Given an array of size N, the task is to determine whether its possible to sort the array or not by just one shuffle. In one shuffle, we can shift some contiguous elements from the end of the array and place it in the front of the array.
For eg:
- A = {2, 3, 1, 2}, we can shift {1, 2} from the end of the array to the front of the array to sort it.
- A = {1, 2, 3, 2} since we cannot sort it in one shuffle hence it’s not possible to sort the array.
Examples:
Input: arr[] = {1, 2, 3, 4}
Output: Possible
Since this array is already sorted hence no need for shuffle.
Input: arr[] = {6, 8, 1, 2, 5}
Output: Possible
Place last three elements at the front
in the same order i.e. {1, 2, 5, 6, 8}
Approach:
- Check if the array is already sorted or not. If yes return true.
- Else start traversing the array elements until the current element is smaller than next element. Store that index where arr[i] > arr[i+1].
- Traverse from that point and check if from that index elements are in increasing order or not.
- If above both conditions satisfied then check if last element is smaller than or equal to the first element of given array.
- Print “Possible” if above three conditions satisfied else print “Not possible” if any of the above 3 conditions failed.
Below is the implementation of above approach:
C++
// C++ implementation of above approach#include <bits/stdc++.h>using namespace std;// Function to check if it is possiblebool isPossible(int a[], int n){ // step 1 if (is_sorted(a, a + n)) { cout << "Possible" << endl; } else { // break where a[i] > a[i+1] bool flag = true; int i; for (i = 0; i < n - 1; i++) { if (a[i] > a[i + 1]) { break; } } // break point + 1 i++; // check whether the sequence is // further increasing or not for (int k = i; k < n - 1; k++) { if (a[k] > a[k + 1]) { flag = false; break; } } // If not increasing after break point if (!flag) return false; else { // last element <= First element if (a[n - 1] <= a[0]) return true; else return false; } }}// Driver codeint main(){ int arr[] = { 3, 1, 2, 2, 3 }; int n = sizeof(arr) / sizeof(arr[0]); if (isPossible(arr, n)) cout << "Possible"; else cout << "Not Possible"; return 0;} |
Output:
Possible
Time Complexity: O(n)
Auxiliary Space: O(1)
Please refer complete article on Check if it is possible to sort the array after rotating it for more details!
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