Combination Sum IV

Problem Statement

The task at hand is to determine the number of distinct ways to combine a given set of distinct integers (nums) to sum up to a specific target integer (target). Each integer from the nums array can be used multiple times in the combination. The sequence of numbers in the combinations matters; thus, different permutations of the same numbers are considered as separate combinations. This calculation needs to be efficient to ensure the results fit within a standard 32-bit integer limit.

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

Approach and Intuition

To solve this problem effectively, a dynamic programming approach can be used. Here's the intuition:

1. Initialization: Start by creating an array (dp) where each index represents a target from 0 to target, initialized with zeros. The dp[0] is set to 1 because there is exactly one way to get a sum of zero: by using no elements.

2. Dynamic Array Population:

   • For each number in nums, iterate through the possible sums from the number up to the target.
   • For each sum, update dp[sum] by adding the value from dp[sum - num]. This update reflects the number of ways to achieve sum by adding the current number to the ways previously computed to achieve the sum - num.

3. Result Extraction: After updating the dp array, the value at dp[target] will provide the number of possible combinations that can add up to target.

This dynamic programming approach is efficient as it leverages previous computations in a methodical manner, thereby reducing redundant calculations. It navigates through each unique number and builds upon smaller subproblems to reach the solution.

Examples for Clarity

• In the first example (nums = [1,2,3], target = 4), the approach will calculate ways to reach all sums from 1 to 4 using each num one at a time. Starting from 1, it will count single increments, then pairs, then combinations with 3, summing up all possible permutations.

• For the second example (nums = [9], target = 3), since 9 is greater than the 3, no combinations can yield a sum of 3. Hence, the result is straightforwardly 0.

By applying the above method, problems even at the upper constraints can be addressed efficiently, considering the distinct nature of elements and the range provided by the problem's constraints.

Solutions

class Solution {
    
    public int countCombinations(int[] elements, int goal) {
        int[] combinations = new int[goal + 1];
        combinations[0] = 1;
    
        for (int sum = 1; sum <= goal; sum++) {
            for (int element : elements) {
                if (sum >= element) {
                    combinations[sum] += combinations[sum - element];
                }
            }
        }
        return combinations[goal];
    }
}