# Count of unique palindromic strings of length X from given string

Given a string s and an integer X, our task is to find the number of distinct palindromic strings of length X from the given string.

Examples:

Input: s = “aaa”, X = 2
Output: 1
Explanation:
Here all the characters of the string is the same so we can only make a one different string {aa} of length 2.

Input: s = “aabaabbc”, X = 3
Output: 6
Explanation:
To make a palindromic string of length 3 we have 6 possible strings aaa, aba, aca, bab, bcb, bbb.

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

Naive Approach: The naive method is to generate all the possible subsequence of length X, then check if that subsequence forms a palindrome or not.

Time Complexity: O(2N)
Auxiliary Space: O(X)

Efficient Approach: The idea is to find the frequency of all the characters of the string. Count the different numbers of the characters we can put at the particular position and decrease the number of count by 2 because for palindromic string the position (i) and (X – i) will be the same. If the length of X is odd then we have to put the only single char at that position X/2.

Below is the implementation of the above approach:

## C++

 `// C++ implementation to count different ` `// palindromic string of length X ` `// from the given string S ` ` `  `#include ` `using` `namespace` `std; ` ` `  `// Function to count different ` `// palindromic string of length X ` `// from the given string S ` `long` `long` `findways(string s, ``int` `x) ` `{ ` `    ``// Base case ` `    ``if` `(x > (``int``)s.length()) ` `        ``return` `0; ` ` `  `    ``long` `long` `int` `n = (``int``)s.length(); ` ` `  `    ``// Create the frequency array ` `    ``int` `freq; ` ` `  `    ``// Intitalise frequency array with 0 ` `    ``memset``(freq, 0, ``sizeof` `freq); ` ` `  `    ``// Count the frequency in the string ` `    ``for` `(``int` `i = 0; i < n; ++i) ` `        ``freq[s[i] - ``'a'``]++; ` ` `  `    ``multiset<``int``> se; ` ` `  `    ``for` `(``int` `i = 0; i < 26; ++i) ` `        ``if` `(freq[i] > 0) ` `            ``// Store frequency of the char ` `            ``se.insert(freq[i]); ` ` `  `    ``long` `long` `ans = 1; ` ` `  `    ``for` `(``int` `i = 0; i < x / 2; ++i) { ` `        ``long` `long` `int` `count = 0; ` `        ``for` `(``auto` `u : se) { ` `            ``// check the frequency which ` `            ``// is greater than zero ` `            ``if` `(u >= 2) ` ` `  `                ``// No. of different char we can ` `                ``// put at the position of ` `                ``// the i and x - i ` `                ``count++; ` `        ``} ` ` `  `        ``if` `(count == 0) ` `            ``return` `0; ` ` `  `        ``else` `            ``ans = ans * count; ` ` `  `        ``// Iterator pointing to the ` `        ``// last element of the set ` `        ``auto` `p = se.end(); ` `        ``p--; ` `        ``int` `val = *p; ` `        ``se.erase(p); ` `        ``if` `(val > 2) ` `            ``// decrease the value of the char ` `            ``// we put on the position i and n - i ` `            ``se.insert(val - 2); ` `    ``} ` ` `  `    ``if` `(x % 2 != 0) { ` `        ``long` `long` `int` `count = 0; ` `        ``for` `(``auto` `u : se) ` `            ``// different no of char we can ` `            ``// put at the position x/2 ` `            ``if` `(u > 0) ` `                ``count++; ` ` `  `        ``ans = ans * count; ` `    ``} ` ` `  `    ``// Return total no of ` `    ``// different string ` `    ``return` `ans; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``string s = ``"aaa"``; ` `    ``int` `x = 2; ` `    ``cout << findways(s, x); ` ` `  `    ``return` `0; ` `} `

## Python3

 `# Python3 implementation to count  ` `# different palindromic string of ` `# length X from the given string S ` ` `  `# Function to count different ` `# palindromic string of length X ` `# from the given string S ` `def` `findways(s, x): ` ` `  `    ``# Base case ` `    ``if``(x > ``len``(s)): ` `        ``return` `0` ` `  `    ``n ``=` `len``(s) ` ` `  `    ``# Create the frequency array ` `    ``# Intitalise frequency array with 0 ` `    ``freq ``=` `[``0``] ``*` `26` ` `  `    ``# Count the frequency in the string ` `    ``for` `i ``in` `range``(n): ` `        ``freq[``ord``(s[i]) ``-` `ord``(``'a'``)] ``+``=` `1` ` `  `    ``se ``=` `set``() ` ` `  `    ``for` `i ``in` `range``(``26``): ` `        ``if``(freq[i] > ``0``): ` `             `  `            ``# Store frequency of the char ` `            ``se.add(freq[i]) ` ` `  `    ``ans ``=` `1` `    ``for` `i ``in` `range``(x ``/``/` `2``): ` `        ``count ``=` `0` `        ``for` `u ``in` `se: ` `             `  `            ``# Check the frequency which ` `            ``# is greater than zero ` `            ``if``(u >``=` `2``): ` ` `  `                ``# No. of different char we can ` `                ``# put at the position of ` `                ``# the i and x - i ` `                ``count ``+``=` `1` ` `  `        ``if``(count ``=``=` `0``): ` `            ``return` `0` `        ``else``: ` `            ``ans ``*``=` `count ` ` `  `        ``# Iterator pointing to the ` `        ``# last element of the set ` `        ``p ``=` `list``(se) ` `        ``val ``=` `p[``-``1``] ` `        ``p.pop(``-``1``) ` `        ``se ``=` `set``(p) ` ` `  `        ``if``(val > ``2``): ` `             `  `            ``# Decrease the value of the char ` `            ``# we put on the position i and n - i ` `            ``se.add(val ``-` `2``) ` ` `  `    ``if``(x ``%` `2` `!``=` `0``): ` `        ``count ``=` `0` `        ``for` `u ``in` `se: ` `             `  `            ``# Different no of char we can ` `            ``# put at the position x/2 ` `            ``if``(u > ``0``): ` `                ``count ``+``=` `1` ` `  `        ``ans ``=` `ans ``*` `count ` ` `  `    ``# Return total no of ` `    ``# different string  ` `    ``return` `ans ` ` `  `# Driver code ` `if` `__name__ ``=``=` `'__main__'``: ` ` `  `    ``s ``=` `"aaa"` `    ``x ``=` `2` `     `  `    ``print``(findways(s, x)) ` ` `  `# This code is contributed by Shivam Singh `

Output:

```1
```

Time Complexity: O(N + 26 * X)

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