Combination Sum IV

Identifying Problem Isomorphism

“Combination Sum IV” has an approximate isomorphism “Target Sum”.

In “Combination Sum IV”, you are given an array of distinct integers and a target integer. You’re asked to find the number of possible combinations that add up to the target, where the same number can be chosen from the array any number of times.

In “Target Sum”, you are also given an array of integers and a target integer. However, in this problem, you need to insert a mathematical operator (addition or subtraction) before each number in the array and calculate the sum. Your task is to find the number of different expressions you can create which evaluate to the target.

While the specifics and constraints of the problems are different, both problems share a common theme of exploring combinations of numbers that lead to a certain sum, and require the use of dynamic programming techniques for efficient solutions.

So, understanding the solution for “Combination Sum IV” can help solve “Target Sum”, and vice versa. Even though the problems are not exactly the same, they share enough similarities that strategies and techniques applied to one could be beneficial when dealing with the other.

“Combination Sum” is simpler than “Combination Sum IV”.

Here’s why:

Order of Combinations: In “Combination Sum” (LeetCode problem #39), the order of the numbers in a combination does not matter. This means (2, 2, 3) and (3, 2, 2) are considered the same combination. In “Combination Sum IV” (LeetCode problem #377), the order does matter. This makes “Combination Sum IV” more complex because it requires us to count sequences as well as combinations.
Distinct Numbers: In “Combination Sum”, the numbers in the array can be used multiple times in a combination, but in “Combination Sum IV”, the problem statement specifies that all the elements in the array are distinct. This adds another layer of complexity to the “Combination Sum IV” problem.
Dynamic Programming: Both problems can be solved using a dynamic programming approach. However, the “Combination Sum” problem can be solved using a simpler recursive approach, whereas “Combination Sum IV” generally requires a more advanced understanding of dynamic programming to solve efficiently.
Problem Constraints: The constraints in “Combination Sum IV” are larger than in “Combination Sum”, which requires more efficient solutions.

So, while both problems are challenging and test similar skills, “Combination Sum” is a simpler problem.

10 Prerequisite LeetCode Problems

For Combination Sum IV, the following are a good preparation:

70. Climbing Stairs: This problem is about counting the distinct ways to reach the top of the stairs, given that you can climb 1 or 2 steps at a time. It is relevant because it is a simpler version of counting combinations to reach a target.
198. House Robber: In this problem, you need to calculate the maximum amount of money you can rob given an array of house values. It teaches dynamic programming techniques that are applicable to the Combination Sum IV problem.
62. Unique Paths: This problem involves finding the number of unique paths from the top left to the bottom right of a grid, which is a precursor to counting combinations to reach a target.
322. Coin Change: Here, you need to find the fewest number of coins that you need to make up a certain amount. It introduces the concept of exploring all combinations to reach a target, which is useful for Combination Sum IV.
509. Fibonacci Number: This problem is about finding the Nth Fibonacci number. It is relevant as it’s a straightforward introduction to dynamic programming.
303. Range Sum Query - Immutable: This problem involves creating a data structure to efficiently answer range sum queries, teaching techniques that can be applied to keep track of sums in the Combination Sum IV problem.
121. Best Time to Buy and Sell Stock: This problem involves deciding the best time to buy and sell stock given an array of prices. It’s a good practice for dynamic programming problems where the optimal solution depends on the solution to subproblems.
139. Word Break: This problem requires you to determine if a string can be segmented into dictionary words. The idea of segmenting a larger problem (like a string or a target number) into smaller problems is a useful strategy for the Combination Sum IV problem.
416. Partition Equal Subset Sum: This problem asks if an array can be partitioned into two subsets with equal sum. It is relevant because it involves exploring combinations of numbers, which is a key aspect of the Combination Sum IV problem.
494. Target Sum: In this problem, you’re given a list of numbers and a target, and you need to find the number of ways to use ‘+’ and ‘-’ operations on the numbers to get the target sum. It’s a great warm-up for Combination Sum IV as it also involves counting the number of ways to reach a target sum.

Problem Classification

This problem involves finding all possible combinations of a given set of numbers that add up to a specific target. It falls into the realm of combinatorial algorithms, as it involves counting and combining elements in specific ways. Also, the most efficient way to solve it typically involves using dynamic programming to avoid redundant computation by storing and reusing partial results.

What components:

Input: An array of distinct integers (nums) and a target integer (target).
Output: The number of possible combinations of numbers in nums that add up to target.
Constraints:
- The length of nums is between 1 and 200.
- Each number in nums is a unique integer between 1 and 1000.
- The target is an integer between 1 and 1000.
Task: For each possible combination of numbers in nums, check if the sum of the numbers in the combination equals target. Count and return the number of such combinations.

This is a Counting Problem within the Combinatorial Algorithms domain because the main task involves counting the number of ways something can happen - in this case, counting the number of combinations of numbers that add up to a target value.

It can be considered as an Optimization Problem within the Dynamic Programming domain because dynamic programming techniques are commonly used to solve this type of problem efficiently. Dynamic programming can help optimize the process by breaking the problem down into simpler subproblems and storing the results of these subproblems to avoid redundant computations. In this problem, we can solve smaller instances of the same problem (i.e., find combinations that sum up to smaller targets) and use these results to solve the original problem (i.e., find combinations that sum up to the original target).

Clarification Questions

What should be returned if the target value is less than the smallest value in the array?
Can the target value be zero?
What should be the output if the target cannot be reached by any combination of the integers in the array?
Can the integers in the array be both positive and negative or just positive?
In the case of multiple possible combinations, do we need to return all combinations or just the number of combinations?
Can the same number from the array be used multiple times in a combination?
Can the same combination be used multiple times? For example, if the target is 2, and the array is [1], is the combination (1,1) considered once or twice?
What should be the behavior if the array is empty but the target is not zero?
Can the integers in the array be 0?
What is the expected time complexity for the solution?
Can the order of the array elements affect the output?
Are the numbers in the array guaranteed to be integers?

Problem Analysis and Key Insights

Dynamic Programming: The problem can be broken down into sub-problems, and it seems that it is asking for a way to sum up to a target using the elements of an array. The answer to a larger target can be built from the answers to smaller targets. Therefore, a dynamic programming approach could be a good fit for this problem.
Unbounded Knapsack Problem: This problem is a variant of the Unbounded Knapsack problem where you can take multiple copies of the same item (each number in the array can be used multiple times).
Distinct Combinations: A key insight from the problem is that different sequences are counted as different combinations, i.e., the order matters in this problem. This is a significant point because most combination problems do not count order. For example, the combinations (1,1,2) and (2,1,1) are considered different combinations for this problem.
Unique Elements: The problem statement specifies that all the elements in the nums array are unique. This reduces complexity because we don’t have to handle duplicate numbers.
Integer Constraints: The constraints mentioned in the problem set a clear limit to the input size, which helps to select an appropriate algorithm.
Negative Numbers/Zero: The problem does not mention whether the array can contain negative numbers or zero. In the given constraints, it’s assumed that the elements are positive and greater than zero.
Output Type: The problem asks for the number of combinations, not the combinations themselves. Therefore, we need to devise a solution that counts the combinations, not one that forms them.

Identifying Problem Isomorphism

This problem can be understood in terms of the “Unbounded Knapsack” problem. In the Unbounded Knapsack problem, we are given a set of items, each with a weight and value, and a knapsack with a maximum weight capacity. The task is to maximize the total value of items in the knapsack, and each item can be used multiple times if it doesn’t exceed the maximum capacity.

Analogous to the Unbounded Knapsack problem, in this problem, we are given a set of numbers (items), each with a certain value (equal to its own number). The target number is the maximum capacity of the knapsack, and we are asked to count the number of ways to sum up to the target number (instead of maximizing the value).

The number of ways to achieve a target is analogous to the total value of items in the knapsack. Each number in the given array can be used multiple times, similar to items in the Unbounded Knapsack problem.

In summary, the following mappings form the isomorphism:

Items in the knapsack problem => Numbers in the given array in this problem
Weight of items in the knapsack problem => Value of numbers in this problem (each number has value equal to itself)
Maximum weight capacity of the knapsack => Target number in this problem
Total value of items in the knapsack => Number of ways to sum up to the target in this problem
Maximizing the total value in the knapsack problem => Counting the number of ways to sum up to the target in this problem
Each item can be used multiple times in the knapsack problem => Each number can be used multiple times in this problem

“Coin Change” (LeetCode problem #322) is a typical example of the Unbounded Knapsack problem.

In the “Coin Change” problem, you are given coins of different denominations and a total amount of money. You need to find the fewest number of coins that you need to make up that amount. If that amount of money cannot be made up by any combination of the coins, return -1.

This problem is an instance of the Unbounded Knapsack problem because you can use an unlimited number of each kind of coin (item), each coin has a value (coin denomination), and you are trying to reach an exact target amount (knapsack capacity).

This problem asks for the minimum number of coins rather than the maximum value that can be achieved (as in the classic Unbounded Knapsack problem), but the problem-solving principles are the same.

“Combination Sum IV” (LeetCode problem #377) and “Coin Change” (LeetCode problem #322) are isomorphic to each other as they share similar structures and can be approached with similar problem-solving techniques, specifically dynamic programming.

Here are the subtle differences between these two problems:

In “Combination Sum IV”, we’re asked to calculate the total number of possible combinations that add up to the target, while in “Coin Change”, we’re asked to find the minimum number of coins needed to make the target amount. The objective in “Coin Change” is to minimize, not to count.
While both problems allow using numbers/coins multiple times, the order of numbers matters in “Combination Sum IV” but not in “Coin Change”. In “Coin Change”, (1,2) and (2,1) would be considered the same, but in “Combination Sum IV”, these are counted as separate combinations.

Despite these differences, the “Coin Change” problem can serve as a good foundational understanding for solving the “Combination Sum IV” problem, and vice versa, as they both deal with finding ways to reach a target using given numbers/coins with the option to use each number/coin as many times as needed.

The identification that this problem (“Combination Sum IV”) has a similar structure to the “Coin Change” problem is primarily based on understanding the core problem-solving concept involved, which is dynamic programming, and the scenario in which we’re allowed to use any given number or coin multiple times to reach a target.

Here are the similarities:

Reaching a Target: In both problems, we’re given a target to reach - a sum in “Combination Sum IV” and an amount in “Coin Change”. This target can be reached using any of the given numbers or denominations.
Multiple Uses: In both problems, we can use the same number or coin denomination multiple times. This leads to a very similar dynamic programming state transition.
Dynamic Programming: Both problems can be solved using a dynamic programming approach where we build the solution iteratively. We use a DP array where each index i represents a target, and the value at DP[i] holds the number of combinations for “Combination Sum IV” or the minimum number of coins for “Coin Change” to reach that target.

