programming approach where we build up solutions for smaller amounts to solve larger amounts. This is because we can easily store solutions for all amounts up to the maximum in an array.
Unique Outputs for Specific Cases: The constraints specify two unique outputs for special cases: 0 when the amount is 0 (no coins needed) and -1 when it’s impossible to make up the amount with the given coins. This implies that our solution needs to handle these special cases appropriately.

From these insights, we can conclude that a dynamic programming approach that iterates through the coin denominations and builds up an array of minimum coins needed for each amount up to the total amount would be a promising strategy to solve this problem.

Case Analysis

I’ll provide some examples and test cases to cover different aspects of the problem, including edge and boundary conditions. I’ll also analyze each example to illustrate the key constraints and potential pitfalls.

Minimal Inputs (Edge Case)
Input: coins = [1], amount = 0 Output: 0
In this scenario, the amount we need to make is 0. Regardless of the coin denominations, we always need 0 coins to make an amount of 0.
Single Coin, Positive Amount (Edge Case)
Input: coins = [1], amount = 5 Output: 5
In this case, there is only one coin denomination, and the amount is positive. The fewest number of coins needed will simply be the amount itself because the coin denomination is 1.
Multiple Coins, Exact Match (Typical Case)
Input: coins = [1, 2, 5], amount = 10 Output: 2
Here, there are multiple coin denominations, and there’s a coin that exactly matches the target amount. The output should be 1, because we can use the 10 coin to reach the target.
Multiple Coins, No Exact Match (Typical Case)
Input: coins = [2, 5, 10], amount = 7 Output: -1
In this case, no combination of the given coins can make up the target amount, so the output is -1.
Multiple Coins, Amount Not Reachable (Edge Case)
Input: coins = [2, 4, 6], amount = 7 Output: -1
Here, the target amount is odd, but all coin denominations are even. It’s impossible to sum even numbers to get an odd result, so the output is -1.
Large Inputs (Boundary Case)
Input: coins = [1, 5, 10, 25], amount = 10000 Output: 400
This case is a stress test for the solution to check if it can handle the maximum input size. The minimum number of coins is reached by using as many of the largest denomination (25) as possible, then filling in with smaller denominations as needed.

By testing these different scenarios, we can ensure that the solution is robust and handles all possible input configurations. These cases include the smallest and largest possible inputs, typical scenarios, and special cases where the target amount cannot be reached.

Analyzing different cases of the problem helps in recognizing some key insights and patterns that can guide the solution:

Zero amount: No matter the coin denominations provided, if the target amount is 0, the minimum number of coins needed is always 0. This is because we don’t need any coins to form an amount of 0.
Single coin denomination: When only a single coin denomination is available and the target amount is a multiple of that denomination, the minimum number of coins required is the target amount divided by the denomination. If the target amount isn’t a multiple of the coin denomination, it’s impossible to reach the target, and the output should be -1.
Exact match: If the target amount matches one of the coin denominations exactly, the minimum number of coins needed is 1. This is because we can reach the target amount with a single coin.
No exact match, multiple coin denominations: When multiple coin denominations are available and none of them exactly match the target amount, we need to find the best combination of coins that sum to the target. If no combination can reach the target amount, the output should be -1.
Number parity: The parity (even or odd nature) of the target amount and the coin denominations can impact whether it’s possible to reach the target amount. For example, if the target amount is odd but all coin denominations are even, it’s impossible to reach the target.
Large inputs: For large inputs, the solution needs to be efficient in terms of both time and space complexity. Dynamic programming is an effective strategy in these scenarios, as it builds the solution iteratively and can reuse previously computed results.

By understanding these key insights from analyzing different cases, we can design a solution that covers all possible scenarios and edge cases.

Identification of Applicable Theoretical Concepts

The problem of finding the fewest number of coins to make up a given amount is a classic example of a problem that can be approached using dynamic programming. This approach allows us to break down a complex problem into simpler sub-problems and build up an optimal solution.

