# Check if every index i has an index j such that sum of elements in both directions are equal

Given a circular array of size N. The task is to check if, for every index i starting from 0 to N-1, there exists an index j which is not equal to i such that sum of all the numbers in the clockwise direction from i to j is equal to the sum of all numbers in the anticlockwise direction from i to j.

Examples:

Input: a[] = {1, 4, 1, 4}
Output: Yes
The circular array is 1->4->1->4.
for index 0, j will be 2, then sum of elements from index 0 to 2 in clockwise direction will be 1 + 4 + 1 = 6 and the sum of elements from index 0 to 2 in anticlockwise direction will be 1 + 4 + 1 = 6.

for index 1, j will be 3, then sum of elements from index 1 to 3 in clockwise direction will be 4 + 1 + 4 = 9 and the sum of elements from index 1 to 3 in anticlockwise direction will be 4 + 1 + 4 = 9.

for index 2, j will be 0, then sum of elements from index 2 to 0 in clockwise direction will be 1 + 4 + 1 = 6 and the sum of elements from index 2 to 0 in anticlockwise direction will be 1 + 4 + 1 = 6

for index 3, j will be 1, then sum of elements from index 3 to 1 in clockwise direction will be 4 + 1 + 4 = 9 and the sum of elements from index 3 to 1 in anticlockwise direction will be 4 + 1 + 4 = 9

Input: a[] = {1, 1, 1, 1, 1, 1}
Output: Yes

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

Approach:

• When N is odd, the answer will be “NO” always as it is not possible to find an index j for every index i.
• If N is even, then check if the opposite element is exactly same for every index i, then the answer is “YES”.
• If the any of the index’s opposite element i.e., a[(i+(n/2))] is not equal to a[i], then the answer will be “NO”.

Below is the implementaion of above approach:

## C++

 `// C++ program to check if the ` `// number lies in given range ` `#include ` `using` `namespace` `std; ` ` `  `// Function that returns the maximum element. ` `bool` `check(``int` `a[], ``int` `n) ` `{ ` ` `  `    ``// check for odd ` `    ``if` `(n % 2 == 1) ` `        ``return` `false``; ` ` `  `    ``// check if the opposite element is same ` `    ``// as a[i] ` `    ``for` `(``int` `i = 0; i < n / 2; i++) { ` `        ``if` `(a[i] != a[i + (n / 2)]) ` `            ``return` `false``; ` `    ``} ` ` `  `    ``return` `true``; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `a[] = { 1, 4, 1, 4 }; ` ` `  `    ``int` `n = ``sizeof``(a) / ``sizeof``(a[0]); ` ` `  `    ``if` `(check(a, n)) ` `        ``cout << ``"YES"``; ` `    ``else` `        ``cout << ``"NO"``; ` ` `  `    ``return` `0; ` `} `

## Java

 `//Java program to check if the ` `//number lies in given range ` ` `  `public` `class` `GFG { ` ` `  `    ``//Function that returns the maximum element. ` `    ``static` `boolean` `check(``int` `a[], ``int` `n) ` `    ``{ ` ` `  `     ``// check for odd ` `     ``if` `(n % ``2` `== ``1``) ` `         ``return` `false``; ` ` `  `     ``// check if the opposite element is same ` `     ``// as a[i] ` `     ``for` `(``int` `i = ``0``; i < n / ``2``; i++) { ` `         ``if` `(a[i] != a[i + (n / ``2``)]) ` `             ``return` `false``; ` `     ``} ` ` `  `     ``return` `true``; ` `    ``} ` ` `  `    ``//Driver code ` `    ``public` `static` `void` `main(String[] args) { ` `         `  `        ``int` `a[] = { ``1``, ``4``, ``1``, ``4` `}; ` ` `  `         ``int` `n = a.length; ` ` `  `         ``if` `(check(a, n)) ` `             ``System.out.println(``"YES"``); ` `         ``else` `             ``System.out.println(``"NO"``); ` `    ``} ` `} `

## Python 3

 `# Python 3 Program  to check if the  ` `# number lies in given range  ` ` `  `# Function that returns the maximum element. ` `def` `check(a, n) : ` ` `  `    ``# check for odd  ` `    ``if` `n ``%` `2` `=``=` `1``: ` `        ``return` `False` ` `  `    ``# check if the opposite element is same  ` `    ``# as a[i]  ` `    ``for` `i ``in` `range``(n``/``/``2``) : ` `        ``if` `a[i] !``=` `a[i ``+` `(n``/``/``2``)]: ` `            ``return` `False` ` `  `    ``return` `True` ` `  `# Driver Code ` `if` `__name__ ``=``=` `"__main__"` `: ` `    ``a ``=` `[ ``1``, ``4``, ``1``, ``4``] ` ` `  `    ``n ``=` `len``(a) ` ` `  `    ``if` `check(a, n) : ` `        ``print``(``"YES"``) ` `    ``else` `: ` `        ``print``(``"NO"``) ` `     `  `# This code is contributed by ANKITRAI1 `

## C#

 `// C# program to check if the ` `// number lies in given range ` ` `  `class` `GFG  ` `{ ` ` `  `// Function that returns the ` `// maximum element. ` `static` `bool` `check(``int``[] a, ``int` `n) ` `{ ` ` `  `// check for odd ` `if` `(n % 2 == 1) ` `    ``return` `false``; ` ` `  `// check if the opposite ` `// element is same as a[i] ` `for` `(``int` `i = 0; i < (``int``)n / 2; i++)  ` `{ ` `    ``if` `(a[i] != a[i + (``int``)(n / 2)]) ` `        ``return` `false``; ` `} ` ` `  `return` `true``; ` `} ` ` `  `// Driver code ` `public` `static` `void` `Main() ` `{ ` `    ``int``[] a = ``new` `int``[]{ 1, 4, 1, 4 }; ` ` `  `    ``int` `n = a.Length; ` ` `  `    ``if` `(check(a, n)) ` `        ``System.Console.WriteLine(``"YES"``); ` `    ``else` `        ``System.Console.WriteLine(``"NO"``); ` `} ` `} ` ` `  `// This code is contributed by mits `

## PHP

 `

Output:

```YES
```

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.

My Personal Notes arrow_drop_up

Striver(underscore)79 at Codechef and codeforces D

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.

Article Tags :
Practice Tags :

Be the First to upvote.

Please write to us at contribute@geeksforgeeks.org to report any issue with the above content.