# Check if K palindromic strings can be formed from a given string

Given a string S of size N and an integer K, the task is to find whether the characters of the string can be arranged to make K palindromic strings simultaneously.
Examples:

Input: S = “annabelle”, K = 2
Output: Yes
Explanation:
All characters of string S can be distributed into “elble” and “anna” which are both palindromic.

Input: S =”abcd”, K = 4
Output: Yes
Explanation:
Partition all characters of string as single character.

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

Approach

• If the frequency of every character is even and K lies between 1 and N then it is always possible to form K palindrome strings.
• But if there are some characters (say odd_count) with odd frequency, then K must lie between odd_count and N for K palindromic strings to be possible.

Hence, follow the steps below to solve the problem:

If K exceeds the length of the string, straightaway print “No”.
1. Store the frequency of all character in a Map.
2. Count the number of characters having odd frequency.
3. If the count is less then given K, then print “No”. Otherwise print “Yes”.

Below is the implementation of the above approach:

## C++

 `// C++ program to check ` `// whether the string is ` `// K palindrome or not ` `#include ` `#include ` `using` `namespace` `std; ` ` `  `// function to check ` `// whether the string is ` `// K palindrome or not ` `bool` `can_Construct(string S, ``int` `K) ` `{ ` `    ``// map to frequency of character ` `    ``map<``int``, ``int``> m; ` ` `  `    ``int` `i = 0, j = 0, p = 0; ` ` `  `    ``// Check when k is given ` `    ``// as same as length of string ` `    ``if` `(S.length() == K) { ` `        ``return` `true``; ` `    ``} ` ` `  `    ``// iterator for map ` `    ``map<``int``, ``int``>::iterator h; ` ` `  `    ``// stroing the frequency of every ` `    ``// character in map ` `    ``for` `(i = 0; i < S.length(); i++) { ` `        ``m[S[i]] = m[S[i]] + 1; ` `    ``} ` ` `  `    ``// if K is greater than size ` `    ``// of string then return false ` `    ``if` `(K > S.length()) { ` `        ``return` `false``; ` `    ``} ` ` `  `    ``else` `{ ` ` `  `        ``// check that number of character ` `        ``// having the odd frequency ` `        ``for` `(h = m.begin(); h != m.end(); h++) { ` ` `  `            ``if` `(m[h->first] % 2 != 0) { ` `                ``p = p + 1; ` `            ``} ` `        ``} ` `    ``} ` ` `  `    ``// if k is less than number of odd ` `    ``// frequecny character then it is ` `    ``// again false other wise true ` `    ``if` `(K < p) { ` `        ``return` `false``; ` `    ``} ` ` `  `    ``return` `true``; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``string S = ``"annabelle"``; ` `    ``int` `K = 4; ` ` `  `    ``if` `(can_Construct(S, K)) { ` `        ``cout << ``"Yes"``; ` `    ``} ` `    ``else` `{ ` `        ``cout << ``"No"``; ` `    ``} ` `} `

## Python3

 `# Python3 program to check whether ` `# the is K palindrome or not ` ` `  `# Function to check whether  ` `# the is K palindrome or not ` `def` `can_Construct(S, K): ` `     `  `    ``# map to frequency of character ` `    ``m ``=` `dict``() ` `    ``p ``=` `0` `     `  `    ``# Check when k is given ` `    ``# as same as length of string ` `    ``if` `(``len``(S) ``=``=` `K): ` `        ``return` `True` ` `  `    ``# Stroing the frequency of every ` `    ``# character in map ` `    ``for` `i ``in` `S: ` `        ``m[i] ``=` `m.get(i, ``0``) ``+` `1` ` `  `    ``# If K is greater than size ` `    ``# of then return false ` `    ``if` `(K > ``len``(S)): ` `        ``return` `False` ` `  `    ``else``: ` ` `  `        ``# Check that number of character ` `        ``# having the odd frequency ` `        ``for` `h ``in` `m: ` `            ``if` `(m[h] ``%` `2` `!``=` `0``): ` `                ``p ``=` `p ``+` `1` ` `  `    ``# If k is less than number of odd ` `    ``# frequecny character then it is ` `    ``# again false otherwise true ` `    ``if` `(K < p): ` `        ``return` `False` ` `  `    ``return` `True` ` `  `# Driver code ` `if` `__name__ ``=``=` `'__main__'``: ` `     `  `    ``S ``=` `"annabelle"` `    ``K ``=` `4` ` `  `    ``if` `(can_Construct(S, K)): ` `        ``print``(``"Yes"``) ` `    ``else``: ` `        ``print``(``"No"``) ` ` `  `# This code is contributed by mohit kumar 29 `

Output:

```Yes
```

Time Complexity: O (N).
Auxillary Space: O (N).

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