Here are some of the key concepts involved:

Dynamic Programming (DP): This is a method for solving optimization problems by breaking them down into simpler subproblems and solving each of those subproblems only once, storing their solutions. The main principle behind DP is the idea of overlapping subproblems, i.e., the solution of the main problem depends on the solutions of overlapping subproblems.
In the context of this problem, we can maintain a table where the ith entry stores the minimum number of coins needed to form amount i using the given coin denominations. This table can be filled iteratively, starting from 0 and going up to the target amount.
Greedy algorithm: For some specific cases, a greedy algorithm might seem like a viable option. A greedy algorithm makes the locally optimal choice at each step with the hope that these local choices will lead to a globally optimal solution.
However, in this problem, a greedy algorithm (always choosing the largest possible denomination) does not always produce the optimal solution. For example, if the coin denominations were [1, 3, 4] and the target amount was 6, a greedy algorithm would choose two 3-coin denominations, but the optimal solution is to choose three 2-coin denominations.
Unbounded Knapsack Problem: This problem is a variant of the unbounded knapsack problem. The knapsack problem is a problem in combinatorial optimization: Given a set of items, each with a weight and a value, determine the number of each item to include in a collection so that the total weight is less than or equal to a given limit and the total value is as large as possible.
In this problem, the ‘value’ is the amount we want to reach, the ‘items’ are the coin denominations, and we want to minimize the number of coins (‘weight’).

By recognizing and applying these concepts, we can make the problem more manageable and derive an efficient solution.

Simple Explanation

Imagine you’re playing a game where you need to pay a certain amount to unlock the next level. You have a pouch of coins of different values. The game has a rule: you can use as many coins of each value as you like, but you want to use the smallest number of coins possible.

Now think about this: If you need to pay 15 points to unlock the next level and you have coins of 1, 5, and 10 points, what would you do?

The smart way is to first use a 10-point coin, then a 5-point coin. So you have used just 2 coins to make the payment. You could have also used fifteen 1-point coins, but that would mean using more coins.

Sometimes, you may have to pay an amount that you can’t exactly make with the coins you have. Like if you had to pay 3 points, but you only have 2-point coins. In that case, you just can’t make the payment.

In programming terms, you’re given a list of coin values (like 1, 5, and 10) and an amount to make (like 15). Your task is to find the smallest number of coins you need to use to make that amount. If it’s not possible to make the exact amount with the coins you have, you should return -1.

This problem is about making smart choices to use the fewest coins.

Problem Breakdown and Solution Methodology

To solve this problem, we would approach it as if we were trying to fill a piggy bank to reach a certain amount. Each denomination of coin is like a different size block, and the piggy bank is like a target amount we’re trying to reach.

Here are the detailed steps we would follow to solve the problem:

Setup: First, we need to create an array or list of size amount + 1 to store the fewest number of coins needed to get each possible amount up to the target. We’ll call this dp. Initially, we fill this array with a large value like Infinity to represent that we don’t yet know how to make those amounts. We’ll set dp[0] to 0, as we need zero coins to make the amount zero.
Filling the DP array: For each coin in our coin array, we’ll go through our dp array and try to “fill” each spot using that coin. If the current coin value is less than or equal to the target amount (i.e., the spot in the dp array we’re looking at), we’ll check if using this coin gives us fewer coins than the current value in dp for that amount. If it does, we update the value at that spot in the dp array.
Specifically, if we’re looking at spot i in our dp array and coin c, we’ll check dp[i - c] + 1 (which represents using coin c to make amount i) against the current dp[i]. If dp[i - c] + 1 is less, we update dp[i] with that smaller number.
This process is akin to trying to fill each possible amount with each coin, always keeping track of the fewest coins we can use to do so.
Final Result: After we’ve gone through all our coins and filled up our dp array, we’ll look at dp[amount], which represents the fewest number of coins we need to make up our target amount. If this value is still Infinity, that means we weren’t able to make that amount with the coins we have, and we should return -1. Otherwise, we return dp[amount].

Let’s go through an example. Say we have coins of denominations [1, 2, 5] and our target amount is 11.

We setup our dp array as [0, Infinity, Infinity, Infinity, ..., Infinity] (length = 12).
For coin 1, we can fill every spot, so our dp array becomes [0, 1, 2, 3, ..., 11].
For coin 2, we start at the spot 2 in our dp array and go through to the end, checking if dp[i - 2] + 1 is less than dp[i]. If it is, we update dp[i]. Our dp array now becomes [0, 1, 1, 2, 2, ..., 6].
Finally, for coin 5, we start at spot 5 in our dp array and do the same thing. Our dp array becomes [0, 1, 1, 2, 2, 1, 2, 2, 3, 3, 2, 3].
The final answer is dp[11], which is 3, so we can make the amount 11 with 3 coins.

Note: If the coin array or the target amount change, we would just follow the same steps, filling our dp array with each coin and checking dp[amount] at the end. The complexity of this approach is O(amount * coins.length), as we go through our dp array once for each coin.

