Combination Sum IV

Description

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

Let dp[i] be the number of combinations that add up to i. Then there are n exclusive and exhaustive cases: the combinations with the first number being nums[j], for 0 <= j < n. We can sort nums to prune nums[j] that are larger that i.

class Solution {
public:
    int combinationSum4(vector<int>& nums, int target) {
        vector<int> dp(target + 1);
        sort(nums.begin(), nums.end());
        dp[0] = 1;
        for(int i = 1; i <= target; ++i){
            for(int num : nums){
                if(num > i)
                    break;
                dp[i] += dp[i - num];
            }
        }
        return dp[target];
    }
};

Follow up

Limitation: there are no combinations that add up to 0. Otherwise there are either 0 or infinite number of combinations.