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 clockwise.
Note: A sorted array is not considered as sorted and rotated, i.e., there should be at least one rotation
Examples:
Input: arr[] = { 3, 4, 5, 1, 2 }
Output: YES
Explanation: 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[] = {3, 4, 6, 1, 2, 5}
Output: NO
Check if an array is sorted and rotated by finding the pivot element(minimum element):
Find the minimum element in the array and if the array has been sorted and rotated then the elements before and after the minimum element will be in sorted order only.
Below is the implementation of the above approach:
C++
#include <bits/stdc++.h> using namespace std; int countAdjacentInversions( int arr[], int n) { int inversions = 0; for ( int i = 0; i < n - 1; i++) { if (arr[i] > arr[i + 1]) { inversions++; } } return inversions; } bool checkRotatedAndSorted( int arr[], int n) { int inversions = countAdjacentInversions(arr, n); // If there are no inversions, the array is sorted. if (inversions == 0) { return true ; } // If there is exactly one inversion and the last element is smaller // than the first element, the array is both sorted and rotated. if (inversions == 1 && arr[n - 1] < arr[0]) { return true ; } // Otherwise, the array is not sorted and rotated. return false ; } int main() { int arr[] = {4, 5, 1, 2, 3}; int n = sizeof (arr) / sizeof (arr[0]); // Function Call if (checkRotatedAndSorted(arr, n)) cout << "YES" << endl; else cout << "NO" << endl; return 0; } |
Java
import java.io.*; class GFG { // Function to count adjacent inversions static int countAdjacentInversions( int arr[], int n) { int inversions = 0 ; for ( int i = 0 ; i < n - 1 ; i++) { if (arr[i] > arr[i + 1 ]) { inversions++; } } return inversions; } // Function to check if an array is sorted and rotated by counting adjacent inversions static boolean checkIfSortRotated( int arr[], int n) { int inversions = countAdjacentInversions(arr, n); // If there is only one inversion and the last element is smaller than the first element if (inversions == 1 && arr[n - 1 ] < arr[ 0 ]) { return true ; } return false ; } // Driver code public static void main(String[] args) { int arr[] = { 5 , 1 , 2 , 3 , 4 }; int n = arr.length; // Function Call boolean x = checkIfSortRotated(arr, n); if (x == true ) System.out.println( "YES" ); else System.out.println( "NO" ); } } |
C#
// C# program to check if an // array is sorted and rotated // clockwise using System; class Program { static bool IsIncRotated( int [] arr, int n, int minIndex, int maxIndex) { if (arr[0] < arr[n - 1]) return false ; return IsIncreasing(arr, 0, maxIndex) && IsIncreasing(arr, minIndex, n - 1); } static bool IsDecRotated( int [] arr, int n, int minIndex, int maxIndex) { if (arr[0] > arr[n - 1]) return false ; return IsDecreasing(arr, 0, minIndex) && IsDecreasing(arr, maxIndex, n - 1); } static bool IsIncreasing( int [] arr, int l, int r) { for ( int i = l + 1; i <= r; i++) { if (arr[i - 1] > arr[i]) return false ; } return true ; } static bool IsDecreasing( int [] arr, int l, int r) { for ( int i = l + 1; i <= r; i++) { if (arr[i - 1] < arr[i]) return false ; } return true ; } static void CheckIfSortRotated( int [] arr) { int minIndex = 0; int maxIndex = 0; // Find the index of the minimum and maximum element for ( int i = 0; i < arr.Length; i++) { if (arr[i] < arr[minIndex]) minIndex = i; if (arr[i] > arr[maxIndex]) maxIndex = i; } bool res = false ; if (maxIndex == minIndex - 1) { res = IsIncRotated(arr, arr.Length, minIndex, maxIndex); } if (minIndex == maxIndex - 1) { res = IsDecRotated(arr, arr.Length, minIndex, maxIndex); } if (res) Console.WriteLine( "YES" ); else Console.WriteLine( "NO" ); } static void Main() { int [] arr = { 10,20,14 }; CheckIfSortRotated(arr); } } // This code is contributed // by susmitabhattacharya2023. |
