# Remove even frequency characters from the string

Given a string ‘str’, the task is to remove all the characters from the string that have even frequencies.

Examples:

```Input: str = "aabbbddeeecc"
Output: bbbeee
The characters a, d, c have even frequencies
So, they are removed from the string.

Input: str = "zzzxxweeerr"
Output: zzzweee
```

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

Approach:

• Create a map and store the frequency of each character from the string to the same map.
• Then, traverse the string and find out which characters have even frequencies with the help of the map.
• Ignore all those characters which have even frequencies and store the rest in a new string.
• Finally, display the new string.

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach ` `#include ` `using` `namespace` `std; ` ` `  `// Function that removes the ` `// characters which have even ` `// frequencies in the string ` `void` `solve(string s) ` `{ ` `    ``// create a map to store the ` `    ``// frequency of each character ` `    ``unordered_map<``char``, ``int``> m; ` `    ``for` `(``int` `i = 0; i < s.length(); i++) { ` `        ``m[s[i]]++; ` `    ``} ` ` `  `    ``// to store the new string ` `    ``string new_string = ``""``; ` ` `  `    ``// remove the characters which ` `    ``// have even frequencies ` `    ``for` `(``int` `i = 0; i < s.length(); i++) { ` ` `  `        