# LeetCode 377. Combination Sum IV

## The Problem

Link to original problem on Leetcode.

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.

Examples

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

## My Solution

### Dynamic Programming - Bottom-up

Here we’ve got another problem solved with dynamic programming. We can break apart this problem into smaller sub-problems to get our solution.

In this first variation, we use the bottom-up approach. First, create our cache dp as an array with length of target + 1. We’ll initialize dp[0] = 1, as there’s exactly one combination to reach a target of zero (i.e., use no numbers). Next we loop over the range 0 to target - 1. If we have no value for dp[i], we continue to the next number in the range. Since we initialized dp[0] to 1, we’ll at least run the rest of the loop that first time if target > 0.

Then for each number i in the range, we’ll loop over each num in the nums array. If num + i <= target, then we’ve found a possibly valid path toward the target. So we’ll set dp[num + i] equal to itself plus dp[i]. That is to say, we take the number of combinations we know gets us to dp[i] and add them to any other already known number of combinations that get us to num + i.

Once we’ve finished all our loops, the total number of valid combinations will be at index target in dp, so we return dp[target]. This is true even if the target is 0, since we initialized dp[0] = 1.

function combinationSum4(nums: number[], target: number): number {
const dp = Array.from({ length: target + 1 }, () => 0);
dp[0] = 1;

for (let i = 0; i < target; i++) {
if (!dp[i]) continue;
for (const num of nums) {
if (num + i <= target) {
dp[i + num] += dp[i];
}
}
}

return dp[target];
}

The time complexity is $O(n * m)$, where $n$ is the target and $m$ is the length of nums. Space complexity is $O(n)$.

### Dynamic Programming - Top-down

The top-down approach is nearly identical to the bottom-up approach. Here we loop not from 0 through target - 1, but instead from 1 through target. We don’t need to check if dp[i] exists here before running the inner loop, because we’ll be setting values to dp[i] instead of dp[num + i]. Next we loop again over num of nums. If num <= i (i.e., there is some value less than i and greater than or equal to 0 that, added to num, gets us to target of i), we’ll set dp[i] to itself plus the combinations we’ve observed at dp[i - num]. And again, we return dp[target].

Just like with bottom-up, time complexity is $O(n * m)$, where $n$ is the target and $m$ is the length of nums. Space complexity is $O(n)$.

function combinationSum4(nums: number[], target: number): number {
const dp = Array.from({ length: target + 1 }, () => 0);
dp[0] = 1;

for (let i = 1; i <= target; i++) {
for (const num of nums) {
if (num <= i) {
dp[i] += dp[i - num];
}
}
}

return dp[target];
}