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

Last Updated : 11 Nov, 2021

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:

C++

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;

// Function that returns true if the array
// can be made increasing or decreasing
// after rotating it in any direction
bool 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] and 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] and 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
    // and the maximum value
    int val1 = INT_MAX, mini = -1, val2 = INT_MIN, maxi;
    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
    // and minimum element is right to maximum
    if (flag == 1 and 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
    // and minimum element is left to maximum
    if (flag == 1 and 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
int main()
{
    int a[] = { 4, 5, 6, 2, 3 };
    int n = sizeof(a) / sizeof(a[0]);

    if (isPossible(a, n))
        cout << "Yes";
    else
        cout << "No";

    return 0;
}

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

Python3

# Python3 implementation of the approach
import sys

# Function that returns True if the array
# can be made increasing or decreasing
# after rotating it in any direction
def isPossible(a, n):

    # If size of the array is less than 3
    if (n <= 2):
        return True;

    flag = 0;
    
    # Check if the array is already decreasing
    for i in range(n - 2):
        if (not(a[i] > a[i + 1] and 
                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 i in range(n - 2): 
        if (not(a[i] < a[i + 1] and 
                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
    # and the maximum value
    val1 = sys.maxsize; mini = -1; 
    val2 = -sys.maxsize; maxi = -1;
    for i in range(n): 
        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 i in range(maxi):
        if (a[i] > a[i + 1]): 
            flag = 0;
            break;

    # If the array is increasing upto max index
    # and minimum element is right to maximum
    if (flag == 1 and maxi + 1 == mini): 
        flag = 1;
        
        # Check if array increasing again or not
        for i in range(mini, n - 1):
            if (a[i] > a[i + 1]): 
                flag = 0;
                break;
            
        if (flag == 1):
            return True;
    
    flag = 1;
    
    # Check if we can make array decreasing
    for i in range(mini):
        if (a[i] < a[i + 1]): 
            flag = 0;
            break;
        
    # If the array is decreasing upto min index
    # and minimum element is left to maximum
    if (flag == 1 and maxi - 1 == mini): 
        flag = 1;

        # Check if array decreasing again or not
        for i in range(maxi, n - 1):
            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
a = [ 4, 5, 6, 2, 3 ];
n = len(a);

if (isPossible(a, n)):
    print("Yes");
else:
    print("No");

# This code is contributed by Rajput-Ji

// C# implementation of the approach 
using System;

class GFG 
{ 
        
    // Function that returns true if the array 
    // can be made increasing or decreasing 
    // after rotating it in any direction 
    static bool 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 = int.MaxValue, mini = -1, 
            val2 = int.MinValue, 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() 
    { 
        int []a = { 4, 5, 6, 2, 3 }; 
        int n = a.Length; 
    
        if (isPossible(a, n)) 
            Console.WriteLine( "Yes"); 
        else
            Console.WriteLine( "No"); 
    } 
}

// This code is contributed by AnkitRai01

JavaScript

<script>
// javascript implementation of the approach    
// Function that returns true if the array
    // can be made increasing or decreasing
    // after rotating it in any direction
    function isPossible(a , n) {
        // If size of the array is less than 3
        if (n <= 2)
            return true;

        var flag = 0;

        // Check if the array is already decreasing
        for (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 (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
        var val1 = Number.MAX_VALUE, mini = -1, val2 = Number.MIN_VALUE, maxi = 0;
        for (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 (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 (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 (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 (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
    
        var a = [ 4, 5, 6, 2, 3 ];
        var n = a.length;

        if (isPossible(a, n))
            document.write("Yes");
        else
            document.write("No");

// This code contributed by umadevi9616
</script>

Output

Yes

Time Complexity: O(N)

Auxiliary Space: O(1)

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

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

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