The key difference between the two problems is the specific goal: “Combination Sum IV” asks for the total number of possible combinations that add up to the target (order matters), whereas “Coin Change” seeks the minimum number of coins needed to make the target amount (order does not matter).

Recognizing these similarities and differences requires understanding the fundamental problem-solving concept involved (dynamic programming in this case) and being familiar with common problem patterns.

Problem Boundary

The scope of this problem, as defined by the problem statement and its constraints, includes:

Input: An array of distinct integers (nums) and a target integer (target). The length of nums is between 1 and 200, inclusive. Each integer in nums is between 1 and 1,000, inclusive. The target integer is between 1 and 1,000, inclusive.
Output: The function should return the number of possible combinations of integers from nums that sum up to target. Note that different sequences are considered different combinations. The number of combinations is guaranteed to fit into a 32-bit integer.
Functionality: The solution needs to accurately compute the total number of combinations of integers from nums that sum up to target, considering different sequences as different combinations.
Performance: Given the constraints, the solution should be able to perform efficiently even in the worst-case scenario (i.e., when nums has a length of 200 and all integers in nums as well as target are at their maximum values).
Correctness: The solution should handle edge cases correctly, such as when nums only contains one integer, or when no combinations exist that sum up to target.

It’s worth noting that the problem’s scope is limited to positive integers only, as per the provided constraints. Additionally, the problem does not require the solution to list all possible combinations; it only requires returning the number of these combinations.

The boundary of the problem is established by its input constraints and expected outputs. Here are the boundary conditions for the problem:

Input Constraints:
- Array Length: The length of the input array nums can range from 1 to 200. So, the boundary cases include situations where the array contains only one element and where it contains 200 elements.
- Array Elements: Each element in the array is a distinct positive integer between 1 and 1,000. Boundary cases could include scenarios where all the numbers are the same (and equal to 1 or 1,000) or where they span the entire range from 1 to 1,000.
- Target: The target integer is between 1 and 1,000. The smallest and largest possible target numbers (1 and 1,000) would be boundary cases.
Output Constraints:
- The function needs to return the number of possible combinations that sum up to the target. This number must fit into a 32-bit integer. A boundary case here would be a scenario where the number of possible combinations is maximal.
Boundary Cases:
- Minimum case: A scenario where nums contains only one element and target is equal to 1.
- Maximum case: A scenario where nums contains 200 elements, all with a value of 1,000, and target is also 1,000.
- Edge case: A scenario where nums contains elements that are all larger than target, making it impossible to reach the target sum.
Performance Constraints:
- The solution should perform well even in the worst-case scenarios defined above.

By exploring and testing these boundary conditions, you can ensure that your solution works correctly and efficiently for all possible inputs, as defined by the problem’s constraints.

Distilling the Problem to Its Core Elements

Fundamental Concept: The fundamental concept this problem is based upon is Dynamic Programming. Specifically, it involves finding the number of ways to reach a certain target sum using a given set of numbers, where each number can be used any number of times and different sequences are counted as different combinations. It requires us to store and reuse past calculation results to build up the answer iteratively.
Simplest Description: Imagine you have a set of building blocks, each with a certain height. You’re trying to build towers of a specific height. The catch is that you can use the same block as many times as you want, and you consider towers of the same height but built in a different order to be different. This problem is about figuring out how many different towers you can build.
Core Problem: The core problem we are trying to solve is counting the number of different combinations of numbers that add up to a specific target. Each number can be used multiple times, and sequences are counted as distinct combinations. A simplified problem statement could be: “Given a list of numbers, in how many distinct ways can you add numbers together (with repetition allowed) to sum up to a target?”
Key Components: The key components of the problem are:
- The list of numbers we have at our disposal (nums)
- The target sum we’re trying to reach (target)
- The rules for forming combinations (any number can be used multiple times, and different sequences count as distinct combinations)
Minimal Set of Operations: The minimal set of operations we need to perform to solve this problem include:
- Initializing a dynamic programming (DP) array to store the number of combinations for each possible sum up to the target.
- Iterating through each number in nums and for each, iterating through the DP array from the current number up to the target. For each index in the DP array, add the number of combinations at the index minus the current number to the number of combinations at the current index.
- Finally, return the value at the target index in the DP array, which represents the total number of combinations that add up to the target.

Visual Model of the Problem

One way to visualize the problem statement for “Combination Sum IV” is to imagine a stairway with target number of stairs. The goal is to reach the top of the stairs, and you can climb the stairs using steps of sizes given in the nums array. The problem asks how many distinct ways you can reach the top, given that you can use the same step size multiple times and the order of the steps matters.

For example, let’s take the input nums = [1,2,3] and target = 4. You can visualize it as a stairway with 4 stairs and you have 3 types of steps (1, 2, 3). Here are the distinct ways to reach the top:

Take four steps of size 1: (1, 1, 1, 1)
Take two steps of size 1, and one step of size 2: (1, 1, 2) and (1, 2, 1) and (2, 1, 1)
Take two steps of size 2: (2, 2)
Take one step of size 1, and one step of size 3: (1, 3) and (3, 1)

This visualization as a stairway helps you understand that this problem is a dynamic programming problem where you’re building up the solution (number of distinct ways to reach the top) iteratively. It also helps illustrate why order matters in this problem - (1, 3) and (3, 1) are two distinct ways to climb the stairs, not one.

Draw a vertical line with 4 evenly spaced horizontal lines intersecting it. These horizontal lines represent the steps, and the vertical line represents the progression from the bottom (0) to the target (4).
Next, write out the numbers in the nums array (1, 2, 3) at the side. These represent the different step sizes you can use.
Starting from the bottom, draw different paths to the top using the step sizes. For instance:
- A path using four 1-sized steps would look like four vertical segments connecting each step.
- A path using two 1-sized steps and one 2-sized step could start with a vertical segment from 0 to 1 (a 1-sized step), another from 1 to 2 (another 1-sized step), and then a longer vertical segment from 2 to 4 (a 2-sized step). Draw other paths for the different orderings of these steps.
- Continue this process for the other combination of steps that sum up to the target.
You will end up with a diagram showing all possible paths (combinations) from the bottom to the top of the stairway. The number of these paths is the answer to the problem.

Remember that each distinct path, even if it uses the same step sizes but in a different order, counts as a separate combination.

This visualization will help illustrate why the problem is a dynamic programming problem - you’re finding all the different ways to reach each step, which allows you to build up the total number of ways to reach the target.

Problem Restatement

Let’s distill this problem down to its key elements:

We’re given an array of unique positive integers, and we also have a target number. We are required to find out the total number of unique ways in which we can add up the numbers from the array to reach the target number.

Here are some important details to keep in mind:

We can use any number from the array as many times as we want. Repetition is allowed.
The order in which we use the numbers matters. So, for instance, if we’re trying to reach a target of 3 and we have the numbers 1 and 2 available, then adding up 1, 1, and 1 (three times 1) is considered a different way from adding up 1 and 2, even though both ways reach the same target.
The length of the given array can be between 1 and 200.
Each number in the array, as well as the target number, will be between 1 and 1,000.
The total number of combinations will always be such that it can fit into a 32-bit integer.
The array contains only unique numbers, so there are no repeated elements in the array.

The goal is to find a function that can take in these inputs (the array of numbers and the target) and return the total number of unique combinations that sum up to the target.

Abstract Representation of the Problem

We have a pool of unique objects, each assigned with a certain positive number. There is also a designated sum total we aim to achieve. The challenge lies in identifying how many unique ways we can pick objects (repeating the same object is permitted) from our pool so that the sum of their assigned numbers equals the designated sum. The sequence of picking matters – even when the same objects are chosen, differing sequences are considered unique.

Here, the pool of objects corresponds to our array of numbers (nums), the assigned positive number of each object is the integer value itself, and the designated sum is the target. We are searching for different ways of picking these objects (where picking the same object multiple times is possible) such that the sum of the objects’ numbers is equal to the target.

This general representation removes specific details about numbers, arrays, and targets, and instead highlights the main structure of the problem and the principal task: identifying the number of ways to pick objects from a pool to total up to a designated sum, where sequence matters and repetition is possible.

By formulating our own general representation, we steer clear of ambiguous terms and express our understanding of the problem.

Terminology

There are a few technical terms that are crucial to understanding this problem and its solution:

Array: An array is a data structure that contains a group of elements. In the context of this problem, the array contains distinct integers that can be used to sum up to the target.
Target: In the context of this problem, target refers to the specific sum that we are trying to achieve by using the integers in the given array.
Dynamic Programming (DP): Dynamic Programming is a method for solving complex problems by breaking them down into simpler subproblems, solving each of those subproblems just once, and storing their solutions. In the context of this problem, we can use a dynamic programming approach to count the number of combinations that sum up to each possible value up to the target, building up to the final solution.
Combination: A combination is a selection of items without considering the order. However, in this problem, different sequences are counted as different combinations, which deviates from the traditional definition of a combination.
32-bit integer: This is a type of integer that can hold 32 bits of data. This is mentioned in the problem to indicate the upper limit of the output value.

Understanding these terms and concepts is crucial for formulating and understanding the solution to the problem.

Problem Simplification and Explanation

Let’s consider an analogy: imagine that you have a toolbox with several different tools (the numbers in the nums array), and you’re trying to assemble a piece of furniture (the target). Each tool has a specific use or value (the integer value of each number). You can use the same tool multiple times, and the order in which you use the tools matters.

Your task is to figure out all the different ways you can use your tools to assemble the furniture. For instance, if you’re trying to assemble a chair with a total value of 4, and you have tools of value 1, 2, and 3, you could:

Use the 1-tool four times.
Use the 1-tool twice and the 2-tool once.
Use the 2-tool twice.
Use the 3-tool once and the 1-tool once.

And so on. The order in which you use the tools matters here. So, using the 1-tool first and then the 2-tool is considered different from using the 2-tool first and then the 1-tool.

The key concept here is figuring out all the different ways to combine the tools to reach your goal, where you can use the same tool multiple times and the order of usage matters. This problem can be solved using dynamic programming: you solve the problem for smaller pieces of furniture first (smaller target values), and use these solutions to figure out how to assemble larger pieces of furniture (larger target values).