Inference of Problem-Solving Approach from the Problem Statement

Sure, here are the key terms or concepts in this problem and how they guide the solution:

Integer Array ‘coins’: This represents the various denominations of coins available. Since the problem specifies we have an infinite number of each kind of coin, this directs us to an approach where we consider each denomination independently and in combination with others.
Integer ‘amount’: This represents the total amount of money we need to make up. It’s the target value we’re trying to reach with our combinations of coins.
Fewest Number of Coins: The objective is to use the minimum number of coins to reach the desired amount. This suggests an optimization problem and hints at a dynamic programming solution where we iteratively build up the solution for each amount up to the target.
Infinite Number of Each Kind of Coin: This tells us that for any denomination, there are unlimited coins available. Hence, we can consider the same denomination again while computing the minimum coins for higher amounts.
Return -1: The problem specifies to return -1 if the amount of money cannot be made up by any combination of the coins. This gives us the base case for our approach when we can’t find a valid combination.
Dynamic Programming: Given the problem’s requirement of finding an optimal solution and the overlapping subproblems (as each amount up to the target can be made up of smaller amounts), a dynamic programming approach is suitable here. We use a DP array to keep track of the minimum number of coins required for each amount up to the target.

By identifying these key terms and understanding their implications, we can map this problem to a classic dynamic programming problem - Coin Change, and hence apply a similar strategy to solve it.

The problem of finding the fewest number of coins that you need to make up a certain amount is a classic example of a problem that can be solved with dynamic programming, and there are a few reasons why we can infer this from the problem statement:

Optimization Problem: The problem asks for the “fewest” number of coins, which implies an optimization problem. Dynamic programming is often used for optimization problems because it allows us to avoid redundant computations and break the problem down into simpler, overlapping subproblems.
Overlapping Subproblems: As we try to build up to the total amount, we will repeatedly calculate the fewest coins needed for smaller amounts. This repetition of subproblems is a key feature that makes dynamic programming an effective strategy.
Optimal Substructure: This problem has an optimal substructure property, which means the optimal solution can be constructed from the optimal solutions of its subproblems. For a given amount, the minimum number of coins needed will be 1 plus the minimum number of coins needed for the amount minus a single coin’s denomination. We solve the subproblems and use their results to solve the larger problem.

These features indicate that dynamic programming is likely to be an effective strategy for solving this problem.

Recognizing these properties can be easier by visualizing the problem in the form of a table or a diagram. Here’s how you can do it:

Optimization Problem: If you’re looking for a minimum or maximum value (like the fewest number of coins), this often indicates an optimization problem.
Overlapping Subproblems: Try solving the problem for smaller inputs and see if you’re solving the same problem multiple times. For this coin change problem, suppose we are trying to make an amount of 6 with coin denominations of 1, 2, and 5. If we calculate the minimum coins needed for each amount from 0 to 6, we will see that to calculate the minimum coins for amount 6, we reuse the solutions of smaller problems (amounts 5, 4, and 1).
Optimal Substructure: Draw a decision tree where each level corresponds to a decision of using a particular coin, and the goal is to reach the target amount with a minimum number of coins. You’ll see that the same subproblems (same amounts) appear in different parts of the tree. Moreover, the optimal solution to a parent problem is built from the optimal solutions to its subproblems.

Here’s how you might construct a dynamic programming table for this problem with coins = [1,2,5] and amount = 6:

Amount	Min Coins Needed
0	0
1	1
2	1
3	2
4	2
5	1
6	2

We fill out each cell in the table by using the formula minCoins[i] = min(minCoins[i], 1 + minCoins[i-coin]) where minCoins[i] represents the minimum number of coins needed to make up amount i, and ‘coin’ is each of the available coins.

By doing this, you can see the properties of dynamic programming problems.

Simple Explanation of the Proof

I’ll explain the rationale and proof of correctness for this dynamic programming solution in simple terms.

The fundamental idea of this problem is to make the given amount using the minimum number of coins. Now, this can be thought of as breaking down the problem into smaller subproblems. Here’s the key idea: the minimum number of coins to make up an amount is 1 plus the minimum number of coins needed to make up the amount that is left after you subtract one coin’s denomination.

Let’s walk through an example to illustrate this. Let’s say we have coin denominations of 1, 2, and 5, and we want to make an amount of 11.

We can start from 0 and work our way up to 11.

To make 0, we need 0 coins.
To make 1, we can use a single 1 coin.
To make 2, we can use a single 2 coin.
To make 3, we can use a 2 coin and a 1 coin.

