Problem Description

You're given a list of numbers called nums and a single number called target. Your task is to form a mathematical expression by adding either a '+' (plus sign) or a '-' (minus sign) before each number in nums, then concatenate (join together) these numbers with their signs to form an expression. The end goal is to find out how many different expressions you can create that, once evaluated, equal the target.

For instance, if nums is [2, 1] and your target is 1, you could create the expression "+2-1", which equals 1. You want to find out all such possible expressions that result in the target.

Here's what you need to consider:

• You can only use '+' or '-' before each number.
• You can't reorder the numbers – their order in the expression must match their order in the given nums list.
• The objective is to count the number of valid expressions, not necessarily to generate each one.

Flowchart Walkthrough

Let's use the algorithm flowchart to deduce the right approach for solving LeetCode 494. Target Sum. You can follow along using the Flowchart. Here's a step-by-step walkthrough:

1. Is it a graph?

   • No: The problem is about achieving a target sum using numbers in an array, not about nodes and edges as in graph problems.

2. Need to solve for kth smallest/largest?

   • No: The problem is not about finding the kth smallest or largest element, but about finding ways to reach a target sum.

3. Involves Linked Lists?

   • No: The problem involves arrays, not linked lists.

4. Does the problem have small constraints?

   • Yes: The problem can have constraints small enough that allows considering all subsets of the numbers.

5. Brute force / Backtracking?

   • Yes: The problem requires exploring multiple combinations to find the number of ways to achieve the target sum, making brute force and backtracking suitable methods.

Conclusion: Based on the flowchart analysis, using a backtracking approach is suggested for solving the problem by exploring various combinations to achieve the target sum. This ensures that all potential solutions are considered.

Intuition

Understanding the problem, we see that it resembles a classic problem in computer science known as the subset sum problem, except it allows for both positive and negative summations. The crux is to figure out how to partition nums into two subsets where the difference between the sums of the subsets equals target.

First step is to notice a helpful fact: if we sum all numbers with a '+' and then subtract the sum of numbers with a '-', we should obtain the target. This is basically the same as finding two subsets with a particular sum difference.

Now, how do we solve this with dynamic programming? We can use a technique similar to the 0-1 Knapsack problem. The idea is to iteratively build up an array dp where each index represents a possible sum of numbers, starting from 0 up to some value that depends on nums and target. Each cell in dp will tell us the number of ways to achieve that sum.

But how do we compute the size of dp and its initial values? We define a new variable n which is the sum that we want one subset to have so that the other subset has a sum of n + target (given that the total sum of nums is s). It turns out n should be (s - target) / 2. We only proceed if s - target is even, otherwise, it's not possible to split nums into two such subsets.

Once we have dp setup starting at dp[0] = 1 (there's one way to achieve a sum of 0: by choosing no numbers), we start populating dp by iterating through each number in nums. For each v in nums, we iterate backwards through dp from n down to v (since v is the smallest sum that including v can achieve) and update dp[j] to include the number of ways we can achieve the sum j - v.

In the end, dp[n] gives us the number of different expressions we can form that equal target.

The code provided implements this dynamic programming approach efficiently.

Learn more about Dynamic Programming and Backtracking patterns.

Solution Approach

The provided solution uses a dynamic programming approach to solve the problem efficiently, much like the "0-1 Knapsack" problem. The solution involves some pre-processing steps and then iteratively filling out a dp array to count the number of ways to reach different sums.

Here's a step-by-step breakdown of the implementation:

1. Calculate the sum s of all numbers in nums. This is done to determine the sum of one subset (n) so that the difference with the other subset's sum will be target.

2. Check for two conditions that might make it impossible to create expressions equal to target:

   • If the total sum s is less than target, it's not possible to create any expression equal to target.
   • If s - target is odd, we can't split the array into two subsets with integer sums that have the needed difference.

3. Initialize the dp array of size n + 1 with zeros and set dp[0] to 1 – this signifies that there is one way to create a sum of zero (by choosing no numbers).

4. Iterate over each number v in nums. For each v, iterate backwards over the dp array from n down to v to avoid recomputing values that rely on the current index. This is crucial because we must not count any combination twice.

5. Update the dp[j] value by adding the number of combinations that exist without the current number v (dp[j - v]). This step accumulates the number of ways there are to achieve a sum of j by either including or excluding the current number v.

By the end of the iteration, dp[-1] (which is dp[n]) will hold the number of possible expressions that can be created using nums to equal the target.

Example of dp Array Update

Here’s a hypothetical example to illustrate the dynamic programming update process:

• Suppose nums = [1,2,3] and target = 1. We compute n = (s - target) / 2, which would be 2 in this case.

• Our dp array is [1, 0, 0, 0, 0] initially.

• During the iteration process:

  • When v = 1, we update dp to [1, 1, 0, 0, 0].
  • When v = 2, we update dp for j from 2 to 1, resulting in [1, 1, 1, 0, 0].
  • When v = 3, we update dp for j from 2 to 1, but this time there are no changes as 3 is too large to affect dp[1] or dp[2].

In the end, dp[n] gives us the count of the number of expressions that sum up to target.

The method described balances the need to consider both including and excluding each number in nums while ensuring each sum is only counted once. It utilizes memory efficiently by only maintaining a one-dimensional array rather than a full matrix that would be required in a naive implementation of dynamic programming for this problem.

Example Walkthrough

Let's walk through a small example to illustrate the provided solution approach. Suppose we have nums = [1, 2, 3] and target = 2.

According to the solution approach, our first step is to calculate the sum s of all numbers in nums. Here, the sum is 1 + 2 + 3 = 6.

Then we check if the total sum s is less than target or if s - target is odd. In this case, 6 is not less than 2, and 6 - 2 is even, so we proceed.

Our next step is to determine the size of the dp array, which will help us count the number of ways to make up different sums. We calculate n = (s - target) / 2, which is (6 - 2) / 2 = 2.

Now we initialize our dp array with zeros and set dp[0] to 1. So initially, our dp array is [1, 0, 0].

Next, we iterate over each number v in nums. For each v, we update the dp array from n down to v:

• For v = 1, update dp[j] from j = 2 to 1. The dp array changes as follows:

  • dp[1] gets updated to dp[1] + dp[1 - 1] which is dp[1] = 0 + 1 = 1.
  • The dp array is now [1, 1, 0].

• For v = 2, update dp[j] from j = 2 to 2:

  • dp[2] gets updated to dp[2] + dp[2 - 2] which is dp[2] = 0 + 1 = 1.
  • The dp array is now [1, 1, 1].

• For v = 3, since our dp array size is only 3, we don't need to update it for v = 3 because 3 is larger than n.

Finally, our dp array is [1, 1, 1], and the dp[-1] or dp[2] indicates that there is 1 possible expression that can be created using nums to sum up to target, which is "1 + 2 - 3".

This walkthrough demonstrates how the dynamic programming approach effectively counts the number of expressions equating to the target without redundantly computing combinations.

Solution Implementation

Time and Space Complexity

The time complexity of the algorithm is O(len(nums) * n), where len(nums) is the number of elements in the input list nums, and n is the computed value (sum(nums) - target) // 2. This is because the algorithm consists of a nested loop, where the outer loop runs for each element in nums, and the inner loop runs from n down to the value of the current element v.

The space complexity of the algorithm is O(n + 1), which simplifies to O(n), as there is a one-dimensional list dp of size n + 1 elements used to store the intermediate results for the subsets that sum up to each possible value from 0 to n.

Learn more about how to find time and space complexity quickly using problem constraints.