377. Combination Sum IV

Problem¶

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?

Solve¶

Dynamic programing¶

O(n ** 2)

It really similar to Coins charge problem (a basic DP problem). Also, it available in Leetcode 518. Coin Change II, do already do a brief explanation on the ordering of for loop can be matter.

Recursive formulation:

for (n: nums)
    totalPossible[i] = totalPossible[i-n] 

Final implementation

class Solution:
    def combinationSum4(self, nums: List[int], target: int) -> int:
        # Note that different sequences are counted as different combinations 
        # :)??
        nums.sort()

        cache = [0] * (target+1)
        cache[0] = 1
        for i in range(1, target+1):
            for n in nums:
                if n > i:
                    break
                cache[i] += cache[i-n]
        return cache[-1]