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 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>
Time Complexity: O(N)
Auxiliary Space: O(1)
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