377. Combination Sum IV

Given an integer array with all positive numbers and no duplicates, find the number of possible combinations that add up to a positive integer target.

Example:

nums = [1, 2, 3]
target = 4

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.

Therefore the output is 7.

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?

Solution:

We create a dp array comb with length target + 1.

comb[i] means the number of combinations that sum == i.

comb[0] is initialized to 1 since if target is 0, we only have one combination, which is nothing.

comb[i] =  sum of comb[i - nums[j]] if i >= nums[j], where j is from 0 to nums.length - 1.



Code:

public class Solution {
    public int combinationSum4(int[] nums, int target) {
        int[] comb = new int[target + 1];
        comb[0] = 1;
        for (int i = 1; i < comb.length; i++) {
            for (int j = 0; j < nums.length; j++) {
                if (i >= nums[j]) {
                    comb[i] += comb[i - nums[j]];
                }
            }
        }
        return comb[target];
    }
}