# Count of K-size substrings having palindromic permutations

Given a string str consist of only lowercase alphabets and an integer K, the task is to count the number of substrings of size K such that any permutation of the substring is a palindrome.
Examples:

Input: str = “abbaca”, K = 3
Output:
Explanation:
The substrings of size 3 whose any permutation is palindrome are {“abb”, “bba”, “aca”}.

Input: str = “aaaa”, K = 1
Output:
Explanation:
The substrings of size 1 whose any permutation is palindrome are {‘a’, ‘a’, ‘a’, ‘a’}.

Naive Approach: A naive solution is to run two-loop to generate all substring of size K. For each substring formed, find the frequency of each character of the substring. If at most one character has an odd frequency, then one of its permutations will be a palindrome. Increment the count for the current substring and print the final count after all the operations.

Time Complexity: O(N*K)

Efficient Approach: This problem can be solved efficiently by using Window Sliding Technique and using frequency array of size 26. Below are the steps:

1. Store the frequency of first K elements of the given string in frequency array(say freq[]).
2. Using frequency array, check the count of elements having an odd frequency, if it is less than 2, then increment
count of palindromic permutation.
3. Now, linearly slide the window ahead till it reaches the end.
4. At each iteration, decrease the count of the first element of the window by 1 and increase the count of the next element of the window by 1 and again check the count of elements in frequency array having an odd frequency, if it is less than 2, then increment the count of palindromic permutation.
5. Repeat the above step till we reach the end of the string and print the count of palindromic permutation.

Below is the implementation of the above approach:

## C++

 `// C++ program for the above approach ` `#include ` `using` `namespace` `std; ` ` `  `// To store the frequency array ` `vector<``int``> freq(26); ` ` `  `// Function to check palindromic of ` `// of any substring using frequency array ` `bool` `checkPalindrome() ` `{ ` ` `  `    ``// Initialise the odd count ` `    ``int` `oddCnt = 0; ` ` `  `    ``// Traversing frequency array to ` `    ``// compute the count of characters ` `    ``// having odd frequency ` `    ``for` `(``auto` `x : freq) { ` ` `  `        ``if` `(x % 2 == 1) ` `            ``oddCnt++; ` `    ``} ` ` `  `    ``// Returns true if odd count is atmost 1 ` `    ``return` `oddCnt <= 1; ` `} ` ` `  `// Function to count the total number ` `// substring whose any permutations ` `// are palindromic ` `int` `countPalindromePermutation( ` `    ``string s, ``int` `k) ` `{ ` ` `  `    ``// Computing the frequncy of ` `    ``// first K charcter of the string ` `    ``for` `(``int` `i = 0; i < k; i++) { ` `        ``freq[s[i] - 97]++; ` `    ``} ` ` `  `    ``// To store the count of ` `    ``// palindromic permutations ` `    ``int` `ans = 0; ` ` `  `    ``// Checking for the current window ` `    ``// if it has any palindromic ` `    ``// permutation ` `    ``if` `(checkPalindrome()) { ` `        ``ans++; ` `    ``} ` ` `  `    ``// Start and end point of window ` `    ``int` `i = 0, j = k; ` ` `  `    ``while` `(j < s.size()) { ` ` `  `        ``// Sliding window by 1 ` ` `  `        ``// Decrementing count of first ` `        ``// element of the window ` `        ``freq[s[i++] - 97]--; ` ` `  `        ``// Incrementing count of next ` `        ``// element of the window ` `        ``freq[s[j++] - 97]++; ` ` `  `        ``// Checking current window ` `        ``// character frequency count ` `        ``if` `(checkPalindrome()) { ` `            ``ans++; ` `        ``} ` `    ``} ` ` `  `    ``// Return the final count ` `    ``return` `ans; ` `} ` ` `  `// Driver Code ` `int` `main() ` `{ ` `    ``// Given string str ` `    ``string str = ``"abbaca"``; ` ` `  `    ``// Window of size K ` `    ``int` `K = 3; ` ` `  `    ``// Function Call ` `    ``cout << countPalindromePermutation(str, K) ` `         ``<< endl; ` `    ``return` `0; ` `} `

