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:Â 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:
Java
import java.util.*;
class GFG{
static void rotateArray( int [] arr, int N)
{
int [] v = arr;
Arrays.sort(v);
for ( int i = 1 ; i <= N; ++i) {
int x = arr[N - 1 ];
i = N - 1 ;
while (i > 0 ){
arr[i] = arr[i - 1 ];
arr[ 0 ] = x;
i -= 1 ;
}
if (arr == v) {
System.out.print( "YES" );
return ;
}
}
System.out.print( "NO" );
}
public static void main(String[] args)
{
int [] arr = { 3 , 4 , 5 , 1 , 2 };
int N = arr.length;
rotateArray(arr, N);
}
}
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Time Complexity: O(N2)
Auxiliary Space: O(N)
Please refer complete article on Modify given array to a non-decreasing array by rotation for more details!