**Quadratic Application - Google Top Interview Questions**

### Problem Statement :

You are given a list of integers nums sorted in ascending order, and integers a, b, and c. Apply the following function for each number x in nums: ax^2 + bx + cax 2 +bx+c and return the resulting list in ascending order. This should be done in \mathcal{O}(n)O(n) time. Constraints n ≤ 100,000 where n is the length of nums Example 1 Input nums = [-2, 3] a = 1 b = -3 c = 2 Output [2, 12] Explanation We have nums[0] = 1*-2**2 + -3*-2 + 2 = 4 + 6 + 2 = 12 nums[1] = 1*3**2 + -3*3 + 2 = 9 + -9 + 2 = 2 After we sort [12, 2], we get [2, 12]

### Solution :

` ````
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)
```

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