## Java

 `// Java program for the above approach ` `import` `java.util.*; ` ` `  `class` `GFG{ ` ` `  `// To store the frequency array ` `static` `int` `[]freq = ``new` `int``[``26``]; ` ` `  `// Function to check palindromic of ` `// of any subString using frequency array ` `static` `boolean` `checkPalindrome() ` `{ ` `     `  `    ``// Initialise the odd count ` `    ``int` `oddCnt = ``0``; ` ` `  `    ``// Traversing frequency array to ` `    ``// compute the count of characters ` `    ``// having odd frequency ` `    ``for``(``int` `x : freq) ` `    ``{ ` `       ``if` `(x % ``2` `== ``1``) ` `           ``oddCnt++; ` `    ``} ` ` `  `    ``// Returns true if odd count ` `    ``// is atmost 1 ` `    ``return` `oddCnt <= ``1``; ` `} ` ` `  `// Function to count the total number ` `// subString whose any permutations ` `// are palindromic ` `static` `int` `countPalindromePermutation(``char` `[]s, ` `                                      ``int` `k) ` `{ ` ` `  `    ``// Computing the frequncy of ` `    ``// first K charcter of the String ` `    ``for``(``int` `i = ``0``; i < k; i++) ` `    ``{ ` `       ``freq[s[i] - ``97``]++; ` `    ``} ` ` `  `    ``// To store the count of ` `    ``// palindromic permutations ` `    ``int` `ans = ``0``; ` ` `  `    ``// Checking for the current window ` `    ``// if it has any palindromic ` `    ``// permutation ` `    ``if` `(checkPalindrome()) ` `    ``{ ` `        ``ans++; ` `    ``} ` ` `  `    ``// Start and end point of window ` `    ``int` `i = ``0``, j = k; ` ` `  `    ``while` `(j < s.length) ` `    ``{ ` ` `  `        ``// Sliding window by 1 ` ` `  `        ``// Decrementing count of first ` `        ``// element of the window ` `        ``freq[s[i++] - ``97``]--; ` ` `  `        ``// Incrementing count of next ` `        ``// element of the window ` `        ``freq[s[j++] - ``97``]++; ` ` `  `        ``// Checking current window ` `        ``// character frequency count ` `        ``if` `(checkPalindrome())  ` `        ``{ ` `            ``ans++; ` `        ``} ` `    ``} ` ` `  `    ``// Return the final count ` `    ``return` `ans; ` `} ` ` `  `// Driver Code ` `public` `static` `void` `main(String[] args) ` `{ ` `     `  `    ``// Given String str ` `    ``String str = ``"abbaca"``; ` ` `  `    ``// Window of size K ` `    ``int` `K = ``3``; ` ` `  `    ``// Function Call ` `    ``System.out.print(countPalindromePermutation( ` `                     ``str.toCharArray(), K) + ``"\n"``); ` `} ` `} ` ` `  `// This code is contributed by Amit Katiyar `

## Python3

 `# Python3 program for the above approach ` ` `  `# To store the frequency array  ` `freq ``=` `[``0``] ``*` `26` ` `  `# Function to check palindromic of  ` `# of any substring using frequency array  ` `def` `checkPalindrome():  ` ` `  `    ``# Initialise the odd count  ` `    ``oddCnt ``=` `0` ` `  `    ``# Traversing frequency array to  ` `    ``# compute the count of characters  ` `    ``# having odd frequency  ` `    ``for` `x ``in` `freq:  ` `        ``if` `(x ``%` `2` `=``=` `1``): ` `            ``oddCnt ``+``=` `1` `     `  `    ``# Returns true if odd count is atmost 1  ` `    ``return` `oddCnt <``=` `1` ` `  `# Function to count the total number  ` `# substring whose any permutations  ` `# are palindromic  ` `def` `countPalindromePermutation(s, k):  ` ` `  `    ``# Computing the frequncy of  ` `    ``# first K charcter of the string  ` `    ``for` `i ``in` `range``(k):  ` `        ``freq[``ord``(s[i]) ``-` `97``] ``+``=` `1` `     `  `    ``# To store the count of  ` `    ``# palindromic permutations  ` `    ``ans ``=` `0` ` `  `    ``# Checking for the current window  ` `    ``# if it has any palindromic  ` `    ``# permutation  ` `    ``if` `(checkPalindrome()):  ` `        ``ans ``+``=` `1` `     `  `    ``# Start and end poof window  ` `    ``i ``=` `0` `    ``j ``=` `k  ` ` `  `    ``while` `(j < ``len``(s)):  ` ` `  `        ``# Sliding window by 1  ` ` `  `        ``# Decrementing count of first  ` `        ``# element of the window  ` `        ``freq[``ord``(s[i]) ``-` `97``] ``-``=` `1` `        ``i ``+``=` `1` ` `  `        ``# Incrementing count of next  ` `        ``# element of the window  ` `        ``freq[``ord``(s[j]) ``-` `97``] ``+``=` `1` `        ``j ``+``=` `1` ` `  `        ``# Checking current window  ` `        ``# character frequency count  ` `        ``if` `(checkPalindrome()):  ` `            ``ans ``+``=` `1` `             `  `    ``# Return the final count  ` `    ``return` `ans  ` ` `  `# Driver Code  ` ` `  `# Given string str  ` `str` `=` `"abbaca"` ` `  `# Window of size K  ` `K ``=` `3` ` `  `# Function call  ` `print``(countPalindromePermutation(``str``, K)) ` ` `  `# This code is contributed by code_hunt `

