{
    stack<int> a1;
    stack<int> b1;
    int val = 0;
    int tmp = 0;
    int t = 0;
    for (int i = a.size() - 1; i >= 0; i--)
    {
        a1.push(a[i]);
    }
    for (int i = b.size() - 1; i >= 0; i--)
    {
        b1.push(b[i]);
    }

    while (tmp <= maxSum && (!a1.empty() || !b1.empty()))
    {
        val = t;

        if (!a1.empty() && (b1.empty() || a1.top() <= b1.top()))
        {
            tmp += a1.top();
            a1.pop();
            t += 1;
        }
        else if (!b1.empty())
        {
            tmp += b1.top();
            b1.pop();
            t += 1;
        }
    }
    return val;
}

what is the problem

You just nailed it. I was looking it for my gaming project.

This question is very poorly described , and its misleading once you go from one stack to other you can't go back but its not mentioned properly

guys this is not a Greedy question where we can pick min of both the stacks. that doesn't work. We need to try all the possibilities. Recursion & Backtracking

Only 2 test cases (0,4)are passing all other are failing , can't think where is it going wrong public static int solve(Stack f,Stack s,int len , int cs,int max){ int n = f.peek()>=s.peek()?s.peek():f.peek(); // System.out.println(f.peek()); // System.out.println(s.peek()); // System.out.println("n:"+n); if(cs+n a, List b) { // Write your code here Stack first = new Stack<>(); for(int i =a.size()-1;i>=0;i--){ first.push(a.get(i)); } // System.out.println;

Stack<Integer> Second = new Stack<>();
for(int i =b.size()-1;i>=0;i--){
    Second.push(b.get(i));
}
int len = 0;
int currsumm = 0;
return solve(first,Second,len,currsumm,maxSum);

}

}