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:

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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?

O(n) time O(n) space memoization solution with HashMap

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

O(n) time O(n) space Top down DP solution

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public class Solution {
    private int[] counts;
    
    public int combinationSum4(int[] nums, int target) {
        counts = new int[target + 1];
        Arrays.fill(counts, -1); // sometimes the combinations of a target is 0. to distinguish this, we need to set the default value to -1.
        counts[0] = 1; // target matches the current num
        return getCombination(nums, target);
    }
    
    public int getCombination(int[] nums, int target) {
        if (counts[target] != -1) return counts[target];
        int count = 0;
        for (int i = 0; i < nums.length; i++) {
            if (target >= nums[i]){
                count += getCombination(nums, target - nums[i]);
            }
        }
        counts[target] = count; // don't forget to update the value
        return count;
    }
}

O(n) time O(n) space Bottom up DP solution

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

Follow-up questions

To allow negative numbers in the array, we need to limit the length of each combinations. Otherwise, the number of combinations can be infinite. Take [1, -1] and target = 1 as an example, we can have an infinite length of [1, -1, 1, -1 … 1] which sums to 1.