给定一个大小为N的数组arr[] ，任务是通过从数组中重复删除一对(例如(arr[i], arr[j])并插入它们的平均值的 Ceil 值)来找到可能的最小剩余数组元素.
例子：
Input: arr[] = { 1, 2, 3 }
Output:
Explanation:
Removing the pair (arr[1], arr[2]) from arr[] and inserting (arr[1] + arr[2] + 1) / 2 into arr[] modifies arr[] to { 1, 2 }.
Removing the pair (arr[0], arr[1]) from arr[] and inserting (arr[0] + arr[1] + 1) / 2 into arr[] modifies arr[] to { 2 }.
Therefore, the required output is 2.
Input: arr[] = { 30, 16, 40 }
Output: 26
Explanation:
Removing the pair (arr[0], arr[2]) from arr[] and inserting (arr[0] + arr[2] + 1) / 2 into arr[] modifies arr[] to { 16, 35 } .
Removing the pair (arr[0], arr[1]) from arr[] and inserting (arr[0] + arr[1] + 1) / 2 into arr[] modifies arr[] to { 26 } .
Therefore, the required output is 26.
方法：该问题可以使用贪心技术解决。这个想法是重复删除第一个和第二大数组元素并插入它们的平均值。最后，打印数组中剩下的最小元素。
- 初始化一个priority_queue，比如PQ ，来存储数组元素，这样最大的元素总是出现在PQ的顶部。
- 遍历数组并将所有数组元素存储在PQ 中。
- 遍历priority_queue的元素，而在priority_queue元件的计数大于1。在每次迭代中，从PQ 中弹出前两个元素并插入它们平均值的Ceil 值。
- 最后，打印PQ 中剩下的元素。
下面是上述方法的实现：
C++
// C++ program to implement
// the above approach
#include
using namespace std;
// Function to find the smallest element
// left in the array by removing pairs
// and inserting their average
int findSmallestNumLeft(int arr[], int N)
{
// Stores array elements such that the
// largest element present at top of PQ
priority_queue PQ;
// Traverse the array
for (int i = 0; i < N; i++) {
// Insert arr[i] into PQ
PQ.push(arr[i]);
}
// Iterate over elements of PQ while count
// of elements in PQ greater than 1
while (PQ.size() > 1) {
// Stores largest element of PQ
int top1 = PQ.top();
// Pop the largest element of PQ
PQ.pop();
// Stores largest element of PQ
int top2 = PQ.top();
// Pop the largest element of PQ
PQ.pop();
// Insert the ceil value of average
// of top1 and top2
PQ.push((top1 + top2 + 1) / 2);
}
return PQ.top();
}
// Driver Code
int main()
{
int arr[] = { 30, 16, 40 };
int N = sizeof(arr)
/ sizeof(arr[0]);
cout << findSmallestNumLeft(
arr, N);
return 0;
} Java
// Java program to implement
// the above approach
import java.util.PriorityQueue;
class GFG{
// Function to find the smallest element
// left in the array by removing pairs
// and inserting their average
static int findSmallestNumLeft(int arr[], int N)
{
// Stores array elements such that the
// largest element present at top of PQ
PriorityQueue PQ = new PriorityQueue((a,b)->b-a);
// Traverse the array
for (int i = 0; i < N; i++)
{
// Insert arr[i] into PQ
PQ.add(arr[i]);
}
// Iterate over elements of PQ while count
// of elements in PQ greater than 1
while (PQ.size() > 1)
{
// Stores largest element of PQ
int top1 = PQ.peek();
// Pop the largest element of PQ
PQ.remove();
// Stores largest element of PQ
int top2 = PQ.peek();
// Pop the largest element of PQ
PQ.remove();
// Insert the ceil value of average
// of top1 and top2
PQ.add((top1 + top2 + 1) / 2);
}
return PQ.peek();
}
// Driver Code
public static void main(String[] args)
{
int arr[] = { 30, 16, 40 };
int N = arr.length;
System.out.print(findSmallestNumLeft(
arr, N));
}
}
// This code is contributed by Amit Katiyar Python3
# Python3 program to implement
# the above approach
# Function to find the smallest element
# left in the array by removing pairs
# and inserting their average
def findSmallestNumLeft(arr, N):
# Stores array elements such that the
# largest element present at top of PQ
PQ = []
# Traverse the array
for i in range(N):
# Insert arr[i] into PQ
PQ.append(arr[i])
# Iterate over elements of PQ while count
# of elements in PQ greater than 1
PQ = sorted(PQ)
while (len(PQ) > 1):
# Stores largest element of PQ
top1 = PQ[-1]
# Pop the largest element of PQ
del PQ[-1]
# Stores largest element of PQ
top2 = PQ[-1]
# Pop the largest element of PQ
del PQ[-1]
# Insert the ceil value of average
# of top1 and top2
PQ.append((top1 + top2 + 1) // 2)
PQ = sorted(PQ)
return PQ[-1]
# Driver Code
if __name__ == '__main__':
arr = [ 30, 16, 40 ]
N = len(arr)
print (findSmallestNumLeft(arr, N))
# This code is contributed by mohit kumar 29C#
// C# program to implement
// the above approach
using System;
using System.Collections.Generic;
class GfG
{
// Function to find the smallest element
// left in the array by removing pairs
// and inserting their average
static int findSmallestNumLeft(int[] arr, int N)
{
// Stores array elements such that the
// largest element present at top of PQ
List PQ = new List();
// Traverse the array
for (int i = 0; i < N; i++) {
// Insert arr[i] into PQ
PQ.Add(arr[i]);
}
PQ.Sort();
PQ.Reverse();
// Iterate over elements of PQ while count
// of elements in PQ greater than 1
while (PQ.Count > 1) {
// Stores largest element of PQ
int top1 = PQ[0];
// Pop the largest element of PQ
PQ.RemoveAt(0);
// Stores largest element of PQ
int top2 = PQ[0];
// Pop the largest element of PQ
PQ.RemoveAt(0);
// Insert the ceil value of average
// of top1 and top2
PQ.Add((top1 + top2 + 1) / 2);
PQ.Sort();
PQ.Reverse();
}
return PQ[0];
}
// Driver code
public static void Main()
{
int[] arr = { 30, 16, 40 };
int N = arr.Length;
Console.Write(findSmallestNumLeft(arr, N));
}
}
// This code is contributed by divyeshrabadiya07. 输出：
26时间复杂度： O(N * log(N))
辅助空间： O(N)
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