Javascript
// Javascript program to check if an // array is sorted and rotated // clockwise function isIncRotated(arr, n, minIndex, maxIndex) { if (arr[0] < arr[n - 1]) { return false ; } return isIncreasing(arr, 0, maxIndex) && isIncreasing(arr, minIndex, n - 1); } function isDecRotated(arr, n, minIndex, maxIndex) { if (arr[0] > arr[n - 1]) { return false ; } return isDecreasing(arr, 0, minIndex) && isDecreasing(arr, maxIndex, n - 1); } function isIncreasing(arr, l, r) { for (let i = l + 1; i <= r; i++) { if (arr[i - 1] > arr[i]) { return false ; } } return true ; } function isDecreasing(arr, l, r) { for (let i = l + 1; i <= r; i++) { if (arr[i - 1] < arr[i]) { return false ; } } return true ; } function checkIfSortRotated(arr) { let minEle = Infinity; let maxEle = -Infinity; let minIndex = 0; let maxIndex = 0; // Find the index of the minimum and maximum element for (let i = 0; i < arr.length; i++) { if (arr[i] < arr[minIndex]) { minIndex = i; } if (arr[i] > arr[maxIndex]) { maxIndex = i; } } let res = false ; if (maxIndex === minIndex - 1) { res = isIncRotated(arr, arr.length, minIndex, maxIndex); } if (minIndex === maxIndex - 1) { res = isDecRotated(arr, arr.length, minIndex, maxIndex); } if (res) { console.log( "YES" ); } else { console.log( "NO" ); } } let arr = [3, 4, 5, 1, 2]; checkIfSortRotated(arr); // This code is contributed by susmitabhattacharya2023. |
Python3
# Python3 program to check if an # array is sorted and rotated clockwise def isIncRotated(arr, n, minIndex, maxIndex): if arr[ 0 ] < arr[n - 1 ]: return False return isIncreasing(arr, 0 , maxIndex) and isIncreasing(arr, minIndex, n - 1 ) def isDecRotated(arr, n, minIndex, maxIndex): if arr[ 0 ] > arr[n - 1 ]: return False return isDecreasing(arr, 0 , minIndex) and isDecreasing(arr, maxIndex, n - 1 ) def isIncreasing(arr, l, r): for i in range (l + 1 , r + 1 ): if arr[i - 1 ] > arr[i]: return False return True def isDecreasing(arr, l, r): for i in range (l + 1 , r + 1 ): if arr[i - 1 ] < arr[i]: return False return True def checkIfSortRotated(arr): minEle = float ( 'inf' ) maxEle = float ( '-inf' ) minIndex = 0 maxIndex = 0 # Find the index of the minimum & maximum element for i in range ( len (arr)): if arr[i] < arr[minIndex]: minIndex = i if arr[i] > arr[maxIndex]: maxIndex = i res = False if maxIndex = = minIndex - 1 : res = isIncRotated(arr, len (arr), minIndex, maxIndex) if minIndex = = maxIndex - 1 : res = isDecRotated(arr, len (arr), minIndex, maxIndex) if res: print ( "YES" ) else : print ( "NO" ) # Driver code arr = [ 3 , 4 , 5 , 1 , 2 ] checkIfSortRotated(arr) # This code is contributed # by susmitabhattacharya2023 |
YES
Time Complexity: O(N)
Auxiliary Space: O (1)
Check if an array is sorted and rotated by counting adjacent inversions:
As the array is sorted and rotated, so the case when the previous element is greater than the current element will occur only once. If this occurs zero times or more than one times then the array is not properly sorted and rotated
Follow the given steps to solve the problem:
- Take two variables to say x and y, initialized as 0
- Now traverse the array
- If the previous element is smaller than the current, increment x by one
- Else increment y by one.
- After traversal, if y is not equal to 1, return false.
