# Check if permutation of a given string can be made palindromic by removing at most K characters

Given a string str and an integer K, the task is to check if a permutation of the given string can be made a palindromic by removing at most K characters from the given string.

Examples:

Input: str = “geeksforgeeks”, K = 2
Output: Yes
Explanation:
Removing (str, str) from the given string makes the remaining string “geeksrgeeks” palindromic. Therefore, the required output is Yes.

Input: str = “coder”, K = 1
Output: No

Approach: The problem can be solved using Hashing. The idea is to iterate over characters of the given string and store the frequency of each distinct character of the given string. If the count of distinct characters of the given string having odd frequency is less than or equal to (K + 1), then print Yes. Otherwise, print No. Follow the steps below to solve the problem:

Below is the implementation of the above approach:

## C++

 `// C++ program to implement` `// the above approach`   `#include ` `using` `namespace` `std;`   `// Function to check if` `// the string satisfies` `// the given conditions or not` `bool` `checkPalinK(string str, ``int` `K)` `{` `    ``// Stores length of` `    ``// given string` `    ``int` `N = str.length();`   `    ``// Stores frequency of` `    ``// each character of str` `    ``int` `cntFreq = { 0 };`   `    ``for` `(``int` `i = 0; i < N;` `         ``i++) {`   `        ``// Update frequency of` `        ``// current character` `        ``cntFreq[str[i]]++;` `    ``}`   `    ``// Stores count of` `    ``// distinct character` `    ``// whose frequency is odd` `    ``int` `cntOddFreq = 0;`   `    ``// Traverse the cntFreq[]` `    ``// array.` `    ``for` `(``int` `i = 0; i < 256;` `         ``i++) {`   `        ``// If frequency of` `        ``// character i is odd` `        ``if` `(cntFreq[i] % 2` `            ``== 1) {`   `            ``// Update cntOddFreq` `            ``cntOddFreq++;` `        ``}` `    ``}`   `    ``// If count of distinct character` `    ``// having odd frequency is <= K + 1` `    ``if` `(cntOddFreq <= (K + 1)) {` `        ``return` `true``;` `    ``}`   `    ``return` `false``;` `}`   `// Driver Code` `int` `main()` `{` `    ``string str = ``"geeksforgeeks"``;` `    ``int` `K = 2;`   `    ``// If str satisfy` `    ``// the given conditions` `    ``if` `(checkPalinK(str, K)) {` `        ``cout << ``"Yes"``;` `    ``}` `    ``else` `{` `        ``cout << ``"No"``;` `    ``}` `}`

## Java

 `// Java program to implement ` `// the above approach ` `import` `java.util.*;`   `class` `GFG{`   `// Function to check if` `// the string satisfies` `// the given conditions or not` `public` `static` `boolean` `checkPalinK(String str, ` `                                  ``int` `K)` `{` `    `  `    ``// Stores length of` `    ``// given string` `    ``int` `N = str.length();`   `    ``// Stores frequency of` `    ``// each character of str` `    ``int` `cntFreq[] = ``new` `int``[``256``];`   `    ``for``(``int` `i = ``0``; i < N; i++)` `    ``{` `        `  `        ``// Update frequency of` `        ``// current character` `        ``cntFreq[str.charAt(i)]++;` `    ``}`   `    ``// Stores count of` `    ``// distinct character` `    ``// whose frequency is odd` `    ``int` `cntOddFreq = ``0``;`   `    ``// Traverse the cntFreq[]` `    ``// array.` `    ``for``(``int` `i = ``0``; i < ``256``; i++) ` `    ``{` `        `  `        ``// If frequency of` `        ``// character i is odd` `        ``if` `(cntFreq[i] % ``2` `== ``1``)` `        ``{` `            `  `            ``// Update cntOddFreq` `            ``cntOddFreq++;` `        ``}` `    ``}`   `    ``// If count of distinct character` `    ``// having odd frequency is <= K + 1` `    ``if` `(cntOddFreq <= (K + ``1``))` `    ``{` `        ``return` `true``;` `    ``}` `    ``return` `false``;` `}`   `// Driver Code` `public` `static` `void` `main(String args[])` `{` `    ``String str = ``"geeksforgeeks"``;` `    ``int` `K = ``2``;`   `    ``// If str satisfy` `    ``// the given conditions` `    ``if` `(checkPalinK(str, K)) ` `    ``{` `        ``System.out.println(``"Yes"``);` `    ``}` `    ``else` `    ``{` `        ``System.out.println(``"No"``);` `    ``}` `}` `}`   `// This code is contributed by hemanth gadarla`

