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, 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

    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.