# 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
3

Input: arr[] = {3, 5, 8, 4, 7, 6, 4, 7}
Output: NO
```

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

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

1. First insert all the elements in a multiset.
2. Remove the first element of the set and store it in a curr variable.
3. 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”.
4. 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 ` `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); ` ` `  `    ``circlePossible(arr, n); ` ` `  `    ``return` `0; ` `} `

Output:
```YES
```

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