Minimum element left from the array after performing given operations

Given an array arr[] of N integers, the task is to remove the elements from both the ends of the array i.e. in a single operation, either the first or the last element can be removed from the current remaining elements of the array. This operation needs to be performed in such a way that the last element left will have the minimum possible value. Print this minimum value.
Examples: 
 

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 
 

Approach: This problem can be solved greedily, the element with the maximum value from either of the end needs to be removed in a single operation. Following this approach until only one element is left in the array will give us the minimum element from the original array in the end.
Below is the implementation of the above approach: 
 

// C++ implementation of the approach
#include <bits/stdc++.h>
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 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 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# 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

<script>
// javascript implementation of the above approach

    // Function to return the minimum possible
    // value of the last element left after
    // performing the given operations
    function getMin(arr, n)
    {
        var minVal = Math.min.apply(Math, arr);
        return minVal;
    }

    // Driver code
        var arr = [ 5, 3, 1, 6, 9 ];
        var n = arr.length;
        document.write(getMin(arr, n));

// This code is contributed by aashish1995
</script>

1

Time Complexity: O(N), as we are using inbuilt min_element function which will cost us O(N) time.
Auxiliary Space: O(1), as we are not using any extra space.