Breaking the problem down like this allows us to see the structure of the problem and understand what we’re trying to accomplish, which can be helpful for coming up with a solution.

Constraints

The problem “Combination Sum IV” has several characteristics and constraints that can be leveraged to find an efficient solution:

Distinct integers in nums array: The nums array contains distinct integers. This eliminates the possibility of redundant computations due to repeated numbers.
The range of integers and the target: Both the elements of the nums array and the target are within 1 to 1000. This limited range allows us to use dynamic programming effectively, as we can create a DP table of size 1000 to store intermediate results.
Unlimited use of integers: We can use any number from the nums array as many times as we want. This unlimited usage suggests that we can solve the problem using the concept of unbounded knapsack, which can be efficiently solved by dynamic programming.
Order matters: The problem considers different sequences as different combinations. This leads to a larger solution space and allows us to apply permutation-related techniques rather than just combination-related ones.
Small size of the nums array: The length of the nums array can be between 1 and 200. The relatively small size of the nums array is useful in that it allows us to iterate over the array multiple times without significant performance concerns.
32-bit integer size of the result: The result fits in a 32-bit integer. This implies that while there may be many possible combinations, the count will not exceed a 32-bit integer, which limits the maximum size of the output.

By recognizing these specific characteristics and constraints, we can design a more efficient solution that capitalizes on these conditions.

Analyzing the constraints of the problem can provide some key insights:

Use of Dynamic Programming (DP): The constraint that all elements of nums and target are within the range of 1 to 1000 suggests that a dynamic programming approach could be feasible. Dynamic programming is often suitable for problems where we have a target value, and we need to consider different combinations or permutations of input values to reach that target.
Order Matters: This problem considers different sequences as different combinations. This insight is important because it changes the way we will approach the problem. This is a different scenario from problems where we are considering combinations and the order does not matter. Here we need to find all permutations of nums that sum to target, not just combinations.
Unbounded Knapsack Problem: The problem states that we can use any number from the array as many times as we want. This hints at the problem being similar to the Unbounded Knapsack problem, where we can use the same item multiple times. This insight can guide us towards a solution.
Use of a Bottom-Up Approach: The constraints of the problem suggest that a bottom-up approach could be more efficient. The reason for this is that a bottom-up approach allows us to start with smaller, more manageable problems (i.e., smaller target sums) and use the solutions to these problems to iteratively build up to the solution for the larger problem (i.e., the total target sum).
Size of the Result: The constraint that the result fits in a 32-bit integer informs us about the maximum possible size of the output. This can help us avoid overflow errors.

These insights guide us to use a dynamic programming approach that iterates over the array and the range up to the target, accumulating the count of combinations for each value.

Case Analysis

Let’s consider several scenarios to cover the wider input space and examine different aspects of the problem:

Example 1: Small Input Size (Smallest Target Value and Array Size)

Input: nums = [1], target = 1 Output: 1 Explanation: In this case, we have the smallest possible target value and only one number in nums which is 1. The only combination we can have is using 1 to sum up to 1. So, the answer is 1.

Example 2: No Combinations Possible (Largest Number in nums)

Input: nums = [1000], target = 999 Output: 0 Explanation: Here we have the largest possible number in nums and there is no way to add up to the target. So, the answer is 0.

Example 3: Multiple Combinations with Duplicates (Middle-Range Target Value and Array Size)

Input: nums = [1, 2, 3], target = 4 Output: 7 Explanation: Here, the possible combination ways are (1,1,1,1), (1,1,2), (1,2,1), (1,3), (2,1,1), (2,2), and (3,1). Even though some combinations contain the same numbers, the order is different, and thus they are counted as separate combinations.

Example 4: Multiple Combinations with Large Target (Large Target Value and Array Size)

Input: nums = [1, 2, 3, ..., 200], target = 1000 Output: This would produce a large number of combinations, too large to be practically calculated manually. Explanation: This case tests the ability of the algorithm to handle the maximum size of input arrays and target values as per the problem constraints.

Example 5: Repetitive Combinations (Small Target Value, Large nums)

Input: nums = [1, 2, 3, ..., 200], target = 2 Output: 2 Explanation: Here, the possible combinations are (1,1) and (2). Even though we have a large number of values in nums, the small target limits the number of combinations.

These test cases cover a variety of scenarios, including edge and boundary conditions. Analyzing these test cases helps us understand different aspects of the problem, the role of order, the effect of duplicate numbers, and the impact of the size of nums and target on the number of combinations.

Visualizing these cases involves imagining how each scenario unfolds in terms of combinations.

Example 1: Small Input Size (Smallest Target Value and Array Size)

Here, nums = [1] and target = 1. We only have one number and the target is also one. This can be visualized as a single step towards the target. We can think of it as a single unit block that exactly measures the target distance.

nums: [1]

target: 1
combinations: 1

Visual: [1]

Example 2: No Combinations Possible (Largest Number in nums)

Here, nums = [1000] and target = 999. The target is less than the single number available in nums. We cannot form the target sum. This can be visualized as having a measurement unit that is too large for the target distance.

nums: [1000]

target: 999
combinations: 0

Visual: No combinations available

Example 3: Multiple Combinations with Duplicates (Middle-Range Target Value and Array Size)

Here, nums = [1, 2, 3] and target = 4. We have several numbers in nums that can be combined in various ways to form the target. This can be visualized as various paths to the target with different step sizes.

nums: [1, 2, 3]

target: 4
combinations: 7

Visual: [1,1,1,1], [1,1,2], [1,2,1], [1,3], [2,1,1], [2,2], [3,1]

Example 4: Multiple Combinations with Large Target (Large Target Value and Array Size)

Here, nums = [1, 2, 3, ..., 200] and target = 1000. This scenario provides a vast number of combinations. Visualizing it could be imagining many possible paths to a distant target with various step sizes.

Example 5: Repetitive Combinations (Small Target Value, Large nums)

Here, nums = [1, 2, 3, ..., 200] and target = 2. Though nums has many numbers, the small target limits the combinations. This can be visualized as only a few short paths to a nearby target despite the availability of many step sizes.

nums: [1, 2, 3, ..., 200]

target: 2
combinations: 2

Visual: [1,1], [2]

These visual representations help to conceptualize how the problem’s constraints influence the possible combinations.

Analyzing different cases provides the following key insights:

Role of the target: The value of the target plays a crucial role in determining the number of combinations. If the target is small, even if there are many numbers in the nums array, there will be fewer combinations. Similarly, if the target is larger, there will be more combinations.
Role of the nums array: If the nums array contains larger values that exceed the target, no combinations can be formed to sum to the target. Similarly, if the nums array contains smaller values, especially 1, we can always form the target by adding 1s.
Order of the numbers: In this problem, the order of the numbers in the combination matters. So, even with the same numbers, changing the order creates a new combination. For instance, [1,2] and [2,1] are considered different combinations. This significantly impacts the count of combinations.
Effect of duplicates: Since nums only contains unique numbers, we don’t have to deal with duplicates in nums. However, the same number can be used multiple times in a combination. This feature allows for a wider range of combinations and increases the overall count of combinations.
Handling large inputs: The problem has constraints allowing for a large nums array (up to 200 elements) and a large target value (up to 1000). Any proposed solution needs to handle these larger inputs efficiently.

These insights will help us devise a strategy to tackle the problem effectively and efficiently. In particular, they suggest that dynamic programming can be a suitable approach due to the repetitive sub-problem structure and the need to track multiple possible combinations for each target sum up to the given target.

Identification of Applicable Theoretical Concepts

This problem is a perfect candidate for Dynamic Programming (DP), which is a method for solving complex problems by breaking them down into simpler overlapping subproblems. It is particularly useful when the solution of a problem relies on solutions to smaller instances of the same problem.

The problem of finding the number of combinations that add up to a target can be solved by building up an array (or a DP table) where each index represents a possible target value and the value at each index represents the number of ways to reach that target. This is an application of the bottom-up DP approach.

Here are the key concepts related to this problem:

Permutations and Combinations: The problem is essentially asking for the total number of permutations of nums that can add up to the target. Understanding the differences between permutations and combinations helps solve this problem because different sequences are counted as different combinations.
Unbounded Knapsack Problem: This problem is similar to the Unbounded Knapsack problem, where we can pick an item (number) as many times as we want, and we aim to maximize or minimize a certain attribute (in this case, the number of combinations that sum to the target).
Recursion: The problem can be solved using a recursive approach where we reduce the target by the current number and then solve the sub-problem for the reduced target. However, due to overlapping subproblems, this approach would be inefficient without applying DP to store and reuse the results of already solved subproblems.
Space and Time Complexity Analysis: Given the constraints of the problem (1 <= nums.length <= 200, 1 <= nums[i], target <= 1000), understanding space and time complexity is crucial to designing an efficient solution.

Applying these mathematical and algorithmic concepts can simplify the problem and make it more manageable.

Simple Explanation

Let’s take a fun and simple example.

Imagine you have different types of LEGO blocks, each of a different length. Let’s say you have LEGO blocks of lengths 1, 2, and 3 units. Your task is to build a LEGO tower of a certain height, say 4 units.

Now, you’re asked how many different ways you can build this LEGO tower of height 4. Remember, you can use as many blocks of each type as you want, and the order in which you place the blocks matters. So, if you first use a block of length 2 and then a block of length 1, that’s a different combination than first using a block of length 1 and then a block of length 2.

Let’s see some examples:

You can stack four 1-unit blocks to make a tower of height 4.
You could also stack two 2-unit blocks.
You could stack a 1-unit block, a 2-unit block, and then another 1-unit block.
And so on…

So, the question is like asking: “How many different combinations of LEGO blocks can you use to create a tower of a certain height?” The blocks are like the numbers in our list, and the height of the tower is like our target sum.

In coding terms, we have a list of numbers and a target sum, and we need to find out how many different combinations of those numbers add up to the target sum. The key here is that the same number can be used multiple times, and the order matters – just like how you can use the same type of LEGO block multiple times and the order you stack them matters.

Problem Breakdown and Solution Methodology

Let’s break down the approach to solving this problem.

We can solve this problem using a technique called Dynamic Programming (DP), which is used when the solution to a problem can be found by solving smaller versions of the same problem.

