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Find last two remaining elements after removing median of any 3 consecutive elements repeatedly

  • Last Updated : 10 May, 2021

Given a sequence A1, A2, A3, … An of distinct integers. The task is to find the last 2 remaining elements after removing the median of any 3 consecutive elements repeatedly from the sequence.
Examples: 
 

Input: A[] = {2, 5, 3} 
Output: 2 5 
Median of {2, 5, 3} is 3, after removing 
it the remaining elements are {2, 5}.
Input: A[] = {38, 9, 102, 10, 96, 7, 46, 28, 88, 13} 
Output: 7 102 
 

 

Approach: For every operation, the median element is the element which is neither the maximum nor the minimum. So, after applying the operation, neither the minimum nor the maximum element is affected. After generalizing this, it can be seen that the final array shall contain only the minimum and the maximum element from the initial array.
Below is the implementation of the above approach: 
 

CPP




// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to find the last
// two remaining elements
void lastTwoElement(int A[], int n)
{
 
    // Find the minimum and the maximum
    // element from the array
    int minn = *min_element(A, A + n);
    int maxx = *max_element(A, A + n);
 
    cout << minn << " " << maxx;
}
 
// Driver code
int main()
{
    int A[] = { 38, 9, 102, 10, 96,
                7, 46, 28, 88, 13 };
    int n = sizeof(A) / sizeof(int);
 
    lastTwoElement(A, n);
 
    return 0;
}

Java




// Java implementation of the approach
class GFG
{
     
    static int min_element(int A[], int n)
    {
        int min = A[0];
        for(int i = 0; i < n; i++)
            if (A[i] < min )
                min = A[i];
                 
        return min;
         
    }
     
    static int max_element(int A[], int n)
    {
        int max = A[0];
        for(int i = 0; i < n; i++)
            if (A[i] > max )
                max = A[i];
                 
        return max;
    }
     
    // Function to find the last
    // two remaining elements
    static void lastTwoElement(int A[], int n)
    {
     
        // Find the minimum and the maximum
        // element from the array
        int minn = min_element(A, n);
        int maxx = max_element(A, n);
     
        System.out.println(minn + " " + maxx);
    }
     
    // Driver code
    public static void main (String[] args)
    {
        int A[] = { 38, 9, 102, 10, 96,
                    7, 46, 28, 88, 13 };
                     
        int n = A.length;
     
        lastTwoElement(A, n);
    }
}
 
// This code is contributed by AnkitRai01

Python




# Python3 implementation of the approach
 
# Function to find the last
# two remaining elements
def lastTwoElement(A, n):
 
    # Find the minimum and the maximum
    # element from the array
    minn = min(A)
    maxx = max(A)
 
    print(minn, maxx)
 
# Driver code
A = [38, 9, 102, 10, 96,7, 46, 28, 88, 13]
n = len(A)
 
lastTwoElement(A, n)
 
# This code is contributed by mohit kumar 29

C#




// C# implementation of the approach
using System;
 
class GFG
{
    static int min_element(int []A, int n)
    {
        int min = A[0];
        for(int i = 0; i < n; i++)
            if (A[i] < min )
                min = A[i];
                 
        return min;
    }
     
    static int max_element(int []A, int n)
    {
        int max = A[0];
        for(int i = 0; i < n; i++)
            if (A[i] > max )
                max = A[i];
                 
        return max;
    }
     
    // Function to find the last
    // two remaining elements
    static void lastTwoElement(int []A, int n)
    {
     
        // Find the minimum and the maximum
        // element from the array
        int minn = min_element(A, n);
        int maxx = max_element(A, n);
     
        Console.WriteLine(minn + " " + maxx);
    }
     
    // Driver code
    public static void Main ()
    {
        int []A = { 38, 9, 102, 10, 96,
                    7, 46, 28, 88, 13 };
                     
        int n = A.Length;
     
        lastTwoElement(A, n);
    }
}
 
// This code is contributed by AnkitRai01

Javascript




<script>
 
 
// Java Script implementation of the approach
function min_element(A,n)
    {
        let min = A[0];
        for(let i = 0; i < n; i++)
            if (A[i] < min )
                min = A[i];
                 
        return min;
         
    }
     
    function max_element(A,n)
    {
        let max = A[0];
        for(let i = 0; i < n; i++)
            if (A[i] > max )
                max = A[i];
                 
        return max;
    }
     
    // Function to find the last
    // two remaining elements
    function lastTwoElement(A,n)
    {
     
        // Find the minimum and the maximum
        // element from the array
        let minn = min_element(A, n);
        let maxx = max_element(A, n);
     
        document.write(minn + " " + maxx);
    }
     
    // Driver code
     
        let A = [ 38, 9, 102, 10, 96,
                    7, 46, 28, 88, 13 ];
                     
        let n = A.length;
     
        lastTwoElement(A, n);
     
// This code is contributed by sravan kumar Gottumukkala
</script>
Output: 



7 102

 

Time Complexity: O(n)
 

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