Given an integer array arr of size N which contains distinct elements from 1 to N. The task is to check if a position in the array can be found such that all numbers from 1 to N can be counted in a clockwise direction or counterclockwise direction.
Examples:
Input: arr[] = [2, 3, 4, 5, 1]
Output: YES
Explanation:
1 2
5 3
4
If counting is start at index 4
then all numbers can be counted
from 1 to N in a clockwise order.
Input: arr[] = {1, 2, 3, 5, 4]
Output: NO
Explanation:
There is no any index in array
from which given array can count
1 to N in clockwise order or
counterclockwise order.
Approach: The above problem can be solved by observation and analysis.
- An index in the array can be found only if the count of the absolute difference between the consecutive elements greater than 1 is exactly 1, because then only it will be possible to count from 1 to N in clockwise or counterclockwise order.
- If the count of the absolute difference between the adjacent elements is more than 1 Then it will be impossible to count from 1 to N in clockwise or counterclockwise order.
Below is the basic implementation of the above approach:
C++
// C++ program to check Clockwise or// counterclockwise order in an array#include <bits/stdc++.h>using namespace std;bool check_order(vector<int> arr){ int cnt = 0; for (int i = 0; i < arr.size() - 1; i++) { if (abs(arr[i + 1] - arr[i]) > 1) cnt++; } // Comparing the first and last // value of array if (abs(arr[0] - arr[arr.size() - 1]) > 1) cnt++; // If the Count is greater // than 1 then it can't be // represented in required order if (cnt > 1) return false; return true;}// Driver functionint main(){ vector<int> arr = { 2, 3, 4, 5, 1 }; if (check_order(arr)) cout << "YES"; else cout << "NO"; return 0;} |
Java
// Java program to check clockwise or// counterclockwise order in an arrayclass GFG{static boolean check_order(int []arr){ int cnt = 0; for(int i = 0; i < arr.length - 1; i++) { if (Math.abs(arr[i + 1] - arr[i]) > 1) cnt++; } // Comparing the first and last // value of array if (Math.abs(arr[0] - arr[arr.length - 1]) > 1) cnt++; // If the Count is greater // than 1 then it can't be // represented in required order if (cnt > 1) return false; return true;}// Driver codepublic static void main(String[] args){ int []arr = { 2, 3, 4, 5, 1 }; if (check_order(arr)) System.out.print("YES"); else System.out.print("NO");}}// This code is contributed by Amit Katiyar |
Python3
# Python3 program to check clockwise or # counterclockwise order in an array def check_order(arr): cnt = 0 for i in range(len(arr) - 1): if (abs(arr[i + 1] - arr[i]) > 1): cnt += 1 # Comparing the first and last # value of array if (abs(arr[0] - arr[len(arr) - 1]) > 1): cnt += 1 # If the count is greater # than 1 then it can't be # represented in required order if (cnt > 1): return False return True# Driver codearr = [ 2, 3, 4, 5, 1 ]if (check_order(arr)): print("YES")else: print("NO")# This code is contributed by Vishal Maurya. |
C#
// C# program to check clockwise or// counterclockwise order in an arrayusing System;class GFG{static bool check_order(int []arr){ int cnt = 0; for(int i = 0; i < arr.Length - 1; i++) { if (Math.Abs(arr[i + 1] - arr[i]) > 1) cnt++; } // Comparing the first and last // value of array if (Math.Abs(arr[0] - arr[arr.Length - 1]) > 1) cnt++; // If the Count is greater // than 1 then it can't be // represented in required order if (cnt > 1) return false; return true;}// Driver codepublic static void Main(String[] args){ int []arr = { 2, 3, 4, 5, 1 }; if (check_order(arr)) Console.Write("YES"); else Console.Write("NO");}}// This code is contributed by Amit Katiyar |
Javascript
<script>// Javascript program to check clockwise or// counterclockwise order in an array function check_order(arr){ var cnt = 0; for (i = 0; i < arr.length - 1; i++) { if (Math.abs(arr[i + 1] - arr[i]) > 1) cnt++; } // Comparing the first and last // value of array if (Math.abs(arr[0] - arr[arr.length - 1]) > 1) cnt++; // If the Count is greater // than 1 then it can't be // represented in required order if (cnt > 1) return false; return true; } // Driver code var arr = [ 2, 3, 4, 5, 1 ]; if (check_order(arr)) document.write("YES"); else document.write("NO");// This code contributed by gauravrajput1</script> |
Output:
YES
Time Complexity: O(N), as we are using a loop to traverse N times.
Auxiliary Space: O(1), as we are not using any extra space.
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