Here’s the step-by-step approach:

Initialization: We start by initializing an array, dp, of size target + 1, filled with zeros. The indices of this array represent all the possible sums up to the target sum, and the value at each index will eventually represent the number of combinations that add up to that sum. We start by setting dp[0] to 1, because there is exactly one way to make the sum 0 - by choosing no numbers.
Build DP Array: Next, we start two nested loops. The outer loop goes through each number in nums and the inner loop goes through all possible sums from nums[i] to target. For each sum, we add the value of dp[sum - nums[i]] to dp[sum]. This effectively counts all combinations that include the number nums[i].
Count Combinations: After building the dp array, the value of dp[target] will represent the number of combinations that add up to the target.

This process is like building a tower with Lego blocks as explained earlier. Each number in nums is like a different type of Lego block, and the target is the desired height of the tower. We’re figuring out how many different ways we can arrange these Lego blocks to reach the desired height.

Example:

Let’s say we have nums = [1, 2, 3] and target = 4.

After initializing, our dp array will look like this: dp = [1, 0, 0, 0, 0].

We start with the number 1 (nums[0]). For each sum from 1 to 4, we add dp[sum - nums[0]] to dp[sum]. After this, our dp array looks like this: dp = [1, 1, 1, 1, 1].

Next, we move on to the number 2 (nums[1]). Now, our dp array looks like this: dp = [1, 1, 2, 2, 3].

Finally, we do the same for the number 3 (nums[2]). Now, our dp array looks like this: dp = [1, 1, 2, 3, 4].

So there are 4 ways to form the target sum 4 using the numbers in nums.

Effect of Problem’s Parameters:

Increasing the target would result in a larger dp array, increasing both the computation and space required.
Increasing the size of nums would result in more loops in our algorithm, increasing the computation.
Having larger numbers in nums could reduce the number of combinations, as we can’t use larger numbers to form smaller target sums.

This approach ensures that all combinations are accounted for and provides an efficient solution for larger inputs, within the problem constraints.

Inference of Problem-Solving Approach from the Problem Statement

Let’s identify the key terms and concepts in this problem:

Array of distinct integers, nums: This is your set of available numbers (like different types of Lego blocks) that can be used to reach the target.
Target integer, target: This is the specific sum (like the height of the Lego tower) that you’re aiming to reach using the numbers in the nums array.
Return the number of possible combinations: This indicates that we are looking for a count of all possible permutations of nums that can add up to target. The word “combinations” in this problem considers different sequences as different combinations, which is more like permutations in terms of combinatorics.
All the elements of nums are unique: This tells us that we don’t have to worry about duplicate numbers in nums.
1 <= target <= 1000: The bounds on the target value indicate that the problem could potentially require handling large numbers, so efficiency will be important.

In terms of diagrams or tables, a good way to visualize this problem is by using a dynamic programming table or array. Each index in the array corresponds to a potential sum (from 0 up to the target), and the value at that index represents the count of combinations that add up to that sum.

Here’s a simple example to illustrate:

nums = [1, 2, 3], target = 4

We create an array (dp) of size target + 1 and initialize it with zeros:

dp = [0, 0, 0, 0, 0]

We set dp[0] to 1 since there’s exactly one way to sum up to 0, which is choosing no numbers.

dp = [1, 0, 0, 0, 0]

Now, for each number in nums, we loop through the dp array from num to target, and for each of these sums, we add dp[sum - num] to dp[sum]. After we’ve gone through all the numbers in nums, our dp array looks like this:

dp = [1, 1, 2, 3, 4]

So, there are 4 different combinations (permutations) of numbers from nums that can sum up to 4, which is the final result.

This dynamic programming array is a form of tabulation that represents the state of our problem at each potential sum up to the target. It serves as an efficient tool to count the combinations (or permutations in this problem’s context) to reach the target.

Several factors suggest that this problem is a good fit for a dynamic programming solution:

Optimal Substructure: The problem can be broken down into smaller, simpler subproblems, and the solution to the overall problem depends on the solutions to these subproblems. In this case, the total number of combinations that add up to a target sum can be found by summing up the number of combinations that add up to smaller sums.
Overlapping Subproblems: The same subproblems are solved repeatedly. For example, when finding the number of combinations for the target sum, we might need to calculate the number of combinations for smaller sums multiple times.
Ordering: The problem involves sequences or combinations of numbers, and the order in which we process the numbers or build up the sum matters. This is a characteristic of many dynamic programming problems.
Tabulation or Memoization Needed: To avoid recalculating the number of combinations for the same sum multiple times, we can store the number of combinations for each sum in a table (an array in this case) as we calculate them. This technique is known as tabulation in dynamic programming.

By identifying these properties in the problem, we can infer that dynamic programming would be a suitable approach to solving this problem. This problem is particularly similar to other famous dynamic programming problems, like the Unbounded Knapsack problem and the Coin Change problem.

Both combinations and permutations are fundamental ways of counting different arrangements of elements within a set, and they’re used frequently in probability theory and combinatorics. However, they’re used in different contexts and they mean different things:

Permutations:

A permutation is an arrangement of all or part of a set of objects, with the order of the objects being important. For example, if you have three numbers {1, 2, 3}, there are six permutations: 123, 132, 213, 231, 312, and 321.

Combinations:

A combination is also an arrangement of all or part of a set of objects, but in this case, the order of the objects doesn’t matter. With the same set of three numbers {1, 2, 3}, there are three combinations if we are choosing two numbers: {1, 2}, {1, 3}, and {2, 3}. Note that {1, 2} and {2, 1} are considered the same combination, because order doesn’t matter.

Similarities:

Both permutations and combinations are methods of counting arrangements within a set. They’re both key concepts in probability and combinatorics and are often used to solve similar types of problems.

Differences:

The key difference lies in whether the order of arrangement matters or not. In permutations, the order is significant ({1, 2, 3} is different from {3, 2, 1}), but in combinations, the order is not significant ({1, 2} is the same as {2, 1}).

In the context of this problem (“Combination Sum IV”), it’s a bit confusing because the problem actually involves permutations (since different sequences are counted as different combinations), not combinations in the traditional sense. It’s a reminder that sometimes problem names can be a bit misleading, and it’s always important to carefully read and understand the problem statement.

If I have a lock the order matters, so to crack the code I have to try all possible ways of numbers. A real-world example for combinations, where the order does not matter, is the lottery.

Suppose you are playing a lottery game where you need to choose 5 unique numbers from a set of 50. The order in which you choose the numbers does not matter. If the lottery result is {7, 14, 23, 30, 45}, you win the jackpot regardless of whether you picked your numbers in the order {14, 45, 7, 30, 23}, {23, 7, 45, 14, 30}, or any other arrangement of the same numbers.

This is a perfect illustration of combinations because you are essentially selecting a subset of numbers from a larger set, and the order in which these numbers are arranged does not impact the outcome. In other words, every unique selection of five numbers from the larger set of 50 is considered a distinct combination.

Simple Explanation of the Proof

Let’s break it down step by step. We are using dynamic programming here. In simpler terms, dynamic programming is just a fancy way of saying that we are solving larger problems by breaking them down into smaller ones and saving the solutions of these smaller problems so that we don’t have to solve them again.

The problem we’re trying to solve here is “how many combinations of numbers from our list can sum up to a specific target?” This is a perfect scenario for dynamic programming, as it involves overlapping subproblems (smaller sums) and the final answer depends on the combination of these smaller problems.

Here’s an explanation of the solution:

Initialize an array dp: Create an array dp of size target + 1 and fill it with zeros. The idea here is that each index of this array represents a possible sum (from 0 to target), and the value at each index is the number of ways we can get that sum from the numbers available to us.
The base case: We set dp[0] to 1, because there is exactly one way to sum up to 0 - that is by choosing no numbers.
Build up the dp array: We start iterating over our nums array and for each number, we iterate over the dp array from the num index to target. For each of these sums, we add dp[sum - num] to dp[sum].
What we’re doing here is saying that “the number of ways to make a sum (sum) is equal to the number of ways we’ve found to make that sum so far (dp[sum]) plus the number of ways to make a sum that’s smaller by ’num’ (dp[sum - num]).”
The reasoning behind this is that if we have found ‘x’ ways to form a sum ‘y’, and now we have a number ’n’, we can form the sum ‘y+n’ in ‘x’ new ways by adding ’n’ to each of the ‘x’ ways.
Get the answer: After we’ve gone through all the numbers, dp[target] will contain the total number of combinations that can add up to the target.

The key to understanding this algorithm is to see that it systematically builds up to the final solution (the target sum) by solving and using the solutions to smaller problems (smaller sums). That’s the core idea behind dynamic programming.

Stepwise Refinement of our approach to solving this problem:

a. Understanding the Problem: Start by making sure you understand the problem correctly. This problem requires us to count all the combinations of numbers from the given array that can add up to the target. The order of numbers is significant, so [1,2] and [2,1] are considered different combinations.

b. Identifying the Approach: Notice that this problem can be solved by using dynamic programming because it has overlapping subproblems and optimal substructures.

c. Designing the Algorithm: Here’s a step-by-step breakdown:

- **Step 1** - Initialize an array `dp` of size `target + 1` and fill it with zeros. The index of this array represents the sum, and the value at each index is the number of ways we can get that sum.

- **Step 2** - Set `dp[0]` to 1 because there's exactly one way to sum up to zero, i.e., by choosing no numbers.

- **Step 3** - For each number `num` in `nums`, iterate from `num` to `target` on `dp` array and for each sum, add `dp[sum - num]` to `dp[sum]`.

- **Step 4** - After going through all numbers, `dp[target]` will contain the total number of combinations that can sum up to target.

Granular, Actionable Steps:
a. Declare and initialize an array dp of size target + 1 with all elements as 0.
b. Set dp[0] as 1.
c. Start a loop from 0 to target (inclusive), we’ll refer to each element as i.
d. For each i, run a nested loop through the nums array. For each number num in nums, if num is less than or equal to i, increment dp[i] by dp[i - num].
e. Finally, dp[target] will be the answer.
Independent Parts of the Problem:
- Calculating the number of ways to reach each sum is an independent sub-problem. The calculation for one sum does not depend on the calculation for any other sum (although it uses their results), so these calculations can theoretically be done independently.
Repeatable Patterns:
- The problem involves repeating the same action (adding a number to a sum and updating the number of combinations) for each sum and each number. This is a repeatable pattern in the solution.
- Another repeatable pattern is that for every new number, we’re repeating the process of updating the counts for all sums that can be formed with this new number.

Solution Approach and Analysis

Let’s dive into the detailed solution to this problem.

Let’s consider an example, nums = [1,2,3], target = 4. We are asked to find out all possible combinations of numbers from ’nums’ that can add up to the target. Each combination should be considered unique, even if it consists of the same elements in a different order.

