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 elementsvoid 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 codeint 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 elementsdef lastTwoElement(A, n): # Find the minimum and the maximum # element from the array minn = min(A) maxx = max(A) print(minn, maxx)# Driver codeA = [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)
Auxiliary Space: O(1)
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