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# Find the last remaining element after repeated removal of an element from pairs of increasing adjacent array elements

• Last Updated : 14 Sep, 2022

Given an array arr[] consisting of N integers, the task is to print the last remaining array element after repeatedly selecting pairs of increasing adjacent elements (arr[i], arr[i + 1]) and removing any of the elements in the pair. If it is not possible to reduce the array to a single element, then print “Not Possible”.

Examples:

Input: arr[] = {3, 1, 2, 4}
Output: 4
Explanation:
Step 1: Choose a pair (1, 2) and remove 2 from it. Now the array becomes [3, 1, 4].
Step 2: Choose pair (1, 4) and remove 1 from it. Now the array becomes [3, 4].
Step 3: Choose pair (3, 4) and remove 3 from it. Now the array becomes [4].
Hence, the remaining element is 4.

Input: arr[] = {2, 3, 1}
Output: Not Possible

Approach: The given problem can be solved by observing the fact that if the first array element is less than the last array element, then all the elements between them can be removed by performing the given operations. Therefore, simply check if arr[0] < arr[N – 1] or not. If found to be true, print either the first or last array element. Otherwise, print -1.

Below is the implementation of the above approach:

## C++14

 // C++ program for the above approach #include using namespace std;   // Function to print the last remaining // array element after performing // given operations void canArrayBeReduced(int arr[], int N) {     // If size of the array is 1     if (N == 1) {         cout << arr[0];         return;     }       // Check for the condition     if (arr[0] < arr[N - 1]) {         cout << arr[N - 1];     }       // If condition is not satisfied     else         cout << "Not Possible"; }   // Driver Code int main() {     int arr[] = { 6, 5, 2, 4, 1, 3, 7 };     int N = sizeof(arr) / sizeof(arr[0]);       // Function Call     canArrayBeReduced(arr, N);       return 0; }

## Java

 // Java program for the above approach import java.io.*; class GFG {     // Function to print the last remaining   // array element after performing   // given operations   static void canArrayBeReduced(int[] arr, int N)   {       // If size of the array is 1     if (N == 1)     {       System.out.print(arr[0]);       return;     }       // Check for the condition     if (arr[0] < arr[N - 1])     {       System.out.print(arr[N - 1]);     }       // If condition is not satisfied     else       System.out.print("Not Possible");   }     // Driver Code   public static void main(String[] args)   {     int[] arr = { 6, 5, 2, 4, 1, 3, 7 };     int N = arr.length;       // Function Call     canArrayBeReduced(arr, N);   } }   // This code is contributed by Dharanendra L V.

## Python3

 # Python 3 program for the above approach   # Function to print the last remaining # array element after performing # given operations def canArrayBeReduced(arr,  N):       # If size of the array is 1     if (N == 1):         print(arr[0])         return       # Check for the condition     if (arr[0] < arr[N - 1]):         print(arr[N - 1])       # If condition is not satisfied     else:         print("Not Possible")     # Driver Code if __name__ == "__main__":       arr = [6, 5, 2, 4, 1, 3, 7]     N = len(arr)       # Function Call     canArrayBeReduced(arr, N)       # This code is contributed by chitranayal.

## C#

 // C# program for the above approach using System;   public class GFG {     // Function to print the last remaining   // array element after performing   // given operations   static void canArrayBeReduced(int[] arr, int N)   {       // If size of the array is 1     if (N == 1)     {       Console.Write(arr[0]);       return;     }       // Check for the condition     if (arr[0] < arr[N - 1])     {       Console.Write(arr[N - 1]);     }       // If condition is not satisfied     else       Console.Write("Not Possible");   }     // Driver Code   static public void Main()   {     int[] arr = { 6, 5, 2, 4, 1, 3, 7 };     int N = arr.Length;       // Function Call     canArrayBeReduced(arr, N);   } }   // This code is contributed by Dharanendra L V.

## Javascript



Output:

7

Time Complexity: O(1), as we are not using any loops or recursion.

Auxiliary Space: O(1), as we are not using  any extra space.

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