To put it simply, it’s like you have three different kinds of coins (nums array) and you want to find out in how many ways you can add up to a certain amount (target). Here, the order of using the coins matters.

Step 1: We will start by creating an array dp of size target + 1, initialized with zeros. This dp array is like a piggy bank for each amount from 0 to the target (4 in this case), and it will store the count of unique ways we can form the corresponding amount.
So, dp = [0, 0, 0, 0, 0] (for targets 0, 1, 2, 3, 4 respectively).
Step 2: We know that there is exactly one way to sum up to 0 (by choosing no numbers), so we will set dp[0] = 1.
Now, dp = [1, 0, 0, 0, 0].
Step 3: We will iterate over each number in the nums array. For each number, we will iterate from that number to the target on our dp array. For each of these sums, we add dp[sum - num] to dp[sum]. This is like saying, for each coin, if we can make a smaller sum, then we can also make the current sum by using the current coin.
Let’s break it down:
- For num = 1: iterate from 1 to 4, adding dp[sum - 1] to dp[sum].
  - dp = [1, 1, 1, 1, 1] after first pass.
- For num = 2: iterate from 2 to 4, adding dp[sum - 2] to dp[sum].
  - dp = [1, 1, 2, 2, 3] after the second pass.
- For num = 3: iterate from 3 to 4, adding dp[sum - 3] to dp[sum].
  - dp = [1, 1, 2, 3, 4] after the third pass.
- For num = 4 (our target): iterate from 4 to 4, adding dp[sum - 4] to dp[sum].
  - dp = [1, 1, 2, 3, 7] after the final pass.
Step 4: Finally, the value in dp[target] will be our answer, i.e., dp[4] = 7. These are the total number of combinations that can sum up to target.

Note that if we increase the target or the elements of the nums array, the size of the dp array and the time to compute the dp values would increase accordingly, affecting the time complexity of the solution.

Identify Invariant

An invariant in the context of an algorithm is a condition or property that remains true before and after certain operations, or throughout the execution of the algorithm. Identifying the invariants helps to ensure the correctness of the algorithm.

In the “Combination Sum IV” problem, the invariant can be defined as follows:

The value at each index i of the dynamic programming (DP) array dp[] always represents the total number of unique combinations from the nums array that can add up to i, considering the current and all previously processed elements from the nums array.

At the start, the DP array is initialized with zeros, except for dp[0] which is set to 1 because there’s exactly one way to sum up to 0 - by choosing no numbers. As we iterate through the nums array and update the DP array, this invariant condition holds true, because for each number in nums and for each value i from that number to target, we add the count of combinations that make up i - num (which we’ve previously calculated) to the count of combinations for i.

So, the invariant ensures that the DP array always contains the correct counts of combinations for each possible sum up to the target, at any point during the execution of the algorithm.

Identify Loop Invariant

A loop invariant is a condition that is initially true and remains true after each iteration of a loop. It is used to help prove the correctness of an algorithm.

In the context of the “Combination Sum IV” problem, one possible loop invariant can be described as follows:

For the outer loop that iterates over each number in the nums array, the loop invariant is that for each num, all possible combinations that include num and sum up to the value i (where i ranges from num to target) have been correctly calculated and stored in the dp array.

More specifically, within the inner loop where we iterate from num to target, the loop invariant is that dp[i] correctly represents the total number of combinations from the nums array seen so far that add up to i.

These loop invariants help to ensure that, after each iteration of the outer and inner loops, the dp array is accurately updated with the total counts of combinations that can sum up to each value from 0 to target, considering all numbers in the nums array seen so far. These invariants hold true at the start, during the execution, and at the end of the algorithm, thereby helping to prove its correctness.

In the context of the “Combination Sum IV” problem, the invariant and the loop invariant are indeed closely related, as both are focused on the properties of the dynamic programming (DP) array dp[] during the algorithm’s execution. However, they emphasize different aspects and stages of the problem-solving process.

The invariant of the overall algorithm is the property that the value at each index i of the dp[] array always represents the total number of unique combinations from the nums array that can add up to i, at any point during the execution of the algorithm.

The loop invariant, which is specific to the loops in the algorithm, focuses on the fact that after each iteration of the outer loop (for each num in nums[]), and inner loop (for each i from num to target), the dp[i] correctly represents the total number of combinations from the nums array seen so far that add up to i.

So, while they are referring to the same DP array and its properties, they are not exactly the same. The invariant refers to a property of the entire algorithm’s execution, whereas the loop invariant refers to a property maintained through each iteration of the specific loops in the algorithm.

While the concepts of invariants and loop invariants are related, they are not exactly the same.

The term “invariant” generally refers to a condition or property that remains unchanged before and after an operation, or throughout an algorithm’s execution. In the context of this problem, the invariant can be the assertion that the DP array’s value at each index i always represents the total number of unique combinations from the nums array that can add up to i, at any point during the execution of the algorithm.

A “loop invariant”, on the other hand, is a special kind of invariant that applies specifically to loops. It is a condition that is initially true and remains true after each iteration of a loop. For this problem, the loop invariant is that for each num in the outer loop, all possible combinations that include num and sum up to the value i (where i ranges from num to target) have been correctly calculated and stored in the DP array.

So, while both the invariant and the loop invariant contribute to the algorithm’s correctness and maintain certain properties throughout its execution, the loop invariant specifically relates to the properties maintained during each iteration of a loop. In this problem, both the invariant and the loop invariant are important for different parts of the algorithm’s logic.

Thought Process

Let’s start with a step-by-step thought process on how to solve this problem:

Understanding the Problem Statement: The first step in solving any problem is understanding what the problem is asking for. Here, we’re given an array of distinct integers and a target integer. The task is to find the number of possible combinations that add up to the target. It’s also stated that different sequences are counted as different combinations.
Identify the Problem Type: Given that we’re trying to find all possible combinations of the numbers that add up to the target, and considering the possibility of reusing the same number multiple times, this problem can be classified as a dynamic programming problem, specifically an unbounded knapsack problem.
Generate Insights and Formulate a Strategy: Knowing that this is a dynamic programming problem, we can devise a strategy to find the solution. We can start with a dynamic programming array (dp[]) where each index i represents the total number of combinations that can add up to i. We initialize dp[0] to 1, as there is one way to get a sum of 0 (by choosing no numbers).
Algorithm and Coding: We then iterate through each number in the input array and for each number, iterate again from that number up to the target. For each value i in this second iteration, we add the count of combinations that make up i - num (which we’ve previously calculated) to the count of combinations for i. This is because if we already have some combinations that sum up to i - num, we can just add num to each of those combinations to get a sum of i.

Here’s the Python code:

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class Solution:
    def combinationSum4(self, nums: List[int], target: int) -> int:
        dp = [0] * (target + 1)
        dp[0] = 1
        for i in range(1, target + 1):
            for n in nums:
                if i >= n:
                    dp[i] += dp[i - n]
        return dp[target]				

Testing and Verification: After coding the solution, it’s essential to test it with various test cases, including edge cases, to ensure its correctness. Examples include cases where the target is 0, cases where the target is less than the smallest number in nums, cases where nums only contains one number, etc.

The cues in the problem statement suggest that this dynamic programming approach include the need to find all possible combinations, the possibility of using the same number multiple times, and the need to add up to a specific target.

Establishing Preconditions and Postconditions

Parameters:
- The inputs to the method are nums and target.
- nums is a list of integers, and target is an integer.
- In the context of the problem, nums represents the set of distinct numbers we can use to form combinations, and target represents the desired sum of those combinations.
Preconditions:
- Before this method is called, no specific precondition about the state of the program is required.
- The constraints on the input parameters are: 1 <= nums.length <= 200, 1 <= nums[i] <= 1000, all the elements of nums are unique, and 1 <= target <= 1000.
- There is no specific state that the program or some part of it must be in.
Method Functionality:
- This method is expected to find the total number of unique combinations from nums that add up to target.
- It uses the inputs to calculate the return value and does not alter the current state of the program.
Postconditions:
- After the method has been called and has returned, no change to the state of the program or the values of the parameters is made.
- The return value represents the total number of combinations that add up to target.
- There are no side effects, as the method does not change the inputs or any other state of the program.
Error Handling:
- The method does not specifically handle errors. If the preconditions (i.e., constraints) are not met, the behavior will depend on the programming language’s handling of such situations. For instance, if an input is not an integer or if an array element is not within the specified range, a runtime error might occur. In a real-world application, you’d likely want to add error checking to handle such situations gracefully.
- This implementation does not throw exceptions or return a special value in error cases. However, as per the requirements of your particular application or system, such error handling can be incorporated.

Problem Decomposition

Problem Understanding:
- The problem is about finding the total number of distinct combinations that add up to a target number. We are given a list of unique integers and a target sum. A combination can contain any number of any element from the list, and the same combination in different orders is considered distinct.
Initial Breakdown:
- The major stages of the problem are:
  1. Iterating through all possible combinations of the given numbers that add up to the target.
  2. Counting these combinations.
Subproblem Refinement:
- The subproblem of iterating through all possible combinations can be broken down into:
  1. Selecting a number from the list.
  2. Subtracting that number from the target.
  3. Repeating this process until the target reaches 0 (successful combination) or becomes negative (failed combination).
Task Identification:
- The task of subtracting a selected number from the target and repeating this process is performed repeatedly and can be generalized as a single, recursive task.
Task Abstraction:
- The abstracted task would be to recursively attempt to reach zero by subtracting each of the numbers in the list from the current target, and count the successful attempts.
Method Naming:
- This task could be named countCombinations as it counts the successful combinations that add up to the target.
Subproblem Interactions:
- The order of performing tasks is first selecting a number and then performing the countCombinations task. The task is dependent on the selected number and the current target (which is initially the input target and changes as we subtract the selected numbers). The total count is the sum of the successful counts from each number’s countCombinations task.

Since this problem can be solved using dynamic programming, the above steps are the initial steps for understanding and breaking down the problem, and will need to be adjusted for the dynamic programming approach. In a dynamic programming solution, we would build up a table of counts for each possible sum up to the target, reducing the need for recursion and improving efficiency.

From Brute Force to Optimal Solution

Let’s first look at the brute force solution for this problem.

Brute Force Solution:

A simple brute force solution to this problem would be to use recursion to generate all possible combinations of numbers that sum up to the target, and count how many of these combinations exist. We’ll start from the target and subtract each number in the list one by one until we reach zero (a successful combination) or a negative number (an unsuccessful combination).

