Jesse and Cookies


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:

sweetness  Least sweet cookie   2nd least sweet cookie).

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++) {
        int cookie;
        cin >> cookie;
        p.push(cookie * -1);
    }
    
    while (p.top() > k * -1 && p.size() > 1) {
        int cookie1, cookie2, newCookie;
        
        cookie1 = p.top();
        p.pop();
        
        cookie2 = p.top();
        p.pop();
        
        newCookie = cookie1 + 2 * cookie2;
        p.push(newCookie);
        
        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++)
            {
            queue.add(scan.nextInt());
        }
        int count=0;
        while(queue.peek()<k)
            {
            if(queue.size()>=2)
                {
            int x=queue.remove();
            int y=queue.remove();
            y=x+2*y;
            queue.add(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()]
heapq.heapify(cookies)
ops = 0
while (cookies[0] < s):
    try:
        ops += 1
        cookie1 = heapq.heappop(cookies)
        cookie2 = heapq.heappop(cookies)
        newcookie = 1*cookie1 + 2*cookie2
        heapq.heappush(cookies, newcookie)
    except IndexError:
        ops = -1
        break
print(ops)
                        




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