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 :



title-img 


                            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)








View More Similar Problems

Truck Tour

Suppose there is a circle. There are N petrol pumps on that circle. Petrol pumps are numbered 0 to (N-1) (both inclusive). You have two pieces of information corresponding to each of the petrol pump: (1) the amount of petrol that particular petrol pump will give, and (2) the distance from that petrol pump to the next petrol pump. Initially, you have a tank of infinite capacity carrying no petr

View Solution →

Queries with Fixed Length

Consider an -integer sequence, . We perform a query on by using an integer, , to calculate the result of the following expression: In other words, if we let , then you need to calculate . Given and queries, return a list of answers to each query. Example The first query uses all of the subarrays of length : . The maxima of the subarrays are . The minimum of these is . The secon

View Solution →

QHEAP1

This question is designed to help you get a better understanding of basic heap operations. You will be given queries of types: " 1 v " - Add an element to the heap. " 2 v " - Delete the element from the heap. "3" - Print the minimum of all the elements in the heap. NOTE: It is guaranteed that the element to be deleted will be there in the heap. Also, at any instant, only distinct element

View Solution →

Jesse and Cookies

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 t

View Solution →

Find the Running Median

The median of a set of integers is the midpoint value of the data set for which an equal number of integers are less than and greater than the value. To find the median, you must first sort your set of integers in non-decreasing order, then: If your set contains an odd number of elements, the median is the middle element of the sorted sample. In the sorted set { 1, 2, 3 } , 2 is the median.

View Solution →

Minimum Average Waiting Time

Tieu owns a pizza restaurant and he manages it in his own way. While in a normal restaurant, a customer is served by following the first-come, first-served rule, Tieu simply minimizes the average waiting time of his customers. So he gets to decide who is served first, regardless of how sooner or later a person comes. Different kinds of pizzas take different amounts of time to cook. Also, once h

View Solution →