We continue this way up to 11. When we reach 11, we consider each coin:

If we subtract the 1 coin, we need to make 10, which we know from our previous calculations requires 2 coins (two 5 coins). So this choice would require 1 (the 1 coin) + 2 (to make 10) = 3 coins.
If we subtract the 2 coin, we need to make 9, which requires 3 coins (a 5 coin, a 2 coin, and two 1 coins). So this choice would require 1 (the 2 coin) + 3 (to make 9) = 4 coins.
If we subtract the 5 coin, we need to make 6, which requires 2 coins (a 5 coin and a 1 coin). So this choice would require 1 (the 5 coin) + 2 (to make 6) = 3 coins.

So the minimum number of coins to make 11 is 3 (using a 5 coin and making the remaining 6 with a 5 coin and a 1 coin).

The algorithm is correct because it makes optimal choices at each step (choosing the coin that results in the smallest number of coins needed). This is an application of the principle of optimality in dynamic programming, which says that an optimal solution to a problem incorporates optimal solutions to its subproblems.

This is the essence of the dynamic programming approach for this problem - breaking down a complex problem into simpler subproblems, solving those subproblems, and using those solutions to solve the original problem.

Let’s break down our approach to the coin change problem into more granular steps:

Initialization: First, we need to create a dynamic programming (DP) array to store the minimum number of coins needed to make up each amount up to the target amount. We initialize the DP array with a large number (representing infinity) because we don’t know the minimum number of coins yet. We also set DP[0] to 0 because no coins are needed to make up an amount of 0.
Iterate over all coins: For each coin in our denominations array, we perform the next steps.
Update DP array: For each coin, we update the DP array for all amounts greater than or equal to the current coin’s denomination. To do this, we iterate from the value of the current coin to the target amount (inclusive). For each amount, we update the value at DP[amount] to be the minimum between the current value at DP[amount] and the value at DP[amount - coin] + 1. This is because the minimum number of coins needed to make up the current amount is either the minimum number of coins we’ve found so far or the minimum number of coins needed to make up the amount that’s left after subtracting one coin’s denomination plus one coin.
Check Result: After we’ve iterated over all the coins and updated the DP array, we check the value at DP[amount]. If it’s still the large number we initially set, that means it’s impossible to make up the target amount with the given coin denominations, and we return -1. Otherwise, we return the value at DP[amount] as the minimum number of coins needed.

This approach can be divided into independent parts, such as initializing the DP array, iterating over the coins, and updating the DP array for each coin. These steps don’t rely on each other and can be implemented separately.

The repeatable pattern within our solution is the update of the DP array for each coin. For each coin, we go through the DP array and update the minimum number of coins for each amount. This pattern is repeated for each coin in the coins array.

Solution Approach and Analysis

Let’s consider the coin change problem as a real-life situation first to make it more intuitive. Imagine you’re at a shop and need to give some change to a customer. You have coins of different denominations and want to give the customer the least number of coins possible. What would you do? Naturally, you would try to use the largest denomination coins as much as possible before resorting to smaller ones, right?

This same principle is the core of our approach to the problem. However, it’s not always the optimal solution to just use the largest denomination coin, which is why we will use dynamic programming (DP) to try all possible combinations.

Here’s how we approach the problem:

Initialization: Create an array dp of size amount+1 to keep track of the minimum number of coins needed for every amount up to amount. Initialize dp[0] = 0 because no coins are needed for the amount 0. For all other positions, fill with a large number (amount + 1 or Infinity), as we haven’t yet found a way to form these amounts.
Nested Loop Iteration: We iterate over each coin and for each coin, we iterate from that coin value up to the amount. The idea is to check if considering this coin can help reduce the number of coins used to make up the current amount.
DP Update: In the inner loop, for each amount, we check whether using the current coin leads to a smaller number of coins than what we have currently. This is done by comparing dp[i] (current number of coins for amount i) and dp[i - coin] + 1 (new number of coins for amount i if we use current coin). If dp[i - coin] + 1 is smaller, then we update dp[i].
Result: After the double loop ends, dp[amount] will hold the minimum number of coins needed. If dp[amount] is still the large number, that means there’s no combination of coins that sum up to amount, hence we return -1. Otherwise, we return dp[amount].

Let’s apply this process to an example:

Consider coins = [1,2,5] and amount = 11.

Start with dp = [0, Infinity, Infinity, ..., Infinity] (length 12).
For coin = 1, update dp from index 1 to 11. Now dp = [0, 1, 2, 3, ..., 11].
For coin = 2, update dp from index 2 to 11. dp[2] becomes min(2, 1+1) = 1, dp[3] becomes min(3, 1+1) = 2, and so on. After this loop, dp = [0, 1, 1, 2, 2, ..., 6].
For coin = 5, update dp from index 5 to 11. dp[5] becomes min(3, 1+1) = 1, dp[6] becomes min(3, 1+1) = 2, and so on. After this loop, dp = [0, 1, 1, 2, 2, 1, 2, 3, 3, 2, 2, 3].
Return dp[11] which is 3 (11 = 5 + 5 + 1).

