给定一个数组arr[]和一个整数K ，任务是找到所需的合并操作次数，使得数组的所有元素都大于或等于K 。
元素的合并过程——
New Element =
1 * (First Minimum element) +
2 * (Second Minimum element)例子：
Input: arr[] = {1, 2, 3, 9, 10, 12}, K = 7
Output: 2
Explanation:
After the first merge operation elements 1 and 2 is removed,
and the element (1*1 + 2*2 = 5) is inserted into the array
{3, 5, 9, 10, 12}
After the second merge operation elements 3 and 5 is removed,
and the element (3*1 + 5*2 = 13) is inserted into the array
{9, 10, 12, 13}
Thus, 2 operations are required such that all elements are greater than K.
Input: arr[] = {52, 96, 13, 37}, K = 10
Output: 0
Explanation:
All the elements of the array are greater than K already.
Therefore, no merge operation is required.
做法：思路是用Min-Heap来存储元素，然后合并数组的两个最小元素，直到数组的最小元素大于等于K。
下面是上述方法的实现：
C++
// C++ implementation to merge the
// elements of the array untill all
// the array element of the array
// greater than or equal to K
#include
using namespace std;
// Function to find the minimum
// operation required to merge
// elements of the array
int minOperations(int arr[], int K,
int size)
{
int least, second_least,
min_operations = 0,
new_ele = 0, flag = 0;
// Heap to store the elements
// of the array and to extract
// minimum elements of O(logN)
priority_queue,
greater > heap;
// Loop to push all the elements
// of the array into heap
for (int i = 0; i < size; i++) {
heap.push(arr[i]);
}
// Loop to merge the minimum
// elements untill there is only
// all the elements greater than K
while (heap.size() != 1) {
// Condition to check minimum
// element of the array is
// greater than the K
if (heap.top() >= K) {
flag = 1;
break;
}
// Merge the two minimum
// elements of the heap
least = heap.top();
heap.pop();
second_least = heap.top();
heap.pop();
new_ele = (1 * least) +
(2 * second_least);
min_operations++;
heap.push(new_ele);
}
if (heap.top() >= K) {
flag = 1;
}
if (flag == 1) {
return min_operations;
}
return -1;
}
// Driver Code
int main()
{
int N = 6, K = 7;
int arr[] = { 1, 2, 3, 9, 10, 12 };
int size = sizeof(arr) / sizeof(arr[0]);
cout << minOperations(arr, K, size);
return 0;
} Java
// Java implementation to merge the
// elements of the array untill all
// the array element of the array
// greater than or equal to K
import java.util.Collections;
import java.util.PriorityQueue;
class GFG{
// Function to find the minimum
// operation required to merge
// elements of the array
static int minOperations(int arr[], int K,
int size)
{
int least, second_least,
min_operations = 0,
new_ele = 0, flag = 0;
// Heap to store the elements
// of the array and to extract
// minimum elements of O(logN)
PriorityQueue heap = new PriorityQueue<>();
// priority_queue,
// greater > heap;
// Loop to push all the elements
// of the array into heap
for(int i = 0; i < size; i++)
{
heap.add(arr[i]);
}
// Loop to merge the minimum
// elements untill there is only
// all the elements greater than K
while (heap.size() != 1)
{
// Condition to check minimum
// element of the array is
// greater than the K
if (heap.peek() >= K)
{
flag = 1;
break;
}
// Merge the two minimum
// elements of the heap
least = heap.poll();
second_least = heap.poll();
new_ele = (1 * least) +
(2 * second_least);
min_operations++;
heap.add(new_ele);
}
if (heap.peek() >= K)
{
flag = 1;
}
if (flag == 1)
{
return min_operations;
}
return -1;
}
// Driver Code
public static void main(String[] args)
{
int N = 6, K = 7;
int arr[] = { 1, 2, 3, 9, 10, 12 };
int size = arr.length;
System.out.println(minOperations(arr, K, size));
}
}
// This code is contributed by sanjeev2552 Python3
# Python3 implementation to merge the
# elements of the array untill all
# the array element of the array
# greater than or equal to K
# Function to find the minimum
# operation required to merge
# elements of the array
def minOperations(arr, K, size):
least, second_least = 0, 0
min_operations = 0
new_ele, flag = 0, 0
# Heap to store the elements
# of the array and to extract
# minimum elements of O(logN)
heap = []
# Loop to append all the elements
# of the array into heap
for i in range(size):
heap.append(arr[i])
heap = sorted(heap)[::-1]
# Loop to merge the minimum
# elements untill there is only
# all the elements greater than K
while (len(heap) > 0):
# Condition to check minimum
# element of the array is
# greater than the K
if (heap[-1] >= K):
flag = 1
break
# Merge the two minimum
# elements of the heap
least = heap[-1]
del heap[-1]
second_least = heap[-1]
del heap[-1]
new_ele = (1 * least) + (2 * second_least)
min_operations += 1
heap.append(new_ele)
heap = sorted(heap)[::-1]
if (heap[-1] >= K):
flag = 1
if (flag == 1):
return min_operations
return -1
# Driver Code
if __name__ == '__main__':
N, K = 6, 7
arr = [1, 2, 3, 9, 10, 12]
size = len(arr)
print(minOperations(arr, K, size))
# This code is contributed by mohit kumar 29C#
// C# implementation to merge the
// elements of the array untill all
// the array element of the array
// greater than or equal to K
using System;
using System.Collections.Generic;
class GFG
{
// Function to find the minimum
// operation required to merge
// elements of the array
static int minOperations(int[] arr, int K, int size)
{
int least, second_least, min_operations = 0,
new_ele = 0, flag = 0;
// Heap to store the elements
// of the array and to extract
// minimum elements of O(logN)
List heap = new List();
// priority_queue,
// greater > heap;
// Loop to push all the elements
// of the array into heap
for(int i = 0; i < size; i++)
{
heap.Add(arr[i]);
}
heap.Sort();
heap.Reverse();
// Loop to merge the minimum
// elements untill there is only
// all the elements greater than K
while(heap.Count != 1)
{
// Condition to check minimum
// element of the array is
// greater than the K
if(heap[heap.Count - 1] >= K)
{
flag = 1;
break;
}
// Merge the two minimum
// elements of the heap
least = heap[heap.Count - 1];
heap.RemoveAt(heap.Count - 1);
second_least = heap[heap.Count - 1];
heap.RemoveAt(heap.Count - 1);
new_ele = (1 * least) +(2 * second_least);
min_operations++;
heap.Add(new_ele);
heap.Sort();
heap.Reverse();
}
if(heap[heap.Count - 1] >= K)
{
flag = 1;
}
if(flag == 1)
{
return min_operations;
}
return -1;
}
// Driver Code
static public void Main ()
{
int K = 7;
int[] arr = { 1, 2, 3, 9, 10, 12 };
int size = arr.Length;
Console.WriteLine(minOperations(arr, K, size));
}
}
// This code is contributed by avanitrachhadiya2155 输出：
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