# Check if an array has some palindromic subsequence of length at least 3

Given is an array Arr of integers. The task is to determine if the array has any subsequence of at least length 3 that is a palindrome.

Examples:

Input: Arr[] = [1, 2, 1]
Output: YES
Explanation: Here 1 2 1 is a palindrome.

Input: Arr[] = [1, 1, 2, 2, 3, 3, 4, 4, 5, 5]
Output: NO
Explanation: Here no subsequence of length at least 3 exists which is a palindrome.

Approach:

• The idea is to check only for length 3 because if a palindromic subsequence of length greater than 3 exists then there is always a palindromic subsequence of length 3.
• To find the palindromic subsequence of length 3 we just need to find a pair of equal non-adjacent number.

Below is the implementation of the above approach:

## C++

 `// C++ code to check if ` `// palindromic subsequence of` `// length atleast 3 exist or not` `#include` `using` `namespace` `std;`   `string SubPalindrome(``int` `n, ``int` `arr[])` `{` `    ``bool` `ok = ``false``;` `    ``for` `(``int` `i = 0; i < n; i++)` `    ``{` `        ``for``(``int` `j = i + 2; j < n; j++)` `        ``{ ` `            ``if``(arr[i] == arr[j])` `                ``ok = ``true``;` `        ``}` `    ``}` `    ``if``(ok)` `        ``return` `"YES"``;` `    ``else` `        ``return` `"NO"``;` `}` `    `    `// Driver code` `int` `main()` `{`   `    ``// Input values to list` `    ``int` `Arr[] = {1, 2, 2, 3, 2};` `    `  `    ``// Calculating length of array ` `    ``int` `N = ``sizeof``(Arr)/``sizeof``(Arr);` `    `  `    ``cout << SubPalindrome(N, Arr);` `}`   `// This code is contributed by Bhupendra_Singh`

## Java

 `// Java code to check if ` `// palindromic subsequence of` `// length atleast 3 exist or not`   `import` `java.util.*;`   `class` `GFG{` ` `  `static` `String SubPalindrome(``int` `n, ``int` `arr[])` `{` `    ``boolean` `ok = ``false``;` `    ``for` `(``int` `i = ``0``; i < n; i++)` `    ``{` `        ``for``(``int` `j = i + ``2``; j < n; j++)` `        ``{ ` `            ``if``(arr[i] == arr[j])` `                ``ok = ``true``;` `        ``}` `    ``}` `    ``if``(ok)` `        ``return` `"YES"``;` `    ``else` `        ``return` `"NO"``;` `}` `     `  ` `  `// Driver code` `public` `static` `void` `main(String[] args)` `{` ` `  `    ``// Input values to list` `    ``int` `Arr[] = {``1``, ``2``, ``2``, ``3``, ``2``};` `     `  `    ``// Calculating length of array ` `    ``int` `N = Arr.length;` `     `  `    ``System.out.print(SubPalindrome(N, Arr));` `}` `}`   `// This code contributed by sapnasingh4991`

## Python3

 `# Python 3 code to check if ` `# palindromic subsequence of` `# length atleast 3 exist or not` `def` `SubPalindrome (n, arr):` `    ``ok ``=` `False` `    ``for` `i ``in` `range``(n):` `        ``for` `j ``in` `range``(i ``+` `2``, n):` `            ``if` `arr[i] ``=``=` `arr[j]:` `                ``ok ``=` `True` `    ``return``(``'YES'` `if` `ok ``else` `'NO'``)`   `# Driver code`   `# Input values to list` `Arr ``=` `[``1``, ``2``, ``2``, ``3``, ``2``]`   `# Calculating length of the array ` `N ``=` `len``(arr)`   `print` `(SubPalindrome(N, Arr))`

## C#

 `// C# code to check if ` `// palindromic subsequence of` `// length atleast 3 exist or not` `using` `System;`   `public` `class` `GFG{`   `static` `string` `SubPalindrome(``int` `n, ``int` `[]arr)` `{` `    ``bool` `ok = ``false``;` `    ``for``(``int` `i = 0; i < n; i++)` `    ``{` `        ``for``(``int` `j = i + 2; j < n; j++)` `        ``{ ` `            ``if``(arr[i] == arr[j])` `                ``ok = ``true``;` `        ``}` `    ``}` `    ``if``(ok)` `        ``return` `"YES"``;` `    ``else` `        ``return` `"NO"``;` `}` `    `  `// Driver code` `static` `public` `void` `Main ()` `{` `    ``// Input values to list` `    ``int` `[]Arr = { 1, 2, 2, 3, 2 };` `    `  `    ``// Calculating length of array ` `    ``int` `N = Arr.Length;` `    ``Console.WriteLine(SubPalindrome(N, Arr));` `}` `}`   `// This code is contributed by shivanisinghss2110`

## Javascript

 ``

Output:

`YES`

Time Complexity: O(N^2)
Auxiliary Space: O(1)

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