Quadratic Application - Google Top Interview Questions

                        Solution in C++ :

vector<int> solve(vector<int>& nums, int a, int b, int c) {
    auto f = [=](int x) { return (a * x + b) * x + c; };
    int n = nums.size();

    vector<int> ans;
    ans.reserve(n);

    // The remaining input to process is the range of indices [i, j)
    int i = 0, j = n;
    while (j > i) {
        // Our two candidates are always on the end of the range.
        int fi = f(nums[i]);
        int fj = f(nums[j - 1]);

        // If `a >= 0`, then `f` is convex-up. In that case the global max (in the range [i, j))
        // is one of the endpoints, and we will eventually converge upon the global min.
        // Otherwise, we symmetrically take the min so as to reach the global max.
        if (a >= 0 ? fi >= fj : fi <= fj) {
            ans.push_back(fi);
            ++i;
        } else {
            ans.push_back(fj);
            --j;
        }
    }

    // If `a >= 0`, then `ans` is decreasing (since we were repeatedly taking the max), so we need
    // to reverse it.
    if (a >= 0) reverse(ans.begin(), ans.end());

    return ans;
}
                    

                        Solution in Python : 
                            
class Solution:
    def solve(self, nums, a, b, c):
        q = deque()

        def cal(x):
            return a * x ** 2 + b * x + c

        l = 0
        r = len(nums) - 1

        is_pos = a >= 0

        while l <= r:
            lv = cal(nums[l])
            rv = cal(nums[r])

            if is_pos:
                if lv >= rv:
                    q.appendleft(lv)
                    l += 1
                else:
                    q.appendleft(rv)
                    r -= 1
            else:
                if lv < rv:
                    q.append(lv)
                    l += 1
                else:
                    q.append(rv)
                    r -= 1

        return list(q)