## Python3

 `# Python3 program to implement` `# the above approach`   `# Function to check if the ` `# satisfies the given` `# conditions or not` `def` `checkPalinK(``str``, K):` `    `  `    ``# Stores length of` `    ``# given string` `    ``N ``=` `len``(``str``)`   `    ``# Stores frequency of` `    ``# each character of str` `    ``cntFreq ``=` `[``0``] ``*` `256`   `    ``for` `i ``in` `range``(N):` `        `  `        ``# Update frequency of` `        ``# current character` `        ``cntFreq[``ord``(``str``[i])] ``+``=` `1`   `    ``# Stores count of` `    ``# distinct character` `    ``# whose frequency is odd` `    ``cntOddFreq ``=` `0`   `    ``# Traverse the cntFreq[]` `    ``# array.` `    ``for` `i ``in` `range``(``256``):`   `        ``# If frequency of` `        ``# character i is odd` `        ``if` `(cntFreq[i] ``%` `2` `=``=` `1``):`   `            ``# Update cntOddFreq` `            ``cntOddFreq ``+``=` `1`   `    ``# If count of distinct character` `    ``# having odd frequency is <= K + 1` `    ``if` `(cntOddFreq <``=` `(K ``+` `1``)):` `        ``return` `True`   `    ``return` `False`   `# Driver Code` `if` `__name__ ``=``=` `'__main__'``:` `    `  `    ``str` `=` `"geeksforgeeks"` `    ``K ``=` `2`   `    ``# If str satisfy` `    ``# the given conditions` `    ``if` `(checkPalinK(``str``, K)):` `        ``print``(``"Yes"``)` `    ``else``:` `        ``print``(``"No"``)`   `# This code is contributed by mohit kumar 29`

## C#

 `// C# program to implement ` `// the above approach ` `using` `System;`   `class` `GFG{`   `// Function to check if` `// the string satisfies` `// the given conditions or not` `public` `static` `bool` `checkPalinK(String str, ` `                               ``int` `K)` `{` `    `  `    ``// Stores length of` `    ``// given string` `    ``int` `N = str.Length;`   `    ``// Stores frequency of` `    ``// each character of str` `    ``int` `[]cntFreq = ``new` `int``;`   `    ``for``(``int` `i = 0; i < N; i++)` `    ``{` `        `  `        ``// Update frequency of` `        ``// current character` `        ``cntFreq[str[i]]++;` `    ``}`   `    ``// Stores count of` `    ``// distinct character` `    ``// whose frequency is odd` `    ``int` `cntOddFreq = 0;`   `    ``// Traverse the cntFreq[]` `    ``// array.` `    ``for``(``int` `i = 0; i < 256; i++) ` `    ``{` `        `  `        ``// If frequency of` `        ``// character i is odd` `        ``if` `(cntFreq[i] % 2 == 1)` `        ``{` `            `  `            ``// Update cntOddFreq` `            ``cntOddFreq++;` `        ``}` `    ``}`   `    ``// If count of distinct character` `    ``// having odd frequency is <= K + 1` `    ``if` `(cntOddFreq <= (K + 1))` `    ``{` `        ``return` `true``;` `    ``}` `    ``return` `false``;` `}`   `// Driver Code` `public` `static` `void` `Main(String []args)` `{` `    ``String str = ``"geeksforgeeks"``;` `    ``int` `K = 2;`   `    ``// If str satisfy` `    ``// the given conditions` `    ``if` `(checkPalinK(str, K)) ` `    ``{` `        ``Console.WriteLine(``"Yes"``);` `    ``}` `    ``else` `    ``{` `        ``Console.WriteLine(``"No"``);` `    ``}` `}` `}`   `// This code is contributed by Amit Katiyar`

Output:

```Yes

```

Time Complexity: O(N + 256), Where N is the length of the given string.
Auxiliary Space: O(256)

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