# Check if elements of an array can be arranged in a Circle with consecutive difference as 1

Given an array of numbers. The task is to check if it is possible to arrange all the numbers in a circle so that any two neighboring numbers differ exactly by 1. Print “YES” if it is possible to get such arrangement and “NO” otherwise.

**Examples:**

Input:arr[] = {1, 2, 3, 2}Output:YES The circle formed is: 1 2 2 3Input:arr[] = {3, 5, 8, 4, 7, 6, 4, 7}Output:NO

Below is the step by step algorithm to solve this problem:

- First insert all the elements in a multiset.
- Remove the first element of the set and store it in a
*curr*variable. - Traverse until the size of multiset reduced to 0.
- Remove elements that are 1 greater or 1 smaller than the
*curr*value. - If there is a value with difference more than 1 then “no circle possible”.

- Remove elements that are 1 greater or 1 smaller than the
- Check if it’s initial and final values of
*curr*varaible are same, print “YES” if it is, otherwise print “NO”.

Below is the implementaion of above approach:

`// C++ program to check if elements of array ` `// can be arranged in Circle with consecutive ` `// difference as 1 ` ` ` `#include <bits/stdc++.h> ` `using` `namespace` `std; ` ` ` `// Function to check if elements of array can ` `// be arranged in Circle with consecutive ` `// difference as 1 ` `int` `circlePossible(` `int` `arr[], ` `int` `n) ` `{ ` ` ` `multiset<` `int` `> s; ` ` ` ` ` `// Initialize the multiset with array ` ` ` `// elements ` ` ` `for` `(` `int` `i = 0; i < n; i++) ` ` ` `s.insert(arr[i]); ` ` ` ` ` `// Get a pointer to first element ` ` ` `int` `cur = *s.begin(); ` ` ` ` ` `// Store the first element in a temp variable ` ` ` `int` `start = cur; ` ` ` ` ` `// Remove the first element ` ` ` `s.erase(s.begin()); ` ` ` ` ` `// Traverse until multiset is non-empty ` ` ` `while` `(s.size()) { ` ` ` ` ` `// Elements which are 1 greater than the ` ` ` `// current element, remove their first occurrence ` ` ` `// and increment curr ` ` ` `if` `(s.find(cur + 1) != s.end()) ` ` ` `s.erase(s.find(++cur)); ` ` ` ` ` `// Elements which are 1 less than the ` ` ` `// current element, remove their first occurrence ` ` ` `// and decrement curr ` ` ` `else` `if` `(s.find(cur - 1) != s.end()) ` ` ` `s.erase(s.find(--cur)); ` ` ` ` ` `// If the set is non-empty and contains element ` ` ` `// which differs by curr from more than 1 ` ` ` `// then circle is not possible return ` ` ` `else` `{ ` ` ` `cout << ` `"NO"` `; ` ` ` `return` `0; ` ` ` `} ` ` ` `} ` ` ` ` ` `// Finally, check if curr and first differs by 1 ` ` ` `if` `(` `abs` `(cur - start) == 1) ` ` ` `cout << ` `"YES"` `; ` ` ` `else` ` ` `cout << ` `"NO"` `; ` ` ` ` ` `return` `0; ` `} ` ` ` `// Driver code ` `int` `main() ` `{ ` ` ` `int` `arr[] = { 1, 1, 2, 2, 2, 3 }; ` ` ` ` ` `int` `n = ` `sizeof` `(arr) / ` `sizeof` `(arr[0]); ` ` ` ` ` `circlePossible(arr, n); ` ` ` ` ` `return` `0; ` `} ` |

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**Output:**

YES

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