### 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)

## 2D Array-DS

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## Dynamic Array

Create a list, seqList, of n empty sequences, where each sequence is indexed from 0 to n-1. The elements within each of the n sequences also use 0-indexing. Create an integer, lastAnswer, and initialize it to 0. There are 2 types of queries that can be performed on the list of sequences: 1. Query: 1 x y a. Find the sequence, seq, at index ((x xor lastAnswer)%n) in seqList.

## Left Rotation

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## Array Manipulation

Starting with a 1-indexed array of zeros and a list of operations, for each operation add a value to each of the array element between two given indices, inclusive. Once all operations have been performed, return the maximum value in the array. Example: n=10 queries=[[1,5,3], [4,8,7], [6,9,1]] Queries are interpreted as follows: a b k 1 5 3 4 8 7 6 9 1 Add the valu

## Print the Elements of a Linked List

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