### Problem Statement :

```Jesse loves cookies. He wants the sweetness of all his cookies to be greater than value K. To do this, Jesse repeatedly mixes two cookies with the least sweetness. He creates a special combined cookie with:

He repeats this procedure until all the cookies in his collection have a sweetness > =  K.
You are given Jesse's cookies. Print the number of operations required to give the cookies a sweetness > = K . Print -1 if this isn't possible.

Input Format

The first line consists of integers N, the number of cookies and K, the minimum required sweetness, separated by a space.
The next line contains N integers describing the array A  where Ai is the sweetness of the ith cookie in Jesse's collection.

Output Format

Output the number of operations that are needed to increase the cookie's sweetness > = K.
Output -1 if this isn't possible.```

### Solution :

```                            ```Solution in C :

In C ++ :

#include <cmath>
#include <cstdio>
#include <vector>
#include <iostream>
#include <algorithm>

#include <queue>

using namespace std;

int main() {
/* Enter your code here. Read input from STDIN. Print output to STDOUT */

int n, k, operations = 0;;
cin >> n;
cin >> k;

priority_queue<int> p;

for (int a = 0; a < n; a++) {
}

while (p.top() > k * -1 && p.size() > 1) {

p.pop();

p.pop();

operations++;
}

if (p.top() > k * -1) cout << "-1";
else cout << operations;

return 0;
}

In Java :

import java.io.*;
import java.util.*;
import java.text.*;
import java.math.*;
import java.util.regex.*;

public class Solution {

public static void main(String[] args) {

Scanner scan=new Scanner(System.in);
int n=scan.nextInt();
int k=scan.nextInt();
PriorityQueue<Integer> queue=new PriorityQueue<Integer>();
for(int i=0;i<n;i++)
{
}
int count=0;
while(queue.peek()<k)
{
if(queue.size()>=2)
{
int x=queue.remove();
int y=queue.remove();
y=x+2*y;
count++;
}
else
{
count=-1;
break;
}
}
System.out.println(count);
}
}

In C :

#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>

long int n;
long int aa[1000002];
void siftdown(long int *a, long int n, long int start) {
long int r, lch, swap, t;
//printf("##%d %d##\n", start, a[start]);
r = start;
while (r * 2 <= n) {
lch = 2 * r;
swap = r;
if (a[lch] < a[swap]) {
swap = lch;
}
if (lch+1 <= n && a[lch+1] < a[swap]) {
swap = lch+1;
}
if (swap == r) {
return;
} else {
t = a[swap];
a[swap] = a[r];
a[r] = t;
r = swap;
}
}
}

void heapify(long int *a, long int n)
{
long int start;
start = n/2;
while (start >= 1) {
siftdown(a, n, start);
start --;
//  print();
}
}
void print(long int *a) {
long int i;
for (i = 1; i <= n; i++)
printf("%ld ", a[i]);
printf("\n");
}
int main() {
long int k, i, t, *a, tmp;
scanf("%ld%ld", &n, &k);

for(i = 1; i <= n; i++)
scanf("%ld", &aa[i]);
//   print(aa);
t = 0;
heapify(aa, n);
//    print(aa);
a = &aa[0];
while (1) {

//     print();
if (a[1] >= k)
break;
if (n == 2) {
n--;
if (a[1] < a[2]) {
a[1] = a[1] + 2 * a[2];
} else {
a[1] = a[2] + 2 * a[1];
}
t++;

}
if (n == 1)
break;

if (a[2] < a[3]) {
a[2] = a[1] + 2*a[2];
siftdown(a, n, 2);
} else {
a[3] = a[1] + 2*a[3];
siftdown(a, n, 3);

}
a[1] = a[n];
n--;
siftdown(a, n, 1);
t++;
// print(a);

}
if (a[1] < k)
printf("-1\n");
else
printf("%ld\n", t);
// print();

return 0;
}

In python3 :

import heapq
n, s = [int(i) for i in input().split()]
cookies = [int(i) for i in input().split()]
ops = 0
try:
ops += 1
except IndexError:
ops = -1
break
print(ops)```
```

## Reverse a doubly linked list

This challenge is part of a tutorial track by MyCodeSchool Given the pointer to the head node of a doubly linked list, reverse the order of the nodes in place. That is, change the next and prev pointers of the nodes so that the direction of the list is reversed. Return a reference to the head node of the reversed list. Note: The head node might be NULL to indicate that the list is empty.

## Tree: Preorder Traversal

Complete the preorder function in the editor below, which has 1 parameter: a pointer to the root of a binary tree. It must print the values in the tree's preorder traversal as a single line of space-separated values. Input Format Our test code passes the root node of a binary tree to the preOrder function. Constraints 1 <= Nodes in the tree <= 500 Output Format Print the tree's

## Tree: Postorder Traversal

Complete the postorder function in the editor below. It received 1 parameter: a pointer to the root of a binary tree. It must print the values in the tree's postorder traversal as a single line of space-separated values. Input Format Our test code passes the root node of a binary tree to the postorder function. Constraints 1 <= Nodes in the tree <= 500 Output Format Print the

## Tree: Inorder Traversal

In this challenge, you are required to implement inorder traversal of a tree. Complete the inorder function in your editor below, which has 1 parameter: a pointer to the root of a binary tree. It must print the values in the tree's inorder traversal as a single line of space-separated values. Input Format Our hidden tester code passes the root node of a binary tree to your \$inOrder* func

## Tree: Height of a Binary Tree

The height of a binary tree is the number of edges between the tree's root and its furthest leaf. For example, the following binary tree is of height : image Function Description Complete the getHeight or height function in the editor. It must return the height of a binary tree as an integer. getHeight or height has the following parameter(s): root: a reference to the root of a binary

## Tree : Top View

Given a pointer to the root of a binary tree, print the top view of the binary tree. The tree as seen from the top the nodes, is called the top view of the tree. For example : 1 \ 2 \ 5 / \ 3 6 \ 4 Top View : 1 -> 2 -> 5 -> 6 Complete the function topView and print the resulting values on a single line separated by space.