L377. Combination Sum IV

dp

Problem:

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.

Solution:

• 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.

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