# Solving Leetcode Interviews in Seconds with AI: Combination Sum IV


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "377" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like [Chatmagic](https://www.chatmagic.app), we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

	# Problem Statement
	> Given an array of distinct integers nums and a target integer target, return the number of possible combinations that add up to target. The test cases are generated so that the answer can fit in a 32-bit integer.   Example 1:  Input: nums = [1,2,3], target = 4 Output: 7 Explanation: The possible combination ways are: (1, 1, 1, 1) (1, 1, 2) (1, 2, 1) (1, 3) (2, 1, 1) (2, 2) (3, 1) Note that different sequences are counted as different combinations.  Example 2:  Input: nums = [9], target = 3 Output: 0    Constraints:  1 <= nums.length <= 200 1 <= nums[i] <= 1000 All the elements of nums are unique. 1 <= target <= 1000    Follow up: 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? 

	# Explanation
	Here's a solution to the problem, along with explanations:

*   **Dynamic Programming:** We use dynamic programming to store the number of combinations for each sum from 0 up to the target. `dp[i]` stores the number of combinations that sum up to `i`.

*   **Bottom-Up Approach:** We build the `dp` array from the bottom up. The base case is `dp[0] = 1` (there's one way to make a sum of 0: using no numbers). For each number in `nums`, we iterate through the `dp` array and update the number of combinations if the current number can be included.

*   **Consider Sequence:** Because the problem specifies different sequences are considered distinct combinations, the outer loop must iterate over `target` and the inner loop should iterate over `nums`. This ensures that all possible sequences are considered.

*   **Runtime Complexity:** O(target * n), where n is the length of nums. **Storage Complexity:** O(target)

	
	# Code
	```python
	def combinationSum4(nums: list[int], target: int) -> int:
    """
    Finds the number of combinations that add up to the target.

    Args:
        nums: A list of distinct integers.
        target: The target integer.

    Returns:
        The number of possible combinations that add up to the target.
    """

    dp = [0] * (target + 1)
    dp[0] = 1

    for i in range(1, target + 1):
        for num in nums:
            if i - num >= 0:
                dp[i] += dp[i - num]

    return dp[target]
	```
			
				# Summary
				**Follow-up: Negative Numbers**

If negative numbers are allowed, the problem becomes much more complex. Without any constraints, the number of combinations can be infinite. For example, if `nums = [-1, 1]` and `target = 1`, we can have combinations like `[1]`, `[1, -1, 1]`, `[1, -1, 1, -1, 1]`, and so on.

**Limitation:**

To allow negative numbers, we need to add a constraint on the *length* of the combination. We could specify a maximum number of elements that can be used in any combination.  Another approach would be to limit the number of times any particular negative number can appear in a combination. Without such constraints the number of combinations will be unbounded.