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

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

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// 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;
}

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Java

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// 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

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Python

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# 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

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C#

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// 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 

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Output:

7 102

Time Complexity: O(n)



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Improved By : mohit kumar 29, AnkitRai01