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?

Analysis:

Define function F(n) = num of possible combinations that add up to n

therefore

nums= [a0, a1, a2, ....]
then :
F(n)= F(n-a0)+F(n-a1)+F(n-a2)+..

Therefore, we can build our answer in Top-Down (memorization) or Bottom-Up way (Dynamic Programming).

For Top-Down way, we use a HashMap to record the F(k) value if it has been calculated.

For Dynamic Programming approach, we define int[] dp=new int[target+1] where dp[i] represent the number of combinations that add up to i. To calculate dp[i] for i we have to build the answer to number j where j is smaller than i first.

Solution

Memorization

public class Solution {
    Map<Integer, Integer> map = new HashMap<>();
    public int combinationSum4(int[] nums, int target) {
        int count = 0;
        if (nums == null || nums.length ==0 || target < 0 ) return 0;
        if ( target ==0 ) return 1;
        if (map.containsKey(target)) return map.get(target);
        for (int num: nums){
            count += combinationSum4(nums, target-num);
        }
        map.put(target, count);
        return count;
    }
}

DP

public class Solution {
    public int combinationSum4(int[] nums, int target) {
        if(nums==null || nums.length==0) return 0;
        int[] dp=new int[target+1];
        Arrays.sort(nums);
        for(int i=1;i<=target;i++){
            for(int a:nums){
                if(a>i) break;
                if(a==i) dp[i]+=1;
                else     dp[i]+=dp[i-a];
            }
        }
        return dp[target];
    }
}