Question

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

It is simple to figure out a backtracking solution:

class Solution {
    public int combinationSum4(int[] nums, int target) {
        Arrays.sort(nums);
        return backtracking(nums, target);
    }

    private int backtracking(int[] nums, int remaining) {
        if(remaining == 0) return 1;
        else {
            int count = 0;
            for(int n: nums) {
                if(n <= remaining) {
                    count += backtracking(nums, remaining - n);
                } else break;
            }
            return count;
        }
    }
}

However, it hits TLE quite quick. The main reason is that it counts the number of combinations for the partial solutions where remaining < target many times, and we can use a DP array to cache them.

class Solution {
    private int[] dp;
    public int combinationSum4(int[] nums, int target) {
        Arrays.sort(nums);
        dp = new int[target + 1];
        Arrays.fill(dp, -1); // Use -1 as a notion that the value is not computed
        return helper(nums, target);
    }

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

We can convert the recursive version to an iterative version by computing each i (1 <= i <= target) iteratively:

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