# C++ Program to Check if it is possible to make array increasing or decreasing by rotating the array

• Last Updated : 08 Jun, 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:

## C++

 `// C++ implementation of the approach``#include ``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);` `    ``if` `(isPossible(a, n))``        ``cout << ``"Yes"``;``    ``else``        ``cout << ``"No"``;` `    ``return` `0;``}`

Output:

`Yes`

Time Complexity: O(n)

Auxiliary Space: O(1)

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