- Then compare the last element with the first element (0th element as current, and last element as previous.) i.e. if the last element is greater increase y by one else increase x by one
- Again check if y equals one return true. i.e. Array is sorted and rotated. Else return false
Below is the implementation of the above approach:
C++
#include <bits/stdc++.h> using namespace std; bool checkRotatedAndSorted( int arr[], int n) { // Your code here // Initializing two variables x,y as zero. int x = 0, y = 0; // Traversing array 0 to last element. // n-1 is taken as we used i+1. for ( int i = 0; i < n - 1; i++) { if (arr[i] < arr[i + 1]) x++; else y++; } // If till now both x,y are greater than 1 means // array is not sorted. If both any of x,y is zero // means array is not rotated. if (y == 1) { // Checking for last element with first. if (arr[n - 1] < arr[0]) x++; else y++; // Checking for final result. if (y == 1) return true ; } // If still not true then definitely false. return false ; } // Driver Code Starts. int main() { int arr[] = { 4, 5, 1, 2, 3 }; int n = sizeof (arr) / sizeof (arr[0]); // Function Call if (checkRotatedAndSorted(arr, n)) cout << "YES" << endl; else cout << "NO" << endl; return 0; } |
Java
import java.io.*; class GFG { // Function to check if an array is // Sorted and rotated clockwise static boolean checkIfSortRotated( int arr[], int n) { // Initializing two variables x,y as zero. int x = 0 , y = 0 ; // Traversing array 0 to last element. // n-1 is taken as we used i+1. for ( int i = 0 ; i < n - 1 ; i++) { if (arr[i] < arr[i + 1 ]) x++; else y++; } // If till now both x,y are greater // than 1 means array is not sorted. // If both any of x,y is zero means // array is not rotated. if (y == 1 ) { // Checking for last element with first. if (arr[n - 1 ] < arr[ 0 ]) x++; else y++; // Checking for final result. if (y == 1 ) return true ; } // If still not true then definitely false. return false ; } // Driver code public static void main(String[] args) { int arr[] = { 5 , 1 , 2 , 3 , 4 }; int n = arr.length; // Function Call boolean x = checkIfSortRotated(arr, n); if (x == true ) System.out.println( "YES" ); else System.out.println( "NO" ); } } |
C#
using System; public class GFG { // Function to check if an array is // Sorted and rotated clockwise static bool checkIfSortRotated( int [] arr, int n) { // Initializing two variables x,y as zero. int x = 0, y = 0; // Traversing array 0 to last element. // n-1 is taken as we used i+1. for ( int i = 0; i < n - 1; i++) { if (arr[i] < arr[i + 1]) x++; else y++; } // If till now both x,y are greater // than 1 means array is not sorted. // If both any of x,y is zero means // array is not rotated. if (y == 1) { // Checking for last element with first. if (arr[n - 1] < arr[0]) x++; else y++; // Checking for readonly result. if (y == 1) return true ; } // If still not true then definitely false. return false ; } // Driver code public static void Main(String[] args) { int [] arr = { 5, 1, 2, 3, 4 }; int n = arr.Length; // Function Call bool x = checkIfSortRotated(arr, n); if (x == true ) Console.WriteLine( "YES" ); else Console.WriteLine( "NO" ); } } // This code is contributed by umadevi9616 |
Javascript
// JavaScript Function to check if an array is // Sorted and rotated clockwise function checkIfSortRotated(arr, n) { // Initializing two variables x,y as zero. var x = 0, y = 0; // Traversing array 0 to last element. // n-1 is taken as we used i+1. for ( var i = 0; i < n - 1; i++) { if (arr[i] < arr[i + 1]) x++; else y++; } // If till now both x,y are greater // than 1 means array is not sorted. // If both any of x,y is zero means // array is not rotated. if (y == 1) { // Checking for last element with first. if (arr[n - 1] < arr[0]) x++; else y++; // Checking for final result. if (y == 1) return true ; } // If still not true then definitely false. return false ; } // Driver code var arr = [ 5, 1, 2, 3, 4 ]; var n = arr.length; // Function Call var x = checkIfSortRotated(arr, n); if (x == true ) document.write( "YES" ); else document.write( "NO" ); // This code is contributed by shivanisinghss2110 |
Python3
def checkRotatedAndSorted(arr, n): # Your code here # Initializing two variables x,y as zero. x = 0 y = 0 # Traversing array 0 to last element. # n-1 is taken as we used i+1. for i in range (n - 1 ): if (arr[i] < arr[i + 1 ]): x + = 1 else : y + = 1 # If till now both x,y are greater than 1 means # array is not sorted. If both any of x,y is zero # means array is not rotated. if (y = = 1 ): # Checking for last element with first. if (arr[n - 1 ] < arr[ 0 ]): x + = 1 else : y + = 1 # Checking for final result. if (y = = 1 ): return True # If still not true then definitely false. return False # Driver Code Starts. arr = [ 4 , 5 , 1 , 2 , 3 ] n = len (arr) # Function Call if (checkRotatedAndSorted(arr, n)): print ( "YES" ) else : print ( "NO" ) # This code is contributed by shivanisinghss2110 |
YES
Time Complexity: O(N)
Auxiliary Space: O(1)
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