377. Combination Sum IV (Medium)

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 1: recursion TLE

Think about the recurrence relation first. How does the # of combinations of the target related to the # of combinations of numbers that are smaller than the target?

So we know that target is the sum of numbers in the array. Imagine we only need one more number to reach target, this number can be any one in the array, right? So the # of combinations of target, comb[target] = sum(comb[target - nums[i]]), where 0 <= i < nums.length, and target >= nums[i].

In the example given, we can actually find the # of combinations of 4 with the # of combinations of 3(4 - 1), 2(4- 2) and 1(4 - 3). As a result, comb[4] = comb[4-1] + comb[4-2] + comb[4-3] = comb[3] + comb[2] + comb[1].

Then think about the base case. Since if the target is 0, there is only one way to get zero, which is using 0, we can set comb[0] = 1.

EDIT: The problem says that target is a positive integer that makes me feel it's unclear to put it in the above way. Since target == 0 only happens when in the previous call, target = nums[i], we know that this is the only combination in this case, so we return 1.

Now we can come up with at least a recursive solution.

int combinationSum4(vector<int>& nums, int target) {
    if (target == 0) return 1;
    int cnt = 0;
    for (int i = 0; i < nums.size(); ++i) {
        if (target >= nums[i]) cnt += combinationSum4(nums, target-nums[i]);
    }
    return cnt;
}

Solution 2: DP

Now for a DP solution, we just need to figure out a way to store the intermediate results, to avoid the same combination sum being calculated many times. We can use an array to save those results, and check if there is already a result before calculation. We can fill the array with -1 to indicate that the result hasn't been calculated yet. 0 is not a good choice because it means there is no combination sum for the target.

version 1: bottom-up 0ms

int combinationSum4(vector<int>& nums, int target) {
    int dp[target+1] = {0};
    dp[0] = 1;

    for (int i = 1; i <= target; ++i) {
        for (int j = 0; j < nums.size(); ++j) {
            if (i >= nums[j]) dp[i] += dp[i-nums[j]];
        }
    }
    return dp[target];
}

version 2: recursive + dp

class Solution {
    int DP(vector<int>& nums, int target, int *dp) {
        if (dp[target] != -1) return dp[target];
        int res = 0;
        for (int i = 0; i < nums.size(); ++i) {
            if (target >= nums[i]) res += DP(nums, target-nums[i], dp);
        }
        dp[target] = res;
        return res;
    }
public:
    int combinationSum4(vector<int>& nums, int target) {
        int dp[target+1];
        dp[0] = 1;
        for (int i = 1; i <= target; ++i) {
            dp[i] = -1;
        }
        return DP(nums, target, dp);
    }
};