C++ ans

vector stones(int n, int a, int b) { if (n <= 1) { return {0}; }

set<int> current;

for (int i = 0; i < n; i++) {
    int value = i * a + (n - 1 - i)*b; // i times a and (n-1-i) times b
    current.insert(value);
}
vector<int> result(current.begin(), current.end());

return result;

}

My solution in C language :

int* stones(int n, int a, int b, int* result_count)
{
    int i = 0, c = (a <= b) ? a : b, d = abs(a - b), t = (a != b) ? n : 1;
    
    int * answer = malloc(t * sizeof(int));
        
    *(answer+0) = 0 + c * (n-1);
        
    for (i = 1; i < t; ++i)
        *(answer+i) = *(answer+i-1) + d;
    
    *result_count = t;
        
    return answer;
}

The solution here is to use the absolute difference between the two differences to increment the initial value (0 + (smallest_value_between_a_and_b * (n-1))) until you reach the expected number of stones (initial value is the first stone).

public static Set<Integer> stones(int num, int first, int second) {
    Set<Integer> finalset = new TreeSet<>(); 
    helper(num, first, second, finalset);
    helper(num, second, first, finalset);
    return finalset;
}

private static void helper(int num, int first, int second, Set<Integer> finalset) {
    int d = num-1;
    int x = d;
    int y = 0;
    for(int i=1; i<=d; i++) {
        List<Integer> lst = new ArrayList<>();
        for(int j = 0; j < x; j++) {
            lst.add(lst.size()>0 ? first + lst.get(lst.size()-1) : first);
        }
        for(int j = 0; j< y; j++) {
            lst.add(lst.size()>0 ? second + lst.get(lst.size()-1) : second);
        }
        x--;
        y++;
        finalset.add(lst.get(lst.size()-1));
    }        
}

We know the output as follow:

• if n=1 output stone will be (0)
• if n=2 output stone will be (a, b)
• if n=3 output stone will be (2a, a+b, 2b)
• if n=4 output stone will be (3a, 2a+b, a+2b, 3b)
• if n=5 output stone will be (4a, 3a+b, 2a+2b, a+3b, 4b) final output can be formulated as (n-i-1)*a + i*b ***where i ranges from 0 to n-1* **

we can iterate over range 'n' and add these values to set remove duplicates.

def stones(n, a, b):
    # Write your code here
    st = set()
    for i in range(n):
        num = (n-i-1)*b + i*a
        st.add(num)

    return sorted(st)

Here is problem solution in python, Java, c++, C and Javascript - https://programmingoneonone.com/hackerrank-manasa-and-stones-problem-solution.html