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 = 9Input:a[] = {1, 1, 1, 1, 1, 1}Output:Yes

**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 implementation of above approach:

## C++

`// C++ program to check if the` `// number lies in given range` `#include <bits/stdc++.h>` `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

`<?php` `// PHP program to check if the` `// number lies in given range` `// Function that returns the` `// maximum element.` `function` `check(` `$a` `, ` `$n` `)` `{` ` ` `// check for odd` ` ` `if` `(` `$n` `% 2 == 1)` ` ` `return` `false;` ` ` `// check if the opposite` ` ` `// element is same as a[i]` ` ` `for` `(` `$i` `= 0; ` `$i` `< ` `$n` `/ 2; ` `$i` `++)` ` ` `{` ` ` `if` `(` `$a` `[` `$i` `] != ` `$a` `[` `$i` `+ (` `$n` `/ 2)])` ` ` `return` `false;` ` ` `}` ` ` `return` `true;` `}` `// Driver code` `$a` `= ` `array` `( 1, 4, 1, 4 );` `$n` `= sizeof(` `$a` `);` `if` `(check(` `$a` `, ` `$n` `))` ` ` `echo` `"YES"` `;` `else` ` ` `echo` `"NO"` `;` `// This code is contributed` `// by Akanksha Rai(Abby_akku)` |

## Javascript

`<script>` `// Javascript program to check if the` `// number lies in given range` `// Function that returns the maximum element.` `function` `check( a, n)` `{` ` ` `// check for odd` ` ` `if` `(n % 2 == 1)` ` ` `return` `false` `;` ` ` `// check if the opposite element is same` ` ` `// as a[i]` ` ` `for` `(` `var` `i = 0; i < parseInt(n / 2); i++) {` ` ` `if` `(a[i] != a[i + parseInt(n / 2)])` ` ` `return` `false` `;` ` ` `}` ` ` `return` `true` `;` `}` `// Driver code` `var` `a = [ 1, 4, 1, 4 ];` `var` `n = a.length;` `if` `(check(a, n))` ` ` `document.write(` `"YES"` `);` `else` ` ` `document.write(` `"NO"` `);` `// This code is contributed by rrrtnx.` `</script>` |

**Output:**

YES

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