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Check if it is possible to sort the array after rotating it

  • Difficulty Level : Easy
  • Last Updated : 22 Mar, 2021

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: 
 

  1. 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.
  2. 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: 
 

  1. Check if the array is already sorted or not. If yes return true.
  2. Else start traversing the array elements until the current element is smaller than next element. Store that index where arr[i] > arr[i+1].
  3. Traverse from that point and check if from that index elements are in increasing order or not.
  4. If above both conditions satisfied then check if last element is smaller than or equal to the first element of given array.
  5. 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 possible
bool 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 code
int 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;
}

Java




// Java implementation of above approach
class solution
{
    //check if array is sorted
static boolean is_sorted(int a[],int n)
{
    int c1=0,c2=0;
    //if array is ascending
    for(int i=0;i<n-1;i++)
    {
        if(a[i]<=a[i+1])
        c1++;
    }
     
    //if array is descending
    for(int i=1;i<n;i++)
    {
        if(a[i]<=a[i-1])
        c2++;
    }
    if(c1==n||c2==n)
    return true;
     
    return false;
}
// Function to check if it is possible
static boolean isPossible(int a[], int n)
{
    // step 1
    if (is_sorted(a,n)) {
        System.out.println("Possible");
    }
   
    else {
   
        // break where a[i] > a[i+1]
        boolean 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;
        }
    }
    return false;
}
   
// Driver code
public static void main(String[] args)
{
   
    int arr[] = { 3, 1, 2, 2, 3 };
    int n = arr.length;
   
    if (isPossible(arr, n))
        System.out.println("Possible");
   
    else
        System.out.println("Not Possible");
   
}
}
//contributed by Arnab Kundu

Python 3




# Python 3 implementation of
# above approach
def is_sorted(a):
    all(a[i] <= a[i + 1]
    for i in range(len(a) - 1))
     
# Function to check if
# it is possible
def isPossible(a, n):
 
    # step 1
    if (is_sorted(a)) :
        print("Possible")
     
    else :
 
        # break where a[i] > a[i+1]
        flag = True
        for i in range(n - 1) :
            if (a[i] > a[i + 1]) :
                break
             
        # break point + 1
        i += 1
 
        # check whether the sequence is
        # further increasing or not
        for k in range(i, n - 1) :
            if (a[k] > a[k + 1]) :
                flag = False
                break
 
        # If not increasing after
        # break point
        if (not flag):
            return False
 
        else :
 
            # last element <= First element
            if (a[n - 1] <= a[0]):
                return True
 
            else:
                return False
 
# Driver code
if __name__ == "__main__":
 
    arr = [ 3, 1, 2, 2, 3 ]
    n = len(arr)
 
    if (isPossible(arr, n)):
        print("Possible")
 
    else:
        print("Not Possible")
 
# This code is contributed
# by ChitraNayal

C#




// C# implementation of above approach
using System;
class GFG
{
// check if array is sorted
static bool is_sorted(int []a, int n)
{
    int c1 = 0, c2 = 0;
     
    // if array is ascending
    for(int i = 0; i < n - 1; i++)
    {
        if(a[i] <= a[i + 1])
        c1++;
    }
     
    // if array is descending
    for(int i = 1; i < n; i++)
    {
        if(a[i] <= a[i - 1])
        c2++;
    }
    if(c1 == n || c2 == n)
    return true;
     
    return false;
}
 
// Function to check if it is possible
static bool isPossible(int []a, int n)
{
    // step 1
    if (is_sorted(a,n))
    {
        Console.WriteLine("Possible");
    }
 
    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;
        }
    }
    return false;
}
 
// Driver code
public static void Main()
{
 
    int []arr = { 3, 1, 2, 2, 3 };
    int n = arr.Length;
 
    if (isPossible(arr, n))
        Console.WriteLine("Possible");
 
    else
        Console.WriteLine("Not Possible");
}
}
 
// This code is contributed by anuj_67

PHP




<?php
// PHP implementation of
// above approach
 
// Function to check if
// it is possible
function is_sorted($a, $n)
{
    $c1 = 0; $c2 = 0;
     
    // if array is ascending
    for($i = 0; $i < $n - 1; $i++)
    {
        if($a[$i] <= $a[$i + 1])
        $c1++;
    }
     
    // if array is descending
    for($i = 1; $i < $n; $i++)
    {
        if($a[$i] <= $a[$i - 1])
        $c2++;
    }
     
    if($c1 == $n || $c2 == $n)
    return true;
     
    return false;
}
 
function isPossible($a, $n)
{
    // step 1
    if (is_sorted($a, $n))
    {
        echo "Possible" . "\n";
    }
 
    else
    {
 
        // break where a[i] > a[i+1]
        $flag = true;
        $i;
        for ($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 ($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 code
$arr = array( 3, 1, 2, 2, 3 );
$n = sizeof($arr);
 
if (isPossible($arr, $n))
    echo "Possible";
else
    echo "Not Possible";
 
// This code is contributed
// by Akanksha Rai(Abby_akku)
?>

Javascript




<script>
// Javascript implementation of above approach 
// check if array is sorted
    function is_sorted(a)
    {
        let c1=0,c2=0;
        // if array is ascending
        for(let i=0;i<n-1;i++)
        {
            if(a[i]<=a[i+1])
            {
                c1++;
            }
        }
        // if array is descending
        for(let i=1;i<n;i++)
        {
            if(a[i]<=a[i-1])
                c2++;
        }
        if(c1==n||c2==n)
        {
            return true;
        }
        return false;
    }
    // Function to check if it is possible
    function isPossible(a,n)
    {
        // step 1
        if (is_sorted(a,n))
        {
            document.write("Possible");   
        }
        else
        {
            // break where a[i] > a[i+1]
            let flag = true;
            let 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 (let 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
            }
        }
             
        return false;
    }
    // Driver code 
    let arr=[3, 1, 2, 2, 3];
    let n = arr.length;
    if(isPossible(arr, n))
        document.write("Possible");
    else
        document.write("Not Possible");
         
    // This code is contributed by avanitrachhadiya2155
</script>
Output: 
Possible

 




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