### 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;
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;
while (1) {

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

}
if (n == 1)
break;

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

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

}
if (a < 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)```
```

## Largest Rectangle

Skyline Real Estate Developers is planning to demolish a number of old, unoccupied buildings and construct a shopping mall in their place. Your task is to find the largest solid area in which the mall can be constructed. There are a number of buildings in a certain two-dimensional landscape. Each building has a height, given by . If you join adjacent buildings, they will form a solid rectangle

## Simple Text Editor

In this challenge, you must implement a simple text editor. Initially, your editor contains an empty string, S. You must perform Q operations of the following 4 types: 1. append(W) - Append W string to the end of S. 2 . delete( k ) - Delete the last k characters of S. 3 .print( k ) - Print the kth character of S. 4 . undo( ) - Undo the last (not previously undone) operation of type 1 or 2,

## Poisonous Plants

There are a number of plants in a garden. Each of the plants has been treated with some amount of pesticide. After each day, if any plant has more pesticide than the plant on its left, being weaker than the left one, it dies. You are given the initial values of the pesticide in each of the plants. Determine the number of days after which no plant dies, i.e. the time after which there is no plan

## AND xor OR

Given an array of distinct elements. Let and be the smallest and the next smallest element in the interval where . . where , are the bitwise operators , and respectively. Your task is to find the maximum possible value of . Input Format First line contains integer N. Second line contains N integers, representing elements of the array A[] . Output Format Print the value

## Waiter

You are a waiter at a party. There is a pile of numbered plates. Create an empty answers array. At each iteration, i, remove each plate from the top of the stack in order. Determine if the number on the plate is evenly divisible ith the prime number. If it is, stack it in pile Bi. Otherwise, stack it in stack Ai. Store the values Bi in from top to bottom in answers. In the next iteration, do the

## Queue using Two Stacks

A queue is an abstract data type that maintains the order in which elements were added to it, allowing the oldest elements to be removed from the front and new elements to be added to the rear. This is called a First-In-First-Out (FIFO) data structure because the first element added to the queue (i.e., the one that has been waiting the longest) is always the first one to be removed. A basic que