Sorting array with reverse around middle
Last Updated :
09 Apr, 2023
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Consider the given array arr[], we need to find if we can sort array with the given operation. The operation is
- We have to select a subarray from the given array such that the middle element(or elements (in case of even
number of elements)) of subarray is also the middle element(or elements (in case of even number of elements)) of
the given array. - Then we have to reverse the selected subarray and place this reversed subarray in the array.
We can do the above operation as many times as we want. The task is to find if we can sort array with the given operation.
Examples:
Input : arr[] = {1, 6, 3, 4, 5, 2, 7}
Output : Yes
We can choose sub-array[3, 4, 5] on
reversing this we get [1, 6, 5, 4, 3, 2, 7]
again on selecting [6, 5, 4, 3, 2] and
reversing this one we get [1, 2, 3, 4, 5, 6, 7]
which is sorted at last thus it is possible
to sort on multiple reverse operation.
Input : arr[] = {1, 6, 3, 4, 5, 7, 2}
Output : No
One solution is we can rotate each element around the center, which gives two possibilities in the array i.e. the value at index ‘i’ or the value at index “length – 1 – i”.
If array has n elements then 2^n combinations possible thus running time would be O(2^n).
Another solution can be make copy of the array and sort the copied array. Then compare each element of the sorted array with equivalent element of original array and its mirror image when pivot around center. Sorting the array takes O(n*logn) and 2n comparisons be required thus running time would be O(n*logn).
Implementation:
C++
// CPP program to find possibility to sort// by multiple subarray reverse operation#include <bits/stdc++.h>using namespace std;bool ifPossible(int arr[], int n){ int cp[n]; // making the copy of the original array copy(arr, arr + n, cp); // sorting the copied array sort(cp, cp + n); for (int i = 0; i < n; i++) { // checking mirror image of elements of sorted // copy array and equivalent element of original // array if (!(arr[i] == cp[i]) && !(arr[n - 1 - i] == cp[i])) return false; } return true;}// driver codeint main(){ int arr[] = { 1, 7, 6, 4, 5, 3, 2, 8 }; int n = sizeof(arr) / sizeof(arr[0]); if (ifPossible(arr, n)) cout << "Yes"; else cout << "No"; return 0;} |
Java
// Java program to find possibility to sort// by multiple subarray reverse operationimport java.util.*;class GFG { static boolean ifPossible(int arr[], int n) { // making the copy of the original array int copy[] = Arrays.copyOf(arr, arr.length); // sorting the copied array Arrays.sort(copy); for (int i = 0; i < n; i++) { // checking mirror image of elements of // sorted copy array and equivalent element // of original array if (!(arr[i] == copy[i]) && !(arr[n - 1 - i] == copy[i])) return false; } return true; } // driver code public static void main(String[] args) { int arr[] = { 1, 7, 6, 4, 5, 3, 2, 8 }; int n = arr.length; if (ifPossible(arr, n)) System.out.println("Yes"); else System.out.println("No"); }} |
Python 3
# Python 3 program to find # possibility to sort by# multiple subarray reverse# operationdef ifPossible(arr, n): cp = [0] * n # making the copy of # the original array cp = arr # sorting the copied array cp.sort() for i in range(0 , n) : # checking mirror image of # elements of sorted copy # array and equivalent element # of original array if (not(arr[i] == cp[i]) and not (arr[n - 1 - i] == cp[i])): return False return True# Driver codearr = [1, 7, 6, 4, 5, 3, 2, 8]n = len(arr) if (ifPossible(arr, n)): print("Yes")else: print("No")# This code is contributed by Smitha |
C#
// C# Program to answer queries on sum // of sum of odd number digits of all // the factors of a numberusing System;class GFG { static bool ifPossible(int []arr, int n) { int []cp = new int[n]; // making the copy of the original // array Array.Copy(arr, cp, n); // sorting the copied array Array.Sort(cp); for (int i = 0; i < n; i++) { // checking mirror image of // elements of sorted copy // array and equivalent element // of original array if (!(arr[i] == cp[i]) && !(arr[n - 1 - i] == cp[i])) return false; } return true; } // Driver code public static void Main() { int []arr = new int[]{ 1, 7, 6, 4, 5, 3, 2, 8 }; int n = arr.Length; if (ifPossible(arr, n)) Console.WriteLine( "Yes"); else Console.WriteLine( "No"); }}// This code is contributed by Sam007 |
PHP
<?php// PHP program to find possibility to sort// by multiple subarray reverse operationfunction ifPossible(&$arr, $n){ $cp = array(); // making the copy of the // original array $cp = $arr; // sorting the copied array sort($cp); for ($i = 0; $i < $n; $i++) { // checking mirror image of elements // of sorted copy array and equivalent // element of original array if (!($arr[$i] == $cp[$i]) && !($arr[$n - 1 - $i] == $cp[$i])) return false; } return true;}// Driver code$arr = array(1, 7, 6, 4, 5, 3, 2, 8);$n = sizeof($arr);if (ifPossible($arr, $n)) echo "Yes";else echo "No";// This code is contributed// by Shivi_Aggarwal?> |
Javascript
<script>// Javascript program to find possibility to sort// by multiple subarray reverse operation function ifPossible(arr, n) { // making the copy of the original array let copy = arr; // sorting the copied array copy.sort(); for (let i = 0; i < n; i++) { // checking mirror image of elements of // sorted copy array and equivalent element // of original array if (!(arr[i] == copy[i]) && !(arr[n - 1 - i] == copy[i])) return false; } return true; }// driver code let arr = [ 1, 7, 6, 4, 5, 3, 2, 8 ]; let n = arr.length; if (ifPossible(arr, n)) document.write("Yes"); else document.write("No");; </script> |
Output
Yes
Time Complexity: O(n log n), where n is the size of the input array. This is because of the sorting operation performed on the copied array.
Auxiliary Space: O(n), where n is the size of the input array. This is because of the copy of the original array created.
