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Java Program for Check if it is possible to make array increasing or decreasing by rotating the array

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  • Last Updated : 26 May, 2022

Given an array arr[] of N distinct elements, the task is to check if it is possible to make the array increasing or decreasing by rotating the array in any direction.
Examples: 
 

Input: arr[] = {4, 5, 6, 2, 3} 
Output: Yes 
Array can be rotated as {2, 3, 4, 5, 6}
Input: arr[] = {1, 2, 4, 3, 5} 
Output: No 
 

 

Approach: There are four possibilities: 
 

  • If the array is already increasing then the answer is Yes.
  • If the array is already decreasing then the answer is Yes.
  • If the array can be made increasing, this can be possible if the given array is first increasing up to the maximum element and then decreasing.
  • If the array can be made decreasing, this can be possible if the given array is first decreasing up to the minimum element and then increasing.

If it is not possible to make the array increasing or decreasing then print No.
Below is the implementation of the above approach: 
 

Java




// Java implementation of the approach
class GFG
{
     
// Function that returns true if the array
// can be made increasing or decreasing
// after rotating it in any direction
static boolean isPossible(int a[], int n)
{
    // If size of the array is less than 3
    if (n <= 2)
        return true;
 
    int flag = 0;
     
    // Check if the array is already decreasing
    for (int i = 0; i < n - 2; i++)
    {
        if (!(a[i] > a[i + 1] &&
              a[i + 1] > a[i + 2]))
        {
            flag = 1;
            break;
        }
    }
 
    // If the array is already decreasing
    if (flag == 0)
        return true;
 
    flag = 0;
     
    // Check if the array is already increasing
    for (int i = 0; i < n - 2; i++)
    {
        if (!(a[i] < a[i + 1] &&
              a[i + 1] < a[i + 2]))
        {
            flag = 1;
            break;
        }
    }
 
    // If the array is already increasing
    if (flag == 0)
        return true;
 
    // Find the indices of the minimum
    // && the maximum value
    int val1 = Integer.MAX_VALUE, mini = -1,
        val2 = Integer.MIN_VALUE, maxi = 0;
    for (int i = 0; i < n; i++)
    {
        if (a[i] < val1)
        {
            mini = i;
            val1 = a[i];
        }
        if (a[i] > val2)
        {
            maxi = i;
            val2 = a[i];
        }
    }
 
    flag = 1;
     
    // Check if we can make array increasing
    for (int i = 0; i < maxi; i++)
    {
        if (a[i] > a[i + 1])
        {
            flag = 0;
            break;
        }
    }
 
    // If the array is increasing upto max index
    // && minimum element is right to maximum
    if (flag == 1 && maxi + 1 == mini)
    {
        flag = 1;
         
        // Check if array increasing again or not
        for (int i = mini; i < n - 1; i++)
        {
            if (a[i] > a[i + 1])
            {
                flag = 0;
                break;
            }
        }
        if (flag == 1)
            return true;
    }
 
    flag = 1;
     
    // Check if we can make array decreasing
    for (int i = 0; i < mini; i++)
    {
        if (a[i] < a[i + 1])
        {
            flag = 0;
            break;
        }
    }
 
    // If the array is decreasing upto min index
    // && minimum element is left to maximum
    if (flag == 1 && maxi - 1 == mini)
    {
        flag = 1;
 
        // Check if array decreasing again or not
        for (int i = maxi; i < n - 1; i++)
        {
            if (a[i] < a[i + 1])
            {
                flag = 0;
                break;
            }
        }
        if (flag == 1)
            return true;
    }
 
    // If it is not possible to make the
    // array increasing or decreasing
    return false;
}
 
// Driver code
public static void main(String args[])
{
    int a[] = { 4, 5, 6, 2, 3 };
    int n = a.length;
 
    if (isPossible(a, n))
        System.out.println( "Yes");
    else
        System.out.println( "No");
}
}
 
// This code is contributed by Arnab Kundu

Output: 

Yes

 

Time Complexity: O(n), where n represents the size of the given array.

Auxiliary Space: O(1), no extra space is required, so it is a constant.

Please refer complete article on Check if it is possible to make array increasing or decreasing by rotating the array for more details!
 


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