Check if an array is sorted and rotated

Given an array of N distinct integers. The task is to write a program to check if this array is sorted and rotated counter-clockwise. A sorted array is not considered as sorted and rotated, i.e., there should at least one rotation.

Examples:

Input : arr[] = { 3, 4, 5, 1, 2 }
Output : YES
The above array is sorted and rotated.
Sorted array: {1, 2, 3, 4, 5}.
Rotating this sorted array clockwise
by 3 positions, we get: { 3, 4, 5, 1, 2}

Input: arr[] = {7, 9, 11, 12, 5}
Output: YES

Input: arr[] = {1, 2, 3}
Output: NO

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach:

• Find the minimum element in the array.
• Now, if the array is sorted and then rotated then all the elements before minimum element will be in increasing order and all elements after the minimum element will also be in increasing order.
• Check if all elements before minimum element are in increasing order.
• Check if all elements before minimum element are in increasing order.
• Check if the last element of the array is smaller than the element just before the minimum element.
• If all of the above three condition satisfies then print YES otherwise print NO.

Below is the implementation of the above idea:

C++

 // CPP program to check if an array is // sorted and rotated clockwise #include #include    using namespace std;    // Function to check if an array is // sorted and rotated clockwise void checkIfSortRotated(int arr[], int n) {     int minEle = INT_MAX;     int maxEle = INT_MIN;        int minIndex = -1;        // Find the minimum element     // and it's index     for (int i = 0; i < n; i++) {         if (arr[i] < minEle) {             minEle = arr[i];             minIndex = i;         }     }        int flag1 = 1;        // Check if all elements before minIndex     // are in increasing order     for (int i = 1; i < minIndex; i++) {         if (arr[i] < arr[i - 1]) {             flag1 = 0;             break;         }     }        int flag2 = 1;        // Check if all elements after minIndex     // are in increasing order     for (int i = minIndex + 1; i < n; i++) {         if (arr[i] < arr[i - 1]) {             flag2 = 0;             break;         }     }        // Check if last element of the array     // is smaller than the element just     // before the element at minIndex     if (flag1 && flag2 && (arr[n - 1] < arr[minIndex - 1]))         cout << "YES";     else         cout << "NO"; }    // Driver code int main() {     int arr[] = { 3, 4, 5, 1, 2 };     int n = sizeof(arr) / sizeof(arr);     checkIfSortRotated(arr, n);     return 0; }

Java

 // Java program to check if an  // array is sorted and rotated  // clockwise  import java.io.*;    class GFG {    // Function to check if an array is  // sorted and rotated clockwise  static void checkIfSortRotated(int arr[],                                 int n)  {      int minEle = Integer.MAX_VALUE;     int maxEle = Integer.MIN_VALUE;        int minIndex = -1;         // Find the minimum element      // and it's index      for (int i = 0; i < n; i++)      {          if (arr[i] < minEle)          {              minEle = arr[i];              minIndex = i;          }      }         boolean flag1 = true;         // Check if all elements before      // minIndex are in increasing order      for (int i = 1; i < minIndex; i++)      {          if (arr[i] < arr[i - 1])          {              flag1 = false;              break;          }      }         boolean flag2 = true;         // Check if all elements after      // minIndex are in increasing order      for (int i = minIndex + 1; i < n; i++)     {          if (arr[i] < arr[i - 1])          {              flag2 = false;              break;          }      }                //check if the minIndex is 0, i.e the first element is the smallest , which is the case when array is sorted but not rotated.     if(minIndex == 0)       {           System.out.print("NO");           return;       }     // Check if last element of the array      // is smaller than the element just      // before the element at minIndex      if (flag1 && flag2 && (arr[n - 1] <                            arr[minIndex - 1]))          System.out.println("YES");      else         System.out.print("NO");  }     // Driver code  public static void main (String[] args) {     int arr[] = { 3, 4, 5, 1, 2 };             int n = arr.length;            checkIfSortRotated(arr, n);  } }    // This code is contributed  // by inder_verma.

Python3

 # Python3 program to check if an  # array is sorted and rotated clockwise import sys    # Function to check if an array is # sorted and rotated clockwise def checkIfSortRotated(arr, n):     minEle = sys.maxsize     maxEle = -sys.maxsize - 1     minIndex = -1        # Find the minimum element     # and it's index     for i in range(n):         if arr[i]< minEle:             minEle = arr[i]             minIndex = i     flag1 = 1        # Check if all elements before      # minIndex are in increasing order     for i in range(1, minIndex):         if arr[i] < arr[i - 1]:             flag1 = 0             break     flag2 = 2        # Check if all elements after      # minIndex are in increasing order     for i in range(minIndex + 1, n) :         if arr[i] < arr[i - 1]:             flag2 = 0             break        # Check if last element of the array     # is smaller than the element just     # before the element at minIndex     if (flag1 and flag2 and          arr[n - 1] < arr[minIndex - 1]):         print("YES")     else:         print("NO")    # Driver code arr = [3, 4, 5, 1, 2 ] n = len(arr) checkIfSortRotated(arr, n)    # This code is contributed  # by Shrikant13

C#

 // C# program to check if an  // array is sorted and rotated  // clockwise  using System;    class GFG {    // Function to check if an array is  // sorted and rotated clockwise  static void checkIfSortRotated(int []arr,                                 int n)  {      int minEle = int.MaxValue;     //int maxEle = int.MinValue;        int minIndex = -1;         // Find the minimum element      // and it's index      for (int i = 0; i < n; i++)      {          if (arr[i] < minEle)          {              minEle = arr[i];              minIndex = i;          }      }         bool flag1 = true;         // Check if all elements before      // minIndex are in increasing order      for (int i = 1; i < minIndex; i++)      {          if (arr[i] < arr[i - 1])          {              flag1 = false;              break;          }      }         bool flag2 = true;         // Check if all elements after      // minIndex are in increasing order      for (int i = minIndex + 1; i < n; i++)     {          if (arr[i] < arr[i - 1])          {              flag2 = false;              break;          }      }         // Check if last element of the array      // is smaller than the element just      // before the element at minIndex      if (flag1 && flag2 && (arr[n - 1] <                            arr[minIndex - 1]))          Console.WriteLine("YES");      else         Console.WriteLine("NO");  }     // Driver code  public static void Main () {     int []arr = { 3, 4, 5, 1, 2 };             int n = arr.Length;            checkIfSortRotated(arr, n);  } }    // This code is contributed  // by inder_verma.

PHP



Output:

YES

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