If the amount or the number of coins increase, the computation time would increase since there are more possibilities to check. But the core process remains the same. The problem is scalable to larger inputs due to the nature of dynamic programming.

Identify Invariant

In the context of algorithms, an invariant is a condition or set of conditions that remain unchanged throughout the execution of an algorithm.

In the coin change problem, the invariant is the dp array. Here’s why:

The dp[i] value represents the minimum number of coins needed to make the amount i using any of the available coins. As we iterate over each coin and then over each amount from that coin value up to the amount, we are essentially trying out all possible combinations of coins for each amount.

The invariant property here is that once a value dp[i] is calculated, it represents the optimal (minimum) number of coins needed to form the amount i. No matter how the iteration continues or what other coins we consider, the dp[i] value won’t change unless we find a better (smaller) number of coins to form the same amount i.

This property of dp[i] holds true throughout the execution of the algorithm and ensures we ultimately find the correct, optimal answers.

It’s worth noting that while dp[i] represents the optimal solution for amount = i, the array dp itself is not fully optimal until the algorithm finishes. The optimality of the complete dp array is a result of the optimal substructure property, which is key to dynamic programming solutions.

Identify Loop Invariant

A loop invariant is a condition [among program variables] that is necessarily true immediately before and immediately after each iteration of a loop.

In the case of the coin change problem, the loop invariant within the dynamic programming solution involves our dp array.

Here is the outer loop where we iterate over all the coins:

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for(let coin of coins) { 
  ...
}

And the inner loop where we iterate over all amounts from the coin’s value to the total amount:

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for(let i = coin; i <= amount; i++) { 
  ...
}

The loop invariant here is that at the start and end of each iteration of the inner loop, dp[i] is the minimum number of coins that we can use to make up the amount i using the currently considered coin and all previously considered coins.

This is because, in each iteration of the inner loop, we update dp[i] to be the minimum between its current value and 1+dp[i-coin] (which represents using the current coin). This ensures that after each inner loop iteration, dp[i] is the minimum number of coins needed for amount i, considering all the coins up to the current one.

And at the end of the outer loop, when all coins have been considered, the dp array contains the minimum number of coins needed for all amounts up to amount using any possible combination of coins.

So, the loop invariant here helps maintain the correctness of the algorithm.

Thought Process

Here’s a step-by-step process:

Understanding the problem: The problem involves finding the minimum number of coins required to make up a specified amount, given an array of coin denominations. The problem statement mentions that you may use an infinite number of each coin denomination.
Recognizing the problem type: This is a classic problem in dynamic programming. The key insight here is recognizing the ‘optimal substructure’ in the problem - the idea that the optimal solution to the problem can be constructed from optimal solutions to its subproblems.
Defining the approach: We can use a dynamic programming approach, where we iteratively build up to the solution for the final amount from the solutions for smaller amounts.

Here’s how to code this in Python:

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def coinChange(coins, amount):
    # We initialize our dp array. The length is amount + 1 and initially filled with amount + 1.
    # The reason for amount + 1 is because the maximum number of coins we can use to form an amount
    # is amount itself (using 1-cent coins). So amount + 1 can be seen as a form of "Infinity" for this problem.
    dp = [amount + 1] * (amount + 1)
    
    # Base case: The number of coins to make amount 0 is always 0 since we don't need any coins.
    dp[0] = 0
    
    # Loop over all the amounts from 1 to amount
    for i in range(1, amount + 1):
        # For each amount, we go through all the coins
        for coin in coins:
            # If it is possible to form this amount using this coin
            if coin <= i:
                # Then we update the dp value for this amount. 
                # The new dp value would be minimum of the current dp value and 1 + dp value of the amount that we would have if we take this coin.
                dp[i] = min(dp[i], 1 + dp[i - coin])
                
    # If the final dp value for the amount is still amount + 1, it means we have not found any combination of coins that can sum up to the amount
    # Thus we return -1. Otherwise, we return the dp value for the amount which represents the minimum number of coins that can sum up to the amount.
    return dp[amount] if dp[amount] <= amount else -1

As for cues in the problem statement, the key lies in the phrases “Return the fewest number of coins” and “You may assume that you have an infinite number of each kind of coin”. These hints suggest that a dynamic programming approach could be used to iterate over each possible amount and store the minimum number of coins required to make that amount, building up to the final required amount.

