Solution Explanation: Combination Sum IV

This problem asks for the number of distinct combinations of numbers from a given array nums that sum up to a target value target. The key is that different sequences are considered different combinations. This suggests a dynamic programming approach.

Approach: Dynamic Programming

We use dynamic programming to solve this problem efficiently. We create a DP array f of size target + 1, where f[i] represents the number of combinations that sum up to i.

1. Initialization: f[0] = 1. This is because there's one way to achieve a sum of 0 (using no numbers). All other f[i] are initialized to 0.

2. Iteration: We iterate through the DP array from index 1 up to target. For each index i, we iterate through the numbers in nums. If a number x is less than or equal to i, it means we can potentially use x in a combination that sums to i. The number of ways to reach i using x is equal to the number of ways to reach i - x (because we've already used x). Therefore, we add f[i - x] to f[i].

3. Result: After iterating through all indices and numbers, f[target] will contain the total number of combinations that sum up to target.

Time Complexity: O(n * target), where n is the length of nums and target is the target sum. We have nested loops iterating through nums and the f array.

Space Complexity: O(target). The space used is dominated by the DP array f.

Code Examples:

The code examples in multiple languages provided in the original response demonstrate this dynamic programming approach. The core logic remains consistent across all languages:

Python3:

class Solution:
    def combinationSum4(self, nums: List[int], target: int) -> int:
        f = [0] * (target + 1)
        f[0] = 1  # Base case: one way to sum to 0
        for i in range(1, target + 1):
            for x in nums:
                if i >= x:
                    f[i] += f[i - x]
        return f[target]
 

Java:

class Solution {
    public int combinationSum4(int[] nums, int target) {
        int[] dp = new int[target + 1];
        dp[0] = 1;
        for (int i = 1; i <= target; i++) {
            for (int num : nums) {
                if (i >= num) {
                    dp[i] += dp[i - num];
                }
            }
        }
        return dp[target];
    }
}

The other languages (C++, Go, TypeScript, JavaScript, C#) follow a similar structure, only differing in syntax. All implementations achieve the same time and space complexity.