## C#

 `// C# program for the above approach ` `using` `System; ` ` `  `class` `GFG{ ` ` `  `// To store the frequency array ` `static` `int` `[]freq = ``new` `int``; ` ` `  `// Function to check palindromic of ` `// of any subString using frequency array ` `static` `bool` `checkPalindrome() ` `{ ` `     `  `    ``// Initialise the odd count ` `    ``int` `oddCnt = 0; ` ` `  `    ``// Traversing frequency array to ` `    ``// compute the count of characters ` `    ``// having odd frequency ` `    ``foreach``(``int` `x ``in` `freq) ` `    ``{ ` `        ``if` `(x % 2 == 1) ` `            ``oddCnt++; ` `    ``} ` ` `  `    ``// Returns true if odd count ` `    ``// is atmost 1 ` `    ``return` `oddCnt <= 1; ` `} ` ` `  `// Function to count the total number ` `// subString whose any permutations ` `// are palindromic ` `static` `int` `countPalindromePermutation(``char` `[]s, ` `                                      ``int` `k) ` `{ ` `    ``int` `i = 0; ` `     `  `    ``// Computing the frequncy of ` `    ``// first K charcter of the String ` `    ``for``(i = 0; i < k; i++) ` `    ``{ ` `       ``freq[s[i] - 97]++; ` `    ``} ` ` `  `    ``// To store the count of ` `    ``// palindromic permutations ` `    ``int` `ans = 0; ` ` `  `    ``// Checking for the current window ` `    ``// if it has any palindromic ` `    ``// permutation ` `    ``if` `(checkPalindrome()) ` `    ``{ ` `        ``ans++; ` `    ``} ` ` `  `    ``// Start and end point of window ` `    ``int` `j = k; ` `        ``i = 0; ` ` `  `    ``while` `(j < s.Length) ` `    ``{ ` `         `  `        ``// Sliding window by 1 ` ` `  `        ``// Decrementing count of first ` `        ``// element of the window ` `        ``freq[s[i++] - 97]--; ` ` `  `        ``// Incrementing count of next ` `        ``// element of the window ` `        ``freq[s[j++] - 97]++; ` ` `  `        ``// Checking current window ` `        ``// character frequency count ` `        ``if` `(checkPalindrome())  ` `        ``{ ` `            ``ans++; ` `        ``} ` `    ``} ` `     `  `    ``// Return the final count ` `    ``return` `ans; ` `} ` ` `  `// Driver Code ` `public` `static` `void` `Main(String[] args) ` `{ ` `     `  `    ``// Given String str ` `    ``String str = ``"abbaca"``; ` ` `  `    ``// Window of size K ` `    ``int` `K = 3; ` ` `  `    ``// Function Call ` `    ``Console.Write(countPalindromePermutation( ` `                  ``str.ToCharArray(), K) + ``"\n"``); ` `} ` `} ` ` `  `// This code is contributed by Amit Katiyar `

Output:

```3
```

Time Complexity: O(N)
Auxiliary Space: O(26) My Personal Notes arrow_drop_up Check out this Author's contributed articles.

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Improved By : amit143katiyar, code_hunt