Establishing Preconditions and Postconditions

Parameters:
- The method takes two parameters: coins and amount.
- coins is a list of integers, and amount is an integer.
- coins represents different denominations of coins that are available. amount represents the total amount of money that needs to be made up.
Preconditions:
- Before this method is called, there’s no requirement about the state of the program.
- The input parameters coins and amount must be valid. coins must be a list of positive integers and amount must be a non-negative integer. The length of coins must be between 1 and 12, inclusive. Each coin’s value must be less than 2^31. The amount must be less than or equal to 10^4.
- There’s no specific state that the program needs to be in.
Method Functionality:
- The method is expected to return the fewest number of coins needed to make up the amount. If it’s not possible to make up the amount using the provided coins, it should return -1.
- The method does not modify the input or affect any global state.
Postconditions:
- After the method has been called and has returned, the state of the program or the values of the parameters remain unchanged.
- The return value represents the fewest number of coins needed to make up the amount, or -1 if it’s impossible.
- The method does not have any side effects.
Error Handling:
- If the preconditions are not met (i.e., the input parameters are not valid), the behavior of the method is not defined in the problem statement. In a real-world scenario, it would be appropriate to throw an exception in such cases. For the purpose of this problem, we assume that the inputs are always valid.

Problem Decomposition

Problem Understanding:
- The problem involves figuring out the fewest number of coins needed to make up a given amount using coins of specified denominations. If it’s not possible to make up the amount with the given coins, we need to return -1.
Initial Breakdown:
- The major parts of the problem are:
  1. Iterate over all amounts up to the target amount.
  2. For each amount, compute the minimum number of coins required to make up that amount using the available coins.
Subproblem Refinement:
- For each amount, we further need to:
  1. Iterate over all coin denominations.
  2. For each coin, check if it’s possible to use the coin to make up the current amount by seeing if it’s less than or equal to the current amount.
Task Identification:
- Identifying the minimum number of coins required to make up an amount given the previous amounts is a repeated task.
Task Abstraction:
- The task can be abstracted into a method that, given a list of minimum coins needed for previous amounts and a current amount, identifies the minimum number of coins required to make up the current amount.
Method Naming:
- A possible name for this method could be minimumCoinsNeeded.
Subproblem Interactions:
- Each amount relies on the minimum number of coins required for previous amounts. Therefore, these subproblems need to be solved in increasing order of amounts.
- The solution for each amount is dependent on the solutions for amounts that are smaller by each available coin denomination. Therefore, for each amount, we need to consider each coin denomination.

From Brute Force to Optimal Solution

1. Brute Force Solution

The most straightforward, naive approach to this problem would be to try all possible combinations of coins until we reach the desired amount, or determine that it’s impossible to do so. This can be done with a recursive function that subtracts each coin denomination from the target amount until it’s either 0 (we’ve found a valid combination of coins) or negative (we’ve overspent and need to backtrack). We would keep track of the smallest number of coins used in any valid combination, and that would be our solution.

The main problem with this approach is that it’s very inefficient. This approach has a time complexity of O(n^m), where n is the amount to be made, and m is the number of coin denominations. This is because, in the worst-case scenario, we are trying all combinations of all coins. The space complexity is O(m), as we’re keeping a recursive stack for each coin denomination.

2. Optimization Process

Step 1: Avoid Re-computation with Memoization

The first step to optimizing our brute force solution is to notice that we’re doing a lot of re-computation. Specifically, we’re computing the minimum number of coins needed for the same amount multiple times. We can avoid this by storing our previously computed results in a table (an array or a dictionary), which we’ll refer to as “dp” (short for “dynamic programming”). This technique is called memoization.

With memoization, before we compute the minimum number of coins for an amount, we first check if we’ve already computed it before. If we have, we just use the stored result. If we haven’t, we compute it and then store the result in our dp table for future use. This reduces the time complexity to O(n*m) as we only compute the result for each amount once.

Step 2: Bottom-Up Dynamic Programming

While the memoization approach greatly improves the time complexity, we can further optimize the space complexity. The recursion in the memoization approach results in O(m) space complexity due to the recursive call stack. However, if we fill up our dp table iteratively in a bottom-up manner (i.e., starting from amount 0 up to the target amount), we can avoid recursion and thus bring the space complexity down to O(n).

In the bottom-up dynamic programming approach, we initialize our dp table to have “infinity” at all indices except for the 0th index, which we initialize to 0 (as it takes 0 coins to make 0 amount). Then, for each amount from 1 to the target amount, we subtract each coin denomination from the current amount, and update the dp value for the current amount to be the minimum of its current value and the dp value of the amount we get after subtracting the coin denomination plus one. The “plus one” is for the coin denomination we subtract.

