Compruebe si es posible hacer que la array aumente o disminuya girando la array

Dada una array arr[] de N elementos distintos, la tarea es verificar si es posible hacer que la array aumente o disminuya rotando la array en cualquier dirección.
Ejemplos: 
 

Entrada: arr[] = {4, 5, 6, 2, 3} 
Salida: Sí  La
array se puede rotar como {2, 3, 4, 5, 6}
Entrada: arr[] = {1, 2, 4, 3 , 5} 
Salida: No 
 

Enfoque: Hay cuatro posibilidades: 
 

• Si la array ya está aumentando , la respuesta es Sí .
• Si la array ya está disminuyendo , la respuesta es Sí .
• Si la array se puede hacer creciente, esto puede ser posible si la array dada primero aumenta hasta el elemento máximo y luego disminuye.
• Si la array se puede hacer decreciente, esto puede ser posible si la array dada primero disminuye hasta el elemento mínimo y luego aumenta.

Si no es posible hacer que la array aumente o disminuya, imprima No.
A continuación se muestra la implementación del enfoque anterior: 
 

// 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 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 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

<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>

Yes

Complejidad de tiempo: O(N)

Espacio Auxiliar: O(1)