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?

Credits:
Special thanks to @pbrother for adding this problem and creating all test cases.

Solution

Recursion with memo

由于本题是顺序相关的，其实应该是permutation而非combination，用memo以避免重复计算子问题。

用DP也可以解。

class Solution {
    public int combinationSum4(int[] nums, int target) {
        return combinationSum4Recur(nums, target, new HashMap<>());
    }
    
    public int combinationSum4Recur(int[] nums, int target, Map<Integer, Integer> map) {
        if (target == 0) {
            return 1;
        }
        if (target < 0) {
            return 0;
        }
        if (map.containsKey(target)) {
            return map.get(target);
        }
        
        int count = 0;
        
        for (int n : nums) {
            count += combinationSum4Recur(nums, target - n, map);
        }
        
        map.put(target, count);
        return count;
    }
}

Follow-up

如果array中有负数，则需要给定一个combination maximum length，用来终止递归。