3. Time and Space Complexity

The optimized dynamic programming solution has a time complexity of O(n*m) and a space complexity of O(n), where n is the target amount and m is the number of coin denominations. This is much more efficient than the brute force solution. It’s worth noting that the time complexity is still not optimal in the worst case, as we’re still considering all coin denominations for each amount up to the target amount, but this is the best we can do given that we need to consider all possible combinations of coins.

How to recognize whether we need a 1 D or 2 D data structure for this problem?

Recognizing whether a problem requires a 1-dimensional or 2-dimensional data structure typically comes down to the nature of the problem and the relationships between its elements.

For this problem, we only need to keep track of the minimal number of coins to make up each amount up to our target amount. The information for each amount depends only on smaller amounts, not on combinations of different amounts. Thus, a 1-dimensional data structure (like an array or list) is sufficient to store the minimal number of coins for each amount, where the index represents the amount and the value represents the minimal number of coins.

If we needed to track relationships between pairs of amounts, or if the minimum number of coins for a given amount depended on more than just smaller amounts, then we might need a 2-dimensional data structure. For instance, in problems involving comparisons or interactions between all pairs of elements (such as finding the longest common subsequence of two strings or the shortest distance between all pairs of cities), a 2D array or matrix is often used.

Remember, the choice between 1D and 2D (or even higher dimensional) data structures is not about the number of input variables, but rather about the relationships between these variables and how they affect the state that you’re trying to track.

Code Explanation and Design Decisions

Let’s go over the code line by line, keeping in mind the broader context of the problem we’re trying to solve – finding the fewest number of coins that make up a given amount. Here’s a Python-based solution using dynamic programming:

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def coinChange(coins, amount):
    dp = [0] + [float('inf')] * amount

    for coin in coins:
        for x in range(coin, amount + 1):
            dp[x] = min(dp[x], dp[x - coin] + 1)
            
    return dp[amount] if dp[amount] != float('inf') else -1

Parameters:
- coins: A list of integers representing the different coin denominations available.
- amount: An integer representing the total amount of money we want to form using the fewest number of coins.
- The significance of these parameters is straightforward based on the problem statement.
Primary Loop:
- The outer loop for coin in coins: iterates over all available coin denominations. Each iteration represents exploring what happens when we include a new coin denomination in our possibilities for making change.
- The inner loop for x in range(coin, amount + 1): iterates from the current coin value up to the target amount. Each iteration here represents attempting to form a specific amount using the current and all previous coin denominations.
Conditions:
- The condition dp[x] = min(dp[x], dp[x - coin] + 1) inside the inner loop determines whether it’s better (i.e., uses fewer coins) to make change for the amount x without using the current coin (dp[x]), or by using the current coin (dp[x - coin] + 1).
Modifications:
- The update dp[x] = min(dp[x], dp[x - coin] + 1) modifies the array dp which stores the minimum number of coins needed to make change for each amount up to amount. This is crucial because it keeps track of our current best guess for the minimum number of coins needed.
Invariant:
- The invariant in this case is the dp array. At any point during execution, for any amount x, dp[x] represents the minimum number of coins needed to make change for x using the coin denominations processed so far.
Final Output:
- The final output dp[amount] if dp[amount] != float('inf') else -1 represents the fewest number of coins that can make up the given amount. If it’s impossible to make change for amount, the output is -1, as per the problem’s requirements.

In essence, this algorithm gradually builds up the solution by iteratively considering each coin denomination and updating our best known way to make change for each amount. It maintains an invariant (the dp array) that encapsulates the problem’s state at each step.

