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Modify given array to a non-decreasing array by rotation
  • Difficulty Level : Medium
  • Last Updated : 03 Mar, 2021
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Given an array arr[] of size N (consisting of duplicates), the task is to check if the given array can be converted to a non-decreasing array by rotating it. If it’s not possible to do so, then print “No“. Otherwise, print “Yes“.

Examples:

Input: arr[] = {3, 4, 5, 1, 2}
Output: Yes
Explanation: After 2 right rotations, the array arr[] modifies to {1, 2, 3, 4, 5}

Input: arr[] = {1, 2, 4, 3}
Output: No

Approach: The idea is based on the fact that a maximum of N distinct arrays can be obtained by rotating the given array and check for each individual rotated array, whether it is non-decreasing or not. Follow the steps below to solve the problem:



  • Initialize a vector, say v, and copy all the elements of the original array into it.
  • Sort the vector v.
  • Traverse the original array and perform the following steps:
    • Rotate by 1 in each iteration.
    • If the array becomes equal to vector v, print “Yes“. Otherwise, print “No“.

Below is the implementation of the above approach:

C++




// C++ program for the above approach
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to check if a
// non-decreasing array can be obtained
// by rotating the original array
void rotateArray(vector<int>& arr, int N)
{
    // Stores copy of original array
    vector<int> v = arr;
 
    // Sort the given vector
    sort(v.begin(), v.end());
 
    // Traverse the array
    for (int i = 1; i <= N; ++i) {
 
        // Rotate the array by 1
        rotate(arr.begin(),
               arr.begin() + 1, arr.end());
 
        // If array is sorted
        if (arr == v) {
 
            cout << "YES" << endl;
            return;
        }
    }
 
    // If it is not possible to
    // sort the array
    cout << "NO" << endl;
}
 
// Driver Code
int main()
{
    // Given array
    vector<int> arr = { 3, 4, 5, 1, 2 };
 
    // Size of the array
    int N = arr.size();
 
    // Function call to check if it is possible
    // to make array non-decreasing by rotating
    rotateArray(arr, N);
}

Java




// Java program for the above approach
import java.util.*;
 
class GFG{
 
  // Function to check if a
  // non-decreasing array can be obtained
  // by rotating the original array
  static void rotateArray(int[] arr, int N)
  {
    // Stores copy of original array
    int[] v = arr;
 
    // Sort the given vector
    Arrays.sort(v);
 
    // Traverse the array
    for (int i = 1; i <= N; ++i) {
 
      // Rotate the array by 1
      int x = arr[N - 1];
      i = N - 1;
      while(i > 0){
        arr[i] = arr[i - 1];
        arr[0] = x;
        i -= 1;
      }
 
      // If array is sorted
      if (arr == v) {
 
        System.out.print("YES");
        return;
      }
    }
 
    // If it is not possible to
    // sort the array
    System.out.print("NO");
  }
 
  // Driver Code
  public static void main(String[] args)
  {
 
    // Given array
    int[] arr = { 3, 4, 5, 1, 2 };
 
    // Size of the array
    int N = arr.length;
 
    // Function call to check if it is possible
    // to make array non-decreasing by rotating
    rotateArray(arr, N);
  }
}
 
// This code is contributed by splevel62.

Python3




# Python 3 program for the above approach
 
 
# Function to check if a
# non-decreasing array can be obtained
# by rotating the original array
def rotateArray(arr, N):
   
    # Stores copy of original array
    v = arr
 
    # Sort the given vector
    v.sort(reverse = False)
 
    # Traverse the array
    for i in range(1, N + 1, 1):
       
        # Rotate the array by 1
        x = arr[N - 1]
        i = N - 1
        while(i > 0):
            arr[i] = arr[i - 1]
            arr[0] = x
            i -= 1
             
        # If array is sorted
        if (arr == v):
            print("YES")
            return
 
    # If it is not possible to
    # sort the array
    print("NO")
 
# Driver Code
if __name__ == '__main__':
   
    # Given array
    arr =  [3, 4, 5, 1, 2]
 
    # Size of the array
    N = len(arr)
 
    # Function call to check if it is possible
    # to make array non-decreasing by rotating
    rotateArray(arr, N)
     
    # This code is contributed by ipg2016107.

C#




// C# program to implement
// the above approach
using System;
class GFG
{
   
  // Function to check if a
  // non-decreasing array can be obtained
  // by rotating the original array
  static void rotateArray(int[] arr, int N)
  {
     
    // Stores copy of original array
    int[] v = arr;
 
    // Sort the given vector
    Array.Sort(v);
 
    // Traverse the array
    for (int i = 1; i <= N; ++i) {
 
      // Rotate the array by 1
      int x = arr[N - 1];
      i = N - 1;
      while(i > 0){
        arr[i] = arr[i - 1];
        arr[0] = x;
        i -= 1;
      }
 
      // If array is sorted
      if (arr == v) {
 
        Console.Write("YES");
        return;
      }
    }
 
    // If it is not possible to
    // sort the array
    Console.Write("NO");
  }
 
 
// Driver code
public static void Main()
{
    // Given array
    int[] arr = { 3, 4, 5, 1, 2 };
 
    // Size of the array
    int N = arr.Length;
 
    // Function call to check if it is possible
    // to make array non-decreasing by rotating
    rotateArray(arr, N);
}
}
 
// This code is contributed by susmitakundugoaldanga.
Output
YES

Time Complexity: O(N2)
Auxiliary Space: O(N)

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