This can be done in Python using the following function:

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def combinationSum4(nums, target):
    if target == 0:
        return 1
    if target < 0:
        return 0
    total = 0
    for num in nums:
        total += combinationSum4(nums, target - num)
    return total

However, this solution is extremely inefficient. It has a time complexity of O(N^M) where N is the length of the nums list and M is the target number. This is because for each function call, we are making N recursive calls until we reach the base case of the recursion (i.e., when the target becomes zero or negative). In the worst case, this leads to an exponential number of function calls, resulting in a huge time complexity.

Optimized Solution:

We can optimize the brute force solution by using dynamic programming to avoid redundant computation. In the brute force solution, we’re computing the same sub-problems over and over again, which is the main source of the inefficiency. By using a dynamic programming approach, we can store the results of sub-problems and reuse them when needed, significantly reducing the number of computations.

Here’s the Python code for the optimized solution using dynamic programming:

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def combinationSum4(nums, target):
    dp = [0] * (target + 1)
    dp[0] = 1
    for i in range(1, target + 1):
        for num in nums:
            if num <= i:
                dp[i] += dp[i - num]
    return dp[target]

In this solution, dp[i] represents the number of combinations that add up to i. We initialize dp[0] to 1 because there is one combination that adds up to zero, which is an empty combination. For each integer i from 1 to target, we update dp[i] by adding the number of combinations that add up to i - num for each num in nums. This is based on the idea that if we have already found all combinations that add up to i - num, we can simply add num to each of these combinations to get a combination that adds up to i.

This dynamic programming solution has a time complexity of O(NM) where N is the length of the nums list and M is the target number, because we have two nested loops: the outer loop runs M times and the inner loop runs N times. This is much more efficient than the brute force solution. The space complexity is O(M) because we use an additional list of length M+1 to store the intermediate results.

Code Explanation and Design Decisions

Initial parameters: There are two parameters - nums and target. nums is a list of integers and target is a single integer. nums represents the numbers that we can use to form combinations, while target is the total sum that we want the combinations to reach.
Primary loop or iteration: The primary loop iterates over the range from 1 to target inclusive. Each iteration i of this loop represents a subproblem of finding the total number of combinations that sum up to i using the numbers in nums. The iteration advances the solution by solving the subproblems incrementally, from smaller sums to larger sums, until it eventually solves the original problem of finding the total number of combinations that sum up to target.
Conditions or branches within the loop: The inner loop iterates over each number num in nums. The condition if num <= i checks if num is less than or equal to the current sum i. This condition is necessary because we can’t use a number num to form a combination that adds up to i if num itself is greater than i.
Updates or modifications to parameters within the loop: Within the inner loop, the statement dp[i] += dp[i - num] updates the total number of combinations that add up to i. This update is necessary because for each num in nums, if num is less than or equal to i, then all combinations that add up to i - num can be extended by num to form a combination that adds up to i. Therefore, the total number of combinations that add up to i is the sum of the total number of combinations that add up to i - num for each num in nums.
Invariant: The invariant in this code is the list dp, where dp[i] represents the total number of combinations that add up to i using the numbers in nums. This invariant is maintained throughout the code by updating dp[i] for each i from 1 to target inclusive. It helps meet the problem’s objectives because it allows us to solve the subproblems incrementally, reuse the solutions to previous subproblems to solve later subproblems, and eventually solve the original problem.
Final output: The final output of the code is dp[target], which represents the total number of combinations that add up to target using the numbers in nums. It satisfies the problem’s requirements because it answers the question asked in the problem statement.

Coding Constructs

High-level problem-solving strategies: The code is using dynamic programming as a high-level strategy. Dynamic programming is a method for solving complex problems by breaking them down into simpler subproblems, solving the subproblems, and storing their solutions to avoid redundant computation. In this case, the code is breaking down the problem of finding the number of combinations that sum up to the target, into smaller subproblems of finding the number of combinations that sum up to smaller sums.
Explaining to a non-programmer: This code is like a super-efficient accountant. Imagine you have a bunch of different coins and you want to find out how many ways you can combine them to add up to a certain amount. This accountant doesn’t just start randomly putting coins together, they start by figuring out how many ways they can make smaller amounts first, and then uses that information to figure out how to make larger amounts. This way, they can efficiently compute the final total without doing any unnecessary work.
Logical elements: The logical constructs used in this code include variables, arrays, loops, and conditions. Variables are used to store data. Arrays, specifically the dp array, is used to store the number of combinations for each sum. Loops are used to iterate over each possible sum and each number in nums. Conditions are used to check whether a number can be used to form a combination that adds up to a particular sum.
Algorithmic approach in plain English: The code starts by creating an array to keep track of the number of combinations for each sum from 0 up to the target. It then loops through each possible sum. For each sum, it loops through each number in nums. If the number is less than or equal to the sum, it updates the number of combinations for the sum by adding the number of combinations for the difference between the sum and the number. Finally, it returns the number of combinations for the target.
Key steps or operations: The key steps in this code are initializing the dp array, iterating over each possible sum, iterating over each number in nums within the loop over the sums, updating the dp array within the inner loop, and returning the final result. These steps are performed to compute the number of combinations for each sum incrementally, from smaller sums to larger sums, reusing the solutions to previous subproblems to avoid redundant computation.
Algorithmic patterns or strategies: The code uses the pattern of dynamic programming, specifically the bottom-up approach. It also uses the strategy of space optimization by using a 1D array to store the number of combinations for each sum instead of a 2D array. Furthermore, it uses the strategy of tabulation to fill up the dp array iteratively.

Language Agnostic Coding Drills

Concept Identification:

The solution to this problem involves several coding concepts:

Variable Declaration and Initialization: This is the process of creating and assigning a starting value to a variable. In this problem, variables are used to store and manipulate data, such as the numbers in the array and the target value.
Arrays: Arrays are a fundamental data structure that stores elements of the same data type. In this problem, an array is used to store the number of combinations for each sum.
Loops: Loops are used to repeatedly execute a block of code. In this problem, loops are used to iterate over each possible sum and each number in the array.
Conditions: Conditions are used to execute different blocks of code based on whether a certain condition is met. In this problem, a condition is used to check whether a number can be used to form a combination that adds up to a particular sum.
Array Indexing and Updating: This concept involves accessing and modifying the values of an array at certain indices. In this problem, array indexing and updating are used to compute and store the number of combinations for each sum.
Dynamic Programming: Dynamic Programming is a strategy for solving complex problems by breaking them down into simpler subproblems and storing the solutions to the subproblems to avoid redundant computation. In this problem, dynamic programming is used to compute the number of combinations for each sum incrementally, reusing the solutions to previous subproblems.

Concepts Listed in Order of Increasing Difficulty:

Variable Declaration and Initialization: This is a fundamental concept that is necessary for nearly all programming tasks, and hence it is at the bottom of the difficulty scale.
Arrays: Understanding the concept of arrays is slightly more challenging, as it requires understanding of how data is stored and accessed in memory.
Loops: Looping constructs introduce a level of complexity as they require understanding how to repeatedly perform an operation, and how to manipulate loop counters.
Conditions: Conditions add complexity because they require understanding how to branch code execution based on logical tests.
Array Indexing and Updating: While this concept builds on the knowledge of arrays, it is a bit more challenging as it requires understanding how to manipulate array data dynamically.
Dynamic Programming: This is the most advanced concept in this problem. It requires a good understanding of problem decomposition, memoization or tabulation, and often involves complex logic.

Problem-Solving Approach:

The problem-solving approach that leads from the problem statement to the final solution can be described as follows:

Begin by understanding the problem statement and identifying that it’s a problem of counting combinations which is a classic use-case for dynamic programming.
Declare and initialize variables to store the input data and the dynamic programming array.
Create the dynamic programming array and initialize it. This array will be used to store the number of combinations for each sum.
Iterate over each possible sum. For each sum, iterate over each number in the input array.
Inside the inner loop, use a condition to check whether the number can be used to form a combination that adds up to the current sum. If it can, update the dynamic programming array to reflect the new number of combinations.
Continue this process until all sums have been computed. The final result is the number of combinations that add up to the target, which is the last element in the dynamic programming array.

Targeted Drills in Python

Coding Drills for General Concepts:

Variable Declaration and Initialization:

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x = 5
print(x)  # Outputs: 5

Arrays:

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numbers = [1, 2, 3, 4, 5]
print(numbers)  # Outputs: [1, 2, 3, 4, 5]

Loops:

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for i in range(5):
    print(i)  # Outputs: 0 1 2 3 4

Conditions:

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x = 10
if x > 5:
    print("x is greater than 5")  # Outputs: x is greater than 5

Array Indexing and Updating:

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numbers = [1, 2, 3, 4, 5]
numbers[0] = 10
print(numbers)  # Outputs: [10, 2, 3, 4, 5]

Coding Drills for Problem-Specific Concepts:

Dynamic Programming:

This concept is more advanced, so we’ll use a simple example of computing Fibonacci numbers.

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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))  # Outputs: 55

Assembling the Drills:

Each of the above drills represents a fundamental operation or concept that can be combined to solve our initial problem. Here’s a rough step-by-step guide to combining these drills:

Begin by declaring and initializing the input array (using the “Arrays” and “Variable Declaration and Initialization” drills).
Declare and initialize the dynamic programming array, initializing the first element to 1 (using the “Arrays” and “Variable Declaration and Initialization” drills).
Iterate over each number in the input array, and for each number, iterate over each sum from that number to the target (using the “Loops” drill).
Within the inner loop, update the dynamic programming array to add the number of combinations excluding the current number to the number of combinations including the current number (using the “Array Indexing and Updating” drill).
After both loops have completed, the last element of the dynamic programming array will contain the total number of combinations that add up to the target (using the “Array Indexing and Updating” drill).

Incorporating the problem-specific dynamic programming concept, we can efficiently compute the number of combinations for each possible sum, reusing previously computed results to avoid redundant computation.

Q&A

[1,1,2] and [2, 1, 1] are two different combinations, never heard this before

In many mathematical contexts, a combination refers to a selection of items where the order of the items does not matter. This means that [1,1,2] and [2,1,1] would be considered the same combination.

However, in some computational contexts (like in this problem), the term combination may refer to a multi-set, where the order does matter, and [1,1,2] and [2,1,1] would indeed be considered two different combinations.

This may seem a bit confusing, but it’s an example of how the same term can have slightly different interpretations in different fields or contexts. The key thing is to understand how the term is being used in the context of the specific problem you’re working on.