Coding Constructs

High-Level Problem-Solving Strategies: The code uses dynamic programming, a technique that solves complex problems by breaking them down into simpler overlapping subproblems. It solves each subproblem only once, stores the result, and reuses it when needed. In the context of this code, the problem of finding the fewest coins to make change is solved by gradually building up the solution for each amount up to the desired amount.
Explaining to a Non-Programmer: Think of it like trying to assemble a specific amount of money using various coin denominations. The goal is to use as few coins as possible. The code works similarly to how you might do it manually: start with the smallest amount you can make with one coin, then see how you can make larger amounts by adding one coin at a time. If you find a better (i.e., smaller) combination of coins for a certain amount, you remember it for future reference.
Logical Elements: The main constructs used in this code, independent of any programming language, are arrays for storing data (to remember the minimum number of coins used for each amount), loops for repeated actions (going through each coin and each amount), and conditions to make decisions (choosing the fewer number of coins for a certain amount).
Algorithmic Approach in Plain English: For each type of coin in the purse, the code looks at every amount up to the target amount. It then asks the question: Is it better to make this amount with the best combination of coins found so far, or is it better to make this amount with the current coin plus the best combination of coins for the remaining amount? It keeps the best answer and continues this process until it has considered every coin for every amount up to the target amount. The best combination for the target amount is the answer.
Key Steps or Operations: The code uses an iterative approach, considering each coin one at a time and seeing how it can form different amounts up to the target amount. For each amount, it keeps track of the fewest coins it can use to make that amount. It does this by comparing the number of coins used for the current amount (excluding the current coin) with the number of coins used for the amount remaining after using the current coin.
Algorithmic Patterns or Strategies: The code follows the pattern of dynamic programming, specifically bottom-up dynamic programming, where solutions to smaller subproblems are calculated first and then used to solve larger subproblems. This pattern is characterized by the creation of a table (in this case, the dp array) that is filled iteratively based on previous computed values, thus avoiding redundant computations.

Language Agnostic Coding Drills

Here are the small units of learning concepts that the above code snippet involves, which are applicable to most modern languages. These are also arranged in the increasing order of difficulty:

Variable Declaration and Initialization: Understanding how to declare and initialize a variable.
List Declaration and Initialization: Learning how to create a list and initializing its values.
Infinite Value Initialization: Understanding how to initialize a variable with an infinite value.
For Loops: Understanding the basic syntax and working of a for loop.
List Indexing: Understanding how to access elements in a list using their indices.
Conditional Statements (If-Else): Understanding how to use if-else conditions to control the flow of code.
Arithmetic Operations: Understanding the usage of basic arithmetic operations such as addition and subtraction.
Comparison Operators: Understanding how to compare two values using operators like <, >, ==, etc.
Nested Loops: Learning to use loops inside other loops.
Updating List Elements: Understanding how to update the value of elements in a list.
Return Statement: Learning how to use the return statement to send back an output from a function.
Dynamic Programming: Understanding the concept of dynamic programming and how to apply it to solve problems efficiently.

You can practice these concepts individually using simple exercises and gradually increase the complexity as you get comfortable with each concept. For instance, you can start by writing simple for loops, then move on to nested loops, and finally try solving simple dynamic programming problems before tackling problems like the one in your code snippet.

Targeted Drills in Python

Based on the breakdown, here are the Python drills targeting each concept:

Variable Declaration and Initialization

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# Declare a variable 'var' and initialize it with 5
var = 5
print(var)

List Declaration and Initialization

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# Declare a list 'lst' and initialize it with integers from 1 to 5
lst = [1, 2, 3, 4, 5]
print(lst)

Infinite Value Initialization

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# Import the math module and initialize a variable 'inf_val' with positive infinity
import math
inf_val = math.inf
print(inf_val)

For Loops

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# Use a for loop to print integers from 1 to 5
for i in range(1, 6):
    print(i)

List Indexing

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# Declare a list 'lst' with integers and print the value at index 2
lst = [1, 2, 3, 4, 5]
print(lst[2])

Conditional Statements (If-Else)

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# Write an if-else statement that prints 'Even' if a variable 'num' is even, else prints 'Odd'
num = 4
if num % 2 == 0:
    print('Even')
else:
    print('Odd')

Arithmetic Operations

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# Declare two variables 'a' and 'b', perform addition, subtraction, and print the results
a = 5
b = 2
print("Addition:", a+b)
print("Subtraction:", a-b)

Comparison Operators

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# Declare two variables 'a' and 'b', compare them using > operator and print the result
a = 5
b = 2
print(a > b)

Nested Loops

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# Use nested loops to print a 2D matrix
for i in range(3):
    for j in range(3):
        print(i, j)

Updating List Elements

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# Declare a list 'lst' and update its first element to 10
lst = [1, 2, 3, 4, 5]
lst[0] = 10
print(lst)

Return Statement

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# Define a function 'add' that takes two parameters and returns their sum
def add(a, b):
    return a + b
print(add(3, 2))

Dynamic Programming

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# Write a function to calculate the nth Fibonacci number using dynamic programming
def fibonacci(n):
    fib = [0, 1] + [0]*(n-1)
    for i in range(2, n+1):
        fib[i] = fib[i-1] + fib[i-2]
    return fib[n]
print(fibonacci(10))

For each drill, the learner can first try to write the code by themselves. If they face any difficulties, they can refer to the provided solutions. They should try to understand each line of the provided solutions to improve their coding skills.

Notes