377. Combination Sum IV

Link

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

DFS + memo

   public int combinationSum4(int[] nums, int target) {
        int[] memo = new int[target+1];
        Arrays.fill(memo, -1);
        return dfs(nums, memo, target);  
    }
    
    public int dfs(int[] nums, int[] memo, int target){
        
        if(target == 0) return 1;
        if(memo[target] != -1) return memo[target];
        int count = 0;
        for(int num : nums){
            if(num <= target){
                count += dfs(nums, memo, target-num);
            }
        }
        memo[target] = count;
        return count;
    }

Button up

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