For the problem you’re dealing with, when we say ‘combination’, we mean a sequence of numbers that add up to a target number, where the order of the numbers in the sequence does matter. So, [1,1,2] and [2,1,1] are counted as separate combinations. This is a somewhat looser usage of the term ‘combination’ than you might see in a strict mathematical context.

This is why it’s important to read and understand the problem statement carefully - to make sure you understand the definitions and constraints it’s using.

What if negative numbers are allowed in the given array? How does it change the problem? What limitation we need to add to the question to allow negative numbers?

If negative numbers are allowed in the given array, it significantly changes the dynamics of the problem. With the inclusion of negative numbers, an infinite number of combinations can sum up to the target number. For example, if the target number is 4, you can reach it with the sequence [4, -1, 1], [4, -1, -1, 1, 1], [4, -1, -1, -1, 1, 1, 1] and so on. In other words, you can add and remove the same negative number and its positive counterpart any number of times.

This would pose a problem to your algorithm since it cannot handle an infinite number of combinations. Therefore, if negative numbers were allowed, you’d have to add some limitations or additional rules to keep the problem manageable.

For instance, you could add a rule that each number can only be used once, or a certain fixed number of times. Alternatively, you could specify that the solution should return the combination with the smallest number of elements, or the one that uses the smallest numbers.

The constraints to be added will largely depend on the specifics of the problem you are trying to solve. In most cases, you’ll need to add constraints that prevent infinite loops and make the problem solvable within a reasonable time frame.

Can someone explain why my intuition of using backtrack is wrong?

Backtracking is a very valid approach to many problems, particularly those involving combinations and permutations. It’s actually a common first instinct, and it can definitely be applied here. The idea would be to start with an empty combination and then, for each number in the input array, try adding it to the combination and see if it gets you closer to the target. If it does, you continue down that path; if it doesn’t, or if it overshoots the target, you backtrack and try the next number.

The problem with backtracking in this particular case, however, is that it can be very inefficient. Specifically, it can lead to a lot of repeated work. Imagine, for example, that the array contains the numbers 1 and 2, and the target is 1000. If you’re using backtracking, you might first try adding 1 to your combination a thousand times. Then, you might try adding 1 nine hundred and ninety-nine times and 2 once. Then 1 nine hundred and ninety-eight times and 2 twice. And so on. Each of these attempts involves a lot of repetition - a lot of adding the same numbers over and over again.

Dynamic programming, on the other hand, avoids this repetition by “remembering” previous results. It solves the problem for smaller sub-targets first, and then uses these solutions to build up to the solution for the larger target. So, in the example above, a dynamic programming solution would start by figuring out how many combinations sum to 1, then how many sum to 2, and so on, up to 1000. At each step, it can just look at the previous steps to see how many ways there are to make up the remainder after adding each possible number.

So while backtracking could certainly be used to solve this problem, dynamic programming is generally a more efficient approach.

We can approach the problem by dividing into the subproblem that how many combinations we can make from target-nums[0], target-nums[1]……. and so on. Specifically their addition will be the ans. Some edge cases are that if it goes out of bound(i.e., if target < nums[i]) and if target is 0 (i.e., no. of combinations of 0 is 1(default case) ).

Yes, your approach is indeed the correct one for this problem! You’re applying dynamic programming in an appropriate way to handle this problem.

Let’s make it more explicit with the following steps:

Create an array dp with length of target + 1 and initialize all elements to be 0. This is because for any given target, there might be 0 combinations.
Now, set dp[0] to be 1. This represents our base case where the target is 0, and there is exactly one way to “pick” numbers to add up to 0 (which is picking no numbers).
Now, iterate through the dp array starting from index 1. For each dp[i], calculate how many ways we can pick numbers from nums to sum up to i.
For calculating dp[i], iterate through each number num in nums. For each num, if i - num >= 0, we add dp[i - num] to dp[i]. This is because if num is one of the numbers that we can use to sum up to i, the number of ways we can pick other numbers to sum to i will be the number of ways to sum to i - num, which is dp[i - num].
Once we finish filling up dp, the answer will be dp[target].

Your edge cases are handled by the dynamic programming setup. If target < num for some num in nums, when we try to compute dp[i - num] during our dp array filling, we will see that i - num < 0 and we just skip this num. When target is 0, we have already set dp[0] = 1 as our base case.

This approach uses dynamic programming to avoid recalculating solutions to subproblems, and it builds up to the final solution in a bottom-up manner. It’s a more efficient solution than using a simple recursive (or backtracking) approach which may end up solving the same subproblems multiple times.

For the first question it was a coin change problem .So i can choose element unlimited amount of times.But when the array is populated with negative numbers then we can not choose the numbers multiple time .Because that will generate infinite combination of configuration. So the problem converts from coin change to knapsack. Correct me if i am wrong.

You’re on the right track with your thinking!

You’re correct that when we introduce negative numbers into the problem, it can generate an infinite number of combinations. For example, if the target sum is 0, and we have -1 and 1 in the array, we can add an unlimited number of -1 and 1 pairs to achieve the target sum of 0, leading to infinite combinations.

This change certainly affects our problem and how we would approach it. The crux here is not merely the transition from a “coin change” to a “knapsack” problem, but in defining new constraints to handle the introduction of negative numbers to maintain a meaningful and solvable problem.

One way we could modify the problem to handle negative numbers would be to add a constraint such that each number in the array can only be used once. This turns the problem into a variation of the subset sum problem: we’re looking for subsets of the array that sum up to the target value. This also avoids the issue of infinite combinations since each number can only be used once. This problem can also be solved using dynamic programming, but with a different DP formulation than the one used for the original problem.

What is the difference between the this and coin change 2 problem ?

Let’s call Combination Sum IV as Problem A. It is about finding the number of combinations that sum up to a target. The Coin Change 2 problem (let’s call it Problem B), on the other hand, is also about finding the number of combinations of coins that can make up a certain amount.

At first glance, both problems seem similar because both involve finding combinations to reach a target sum. However, there are some key differences:

Order of selection matters in Problem A but not in Problem B: In Problem A, [1,2] and [2,1] are considered two different combinations because the order in which the numbers are selected matters. But in Problem B, [1,2] and [2,1] are considered the same combination, because the order in which coins are selected doesn’t matter.
Different Definitions of ‘Combination’: Because of the above point, these two problems use different definitions of “combination”. In Problem A, a “combination” is a sequence where order matters. In Problem B, a “combination” is a set where order does not matter.

These differences affect how we approach solving the problems. Even though both can be solved using dynamic programming, the formulation of the DP problem would be different.

How this question is different from unbounded knapsack problem?

The Unbounded Knapsack problem and the Combination Sum IV problem both deal with the selection of items multiple times from a given set. However, there are fundamental differences between these two problems:

Objective: The primary difference lies in what we are trying to maximize or achieve. In the Unbounded Knapsack problem, we’re trying to maximize the total value of items in the knapsack while not exceeding a specific weight limit. On the other hand, in the Combination Sum IV problem, we’re not trying to maximize or minimize anything. Instead, we’re trying to count the number of ways we can sum up to a specific target.
Order Matters: In the Combination Sum IV problem, the order of the elements matters. This means that [1, 2] and [2, 1] are considered two distinct combinations. In contrast, in the Unbounded Knapsack problem, the order of selection doesn’t matter. It’s only the final collection of items (and their total value) that matters.
Output: The output of the Unbounded Knapsack problem is the maximum value that can be carried in the knapsack. The output of the Combination Sum IV problem is the count of distinct combinations that can sum up to the target.

These differences necessitate different strategies for solving each problem. While dynamic programming can be used for both problems, the structure of the dynamic programming table and the way the solutions are built up will be different.

The test cases are generated so that the answer can fit in a 32-bit integer. Yeah, but this is misleading - the intermediate values don’t always fit in 32 bit signed int.

You’re correct, and that’s a good observation. In the context of this problem, the final result (i.e., the count of combinations that sum up to the target) is guaranteed to fit into a 32-bit integer as per the test case constraints.

However, while solving the problem, we use a dynamic programming (DP) approach where we create an array (or table) to store intermediate results. The size of this DP array is determined by the target value, not by the final count of combinations.

If the target value is very large, the DP array’s size (i.e., target + 1) could exceed what can be handled by a 32-bit integer. This could potentially cause an out-of-memory error or integer overflow. But generally, in competitive programming or interview settings, such edge cases are typically omitted to keep the focus on problem-solving strategies and algorithms.

Therefore, it’s always important to understand the constraints of a problem and consider edge cases in a real-world scenario, as it may influence the choice of data structures and algorithms.

why it cannot be solved as a 2d dp problem like on dp on subequences?. like an unbounded knapsack one?

Indeed, this problem could also be approached using a 2D dynamic programming similar to the unbounded knapsack problem. However, the reason we use a 1D DP approach instead of a 2D DP is to optimize space complexity.

In a 2D DP solution, you would create a 2D table where the row index represents the index of nums (let’s say i), and the column index represents all sums from 0 to target (let’s say j). Each cell would contain the count of combinations that can sum to j using numbers from nums up to index i.

However, when you inspect the 2D table, you’ll realize that to fill the cell for i and j, you only need to look at the cell of the previous row (i-1) and the cell nums[i] columns before in the current row (j-nums[i]). This means that the solution for a particular i and j doesn’t depend on any other i and j except for the ones mentioned.

So, essentially, the count of combinations for a specific sum j only depends on the previous smaller sums, not on which numbers we have considered up to this point. This allows us to reduce the problem to a 1D DP solution, using a 1D table where the index j represents all sums from 0 to target, and each cell contains the count of combinations that sum to j.

This reduces our space complexity from O(n * target) in a 2D DP solution to just O(target) in a 1D DP solution, where n is the size of the nums array. Thus, for large inputs, the 1D DP solution would be more space-efficient.

Can somebody identify the error

at the if condition there is a runtime error: signed integer overflow: 2147483647 + 1 cannot be represented in type ‘int’

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int solveTabulation(vector & nums, int target) {
  vectordp(target + 1, 0);
  dp[0] = 1;
  for (int i = 1; i <= target; i++) {
    for (int j = 0; j < nums.size(); j++) {
      if (i - nums[j] >= 0) { // to check valid index
        dp[i] += dp[i - nums[j]];
      }
    }
  }
  return dp[target];
}

The error message you’re seeing, “signed integer overflow: 2147483647 + 1 cannot be represented in type ‘int’”, is a result of trying to store a value larger than the maximum limit of an integer in an integer variable.

