Leetcode #377 Dynamic Programming Logic Question

Question

I have been struggling to comprehend one of the discussion solutions shown from LC #377 . I'm not grasping how result is being incremented. I can see that there is the line:

result += combinations(nums, dp, target - nums[i])

which obviously is incrementing result, but how exactly is result being incremented? All I can see that is result is initialized to 0 then the line is executed if that if condition is satisfied, so wouldn't the line:

dp[target] = result

always be set to 0?. Really struggling comprehending, so would appreciate any help.

Code:

class Solution {
public int combinationSum4(int[] nums, int target) {
    int[] dp = new int[target + 1];
    Arrays.fill(dp, -1);
    dp[0] = 1;
    return combinations(nums, dp, target);
}

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

}

Answer 1

Focusing reply on your question around result being incremented. This is the code you mentioned:

    int result = 0;
    for (int i = 0; i < nums.length; i++) {
        if (target >= nums[i]) {
            result += combinations(nums, dp, target - nums[i]);
        }
    }
    dp[target] = result;

There is a for loop that increments i from 0 to nums.length-1 . For each such i , if this condition is met (target >= nums[i]) , then result is updated by each call to combinations(nums, dp, target - nums[i]) . You can see from the signature that combinations() returns an int value.

When dp[target] = result is reached, the result has sum of the values returned by each of the calls to combinations() .

The following code may help you. It counts the total even numbers between 1 to n . I wrote it to match the flavor of recursion used in your post. Hope it helps.

public class SumDemo {
    int checkEven(int n) {
        return (n % 2 == 0) ? 1 : 0;
    }

    int countEven(int n) {
        int result = 0;
        for (int i = 1; i <=n; i++) {
            result += checkEven(i);
        }
        return result;
    }

    public static void main(String[] args) {
        SumDemo demo = new SumDemo();
        System.out.println(demo.countEven(10));
    }
}

Result:
5

(Since there are 5 even numbers between 1 to 10 (both inclusive)): 2, 4, 6, 8, 10