### 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)


## Self Balancing Tree

An AVL tree (Georgy Adelson-Velsky and Landis' tree, named after the inventors) is a self-balancing binary search tree. In an AVL tree, the heights of the two child subtrees of any node differ by at most one; if at any time they differ by more than one, rebalancing is done to restore this property. We define balance factor for each node as : balanceFactor = height(left subtree) - height(righ

## Array and simple queries

Given two numbers N and M. N indicates the number of elements in the array A[](1-indexed) and M indicates number of queries. You need to perform two types of queries on the array A[] . You are given queries. Queries can be of two types, type 1 and type 2. Type 1 queries are represented as 1 i j : Modify the given array by removing elements from i to j and adding them to the front. Ty

The median M of numbers is defined as the middle number after sorting them in order if M is odd. Or it is the average of the middle two numbers if M is even. You start with an empty number list. Then, you can add numbers to the list, or remove existing numbers from it. After each add or remove operation, output the median. Input: The first line is an integer, N , that indicates the number o

## Maximum Element

You have an empty sequence, and you will be given N queries. Each query is one of these three types: 1 x -Push the element x into the stack. 2 -Delete the element present at the top of the stack. 3 -Print the maximum element in the stack. Input Format The first line of input contains an integer, N . The next N lines each contain an above mentioned query. (It is guaranteed that each

## Balanced Brackets

A bracket is considered to be any one of the following characters: (, ), {, }, [, or ]. Two brackets are considered to be a matched pair if the an opening bracket (i.e., (, [, or {) occurs to the left of a closing bracket (i.e., ), ], or }) of the exact same type. There are three types of matched pairs of brackets: [], {}, and (). A matching pair of brackets is not balanced if the set of bra

## Equal Stacks

ou have three stacks of cylinders where each cylinder has the same diameter, but they may vary in height. You can change the height of a stack by removing and discarding its topmost cylinder any number of times. Find the maximum possible height of the stacks such that all of the stacks are exactly the same height. This means you must remove zero or more cylinders from the top of zero or more of