In C++, the size of an integer (int) is 4 bytes, and it can store values from -2147483648 to 2147483647. If you try to store a value larger than 2147483647 in an integer variable, you get an integer overflow.

In your case, the problem seems to be with this line:

dp[i] += dp[i - nums[j]];

If dp[i - nums[j]] is large enough, adding it to dp[i] can result in a value that exceeds the maximum limit for an integer.

To fix this, you could use a long long int for your dp array instead of int, as long long int can store much larger values. Here is how you can do it:

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vector<long long int> dp(target+1, 0);

If you’re working with very large numbers, you might want to consider using modular arithmetic to avoid overflow. For example, if the problem statement allows, you could return the answer modulo 10^9+7 (a common modulus used in competitive programming). However, it’s important to remember to apply the modulus operation after every addition to prevent overflow:

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dp[i] = (dp[i] + dp[i - nums[j]]) % 1000000007;

Please make sure to check the problem requirements before using this approach.

Could anyone tell me what’s wrong with my code ? I got “Compiler Error”, it works well in other compilers

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class Solution {

func combinationSum4(_ nums: [Int], _ target: Int) -> Int {
    var count = 0
    helper(nums, target, &count)
    return count
}

func helper(_ nums: [Int], _ target: Int, _ count: inout Int) {
    if target == 0 {
        count += 1
        return
    } else if target < 0 {
        return 
    }
    for i in nums {
        helper(nums, target - i, &count)
    }
}
}

The issue with the code seems to be that it is a hybrid of Python and Swift, which would cause a compiler error since the syntax is not fully consistent with either language.

For example, the “class Solution” definition is Python-like, but Python does not use type annotations like “func combinationSum4(_ nums: [Int], _ target: Int) -> Int”.

On the other hand, the “func” keyword and the type annotations suggest Swift, but Swift uses a different syntax for class definitions: it does not require a colon after the class name.

Here is how the code would look in Swift:

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class Solution {
    func combinationSum4(_ nums: [Int], _ target: Int) -> Int {
        var count = 0
        helper(nums, target, &count)
        return count
    }

    func helper(_ nums: [Int], _ target: Int, _ count: inout Int) {
        if target == 0 {
            count += 1
            return
        } else if target < 0 {
            return 
        }
        for i in nums {
            helper(nums, target - i, &count)
        }
    }
}

Please remember to select the right language in your programming environment. The language selected should match the language you’re coding in.

Another similar (but not same) problem is - https://leetcode.com/problems/coin-change-2/solution/.

There is an important difference between the two, the order of the loops are interchanged.

This problem counts all the permutations [1, 1, 2], [1,2,1] and [1, 1, 2] as different solutions to the sum of 4.

In the coin-change-2 problem, [1, 1, 2], [1,2,1] and [1, 1, 2] are counted as one solution.

That’s correct! The two problems are indeed similar but they have a key difference: the order in which elements are combined. In the “Combination Sum IV” problem, the order of the numbers in the combination is significant, meaning that the combinations [1, 1, 2], [1, 2, 1] and [2, 1, 1] would be counted as three separate combinations. This problem is asking for permutations rather than combinations.

On the other hand, in the “Coin Change 2” problem, the order is not important, meaning [1, 1, 2], [1, 2, 1] and [2, 1, 1] would be counted as one solution. This problem is asking for combinations, not permutations.

Therefore, the looping structure of your solution will differ between the two problems. In “Combination Sum IV”, the outer loop is over the target sum, and the inner loop is over the array of numbers. This gives all permutations of numbers that can add up to the target sum. For “Coin Change 2”, the outer loop is over the array of numbers, and the inner loop is over the target sum. This gives all combinations of numbers that can add up to the target sum.

not a good idea to memorize this way which loop comes first.

I understand your concern. Memorizing the order of loops can indeed seem arbitrary and potentially confusing, particularly when there are closely related problems that require the loops in a different order. In the long run, it’s more important to understand why the loops are in the order they are.

For problems like “Combination Sum IV” and “Coin Change 2”, the order of loops is what determines whether the problem is solved in terms of combinations or permutations. This is because in the “Combination Sum IV”, the outer loop being the target sum allows the same numbers to be reused in different orders, creating different permutations. Conversely, in the “Coin Change 2”, the outer loop being the array of numbers ensures that once a number is used, the same set of numbers isn’t counted again, resulting in combinations.

In essence, rather than memorizing the order of loops, it would be more beneficial to understand the principles that guide the construction of these loops. This understanding can then be applied to solve a wide variety of related problems.

Can you explain about the order of the loops so I don’t have to memorize but understand it deeply?

The order of the loops in these kinds of problems plays a crucial role in determining whether we’re calculating combinations or permutations. The basic difference between the two is:

Combinations: Order doesn’t matter. For example, {1, 2, 3} and {3, 2, 1} are the same.
Permutations: Order matters. For example, {1, 2, 3} and {3, 2, 1} are different.

Consider the two problems - “Combination Sum IV” (where we count permutations) and “Coin Change 2” (where we count combinations).

In “Combination Sum IV”, the outer loop is for ’target’ and the inner loop is for iterating through the array of numbers (’nums’). This means that for each ’target’ value, we’re checking every number in the array, allowing the same number to be reused in different orders, generating different sequences (permutations). This is because we’re exhausting all possible combinations of ’nums’ for a given ’target’ before moving on to the next ’target’ value.

In contrast, in “Coin Change 2”, the outer loop is for iterating through the array of numbers, and the inner loop is for ’target’. This means that for each number in the array, we’re computing its contribution to every possible ’target’ value, without changing the order of the numbers. This approach prevents counting the same set of numbers as different sequences, thus giving us combinations.

So it’s not just the order of loops that matters, but how that order interacts with the problem’s requirements. Understanding this interaction allows you to decide the order based on whether you want combinations or permutations, eliminating the need to memorize the loop order for specific problems.

Well, the way I see it,

The code as this question’s solution is what a normal traversal looks like, we loop range(0, target) first. But the nature of a normal traversal is that we will possibly get duplicated ones (Exactly what will happen in a combination). For coin-change-2, we need to remove duplicated ones. Therefore, we get the coin loop first, in this case, we would fix the order of coins in our “combination”. As a result, we will only count either one of [1,2,…] or [2,1,…], they will not be counted more than once (twice in this case).

That’s a great understanding and observation!

In “Combination Sum IV”, we are essentially asking “given a target, in how many ways can we choose numbers from ’nums’ array to sum up to it?” Therefore, it makes sense to loop through the target first (outer loop) and then go through the ’nums’ array (inner loop) because we want to try all possible permutations of ’nums’ for a given target.

In “Coin Change 2”, we’re instead asking “given a coin, for how many different amounts can we use it?” Therefore, we loop through ’nums’ first (outer loop) and then iterate through targets (inner loop). This way, for each coin, we try to use it to make up all possible target amounts.

So, yes, you’re absolutely right! The order of the loops directly corresponds to the question we are trying to answer in each problem, which determines whether we need to count different orderings (permutations) or not (combinations). Your understanding is correct and well-explained!

The sample solution (bottom-up) could fail for few of the test cases. Use “unsigned long long” type if using C++ to solve this problem

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 int combinationSum4(vector<int>& nums, int target) {
        vector<unsigned long long> dp(target + 1);
        dp[0] = 1;
        for (int i = 1; i <= target; i++) {
            for (int num : nums) {
                if (i >= num) {
                    dp[i] += dp[i - num];
                }
            }
        }
        return dp[target];
    }

The problem with overflow arises when we work with large values and the calculated result exceeds the maximum limit of the used data type. In C++, the maximum limit for int is 2147483647 (or roughly 2*10^9). If a value exceeds this limit, it wraps around from the lowest possible value for int, which results in undefined behavior or incorrect results, a situation commonly referred to as integer overflow.

Using unsigned long long instead of int for the dp vector is a valid solution to avoid overflow issues. This is because unsigned long long can handle much larger values (up to 18446744073709551615 or roughly 1.8*10^19), which can prevent integer overflow in this case.

However, this may not always be the best solution, especially when you are not certain about the size of the values you’ll be dealing with. An even better approach would be to include a condition that checks if the current sum exceeds the maximum limit of the data type, then handles it accordingly.

Also, while it might solve the overflow issue, it’s important to note that the use of larger data types like unsigned long long can lead to increased memory usage and reduced performance, especially in memory-intensive tasks.

So in general, yes, using unsigned long long can help with avoiding overflow issues, but one needs to consider the trade-offs and whether it’s necessary for the particular problem.

Thank you for pointing out this. It works after I follow your code. Could you help to explain why the type cannot be int? I just don’t understand the reason behind this.

In programming, the variables we create are stored in memory, and each type of variable uses a certain amount of memory. Each type also has a range of values it can store. For example, in C++, an int is typically 4 bytes and can store values from -2147483648 to 2147483647.

The error message you were getting, “signed integer overflow,” means that the value you were trying to store exceeded the maximum limit that an int can hold, which is 2147483647.

When you add 1 to 2147483647, the resulting value (2147483648) is larger than the maximum value an int can store, leading to an overflow. This overflow wraps the value around to the negative end of the int range, resulting in an unintended negative value.

By changing int to unsigned long long, you increase the range of values that your variable can store. An unsigned long long in C++ is typically 8 bytes and can store values from 0 to 18446744073709551615. This change allows your code to handle the larger values that are causing the overflow with the int type.

Keep in mind, though, that using larger variable types like unsigned long long uses more memory than int. It’s good practice to use the smallest variable type that can handle all possible values for your use case.

Combination Sum IV

Identifying Problem Isomorphism

10 Prerequisite LeetCode Problems

Problem Classification

Clarification Questions

Problem Analysis and Key Insights

Identifying Problem Isomorphism

Problem Boundary

Distilling the Problem to Its Core Elements

Visual Model of the Problem

Problem Restatement

Abstract Representation of the Problem

Terminology

Problem Simplification and Explanation

Constraints

Case Analysis

Identification of Applicable Theoretical Concepts

Simple Explanation

Problem Breakdown and Solution Methodology

Inference of Problem-Solving Approach from the Problem Statement

Simple Explanation of the Proof

Stepwise Refinement

Solution Approach and Analysis

Identify Invariant

Identify Loop Invariant

Thought Process

Establishing Preconditions and Postconditions

Problem Decomposition

From Brute Force to Optimal Solution

Code Explanation and Design Decisions

Coding Constructs

Language Agnostic Coding Drills

Targeted Drills in Python

Q&A

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