执行给定操作后，数组中剩余的最小元素

给定N个整数的数组arr [] ，任务是从数组的两端删除元素，即在一次操作中，可以从数组的当前剩余元素中删除第一个或最后一个元素。此操作需要以这样一种方式执行，即剩下的最后一个元素将具有最小的可能值。打印此最小值。

例子：

Input: arr[] = {5, 3, 1, 6, 9}
Output: 1
Operation 1: arr[] = {5, 3, 1, 6}
Operation 2: arr[] = {5, 3, 1}
Operation 3: arr[] = {3, 1}
Operation 4: arr[] = {1}

Input: arr[] = {2, 6, 4, 8, 2, 6}
Output: 2

方法：可以贪婪地解决此问题，需要在单个操作中删除两端中最大值的元素。按照这种方法，直到数组中只剩下一个元素为止，最后我们将获得原始数组中的最小元素。

下面是上述方法的实现：

C++

// C++ implementation of the approach
#include 
using namespace std;
  
// Function to return the minimum possible
// value of the last element left after
// performing the given operations
int getMin(int arr[], int n)
{
    int minVal = *min_element(arr, arr + n);
    return minVal;
}
  
// Driver code
int main()
{
    int arr[] = { 5, 3, 1, 6, 9 };
    int n = sizeof(arr) / sizeof(arr[0]);
  
    cout << getMin(arr, n);
  
    return 0;
}

Java

// Java implementation of the above approach
import java.util.*;
class GFG
{
  
// Function to return the minimum possible
// value of the last element left after
// performing the given operations
static int getMin(int arr[], int n)
{
    int minVal = Arrays.stream(arr).min().getAsInt();
    return minVal;
}
  
// Driver code
public static void main(String[] args) 
{
    int arr[] = { 5, 3, 1, 6, 9 };
    int n = arr.length;
  
    System.out.println(getMin(arr, n));
}
} 
  
// This code is contributed by 29AjayKumar

Python3

# Python3 implementation of the approach 
  
# Function to return the minimum possible 
# value of the last element left after 
# performing the given operations 
def getMin(arr, n) : 
  
    minVal = min(arr); 
    return minVal; 
  
# Driver code 
if __name__ == "__main__" : 
  
    arr = [ 5, 3, 1, 6, 9 ]; 
    n = len(arr); 
  
    print(getMin(arr, n)); 
  
# This code is contributed by AnkitRai01

C#

// C# implementation of the above approach
using System;
using System.Linq;
  
class GFG
{
  
// Function to return the minimum possible
// value of the last element left after
// performing the given operations
static int getMin(int []arr, int n)
{
    int minVal = arr.Min();
    return minVal;
}
  
// Driver code
public static void Main(String[] args) 
{
    int []arr = { 5, 3, 1, 6, 9 };
    int n = arr.Length;
  
    Console.WriteLine(getMin(arr, n));
}
}
  
// This code is contributed by Rajput-Ji

输出：

1