Combination Sum IV

Problem

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

DP Recursive Solution

public class Solution {
    public int combinationSum4(int[] nums, int target) {
        int[] dp = new int[target + 1];
        Arrays.fill(dp, -1);
        dp[0] = 1;
        return helper(nums, dp, target);
    }

    private int helper(int[] nums, int[] dp, int target) {
        if (dp[target] != -1) return dp[target];
        int count = 0;
        for (int num : nums) {
            if (target - num >= 0) count += helper(nums, dp, target - num);
        }
        dp[target] = count;
        return count;
    }
}

Analysis

dp[i] means how many combinations we have for the target i
In helper() method, We starts for loop from beginning to the end to include all cases
Notice that init case is dp[0] = 1 (empty combination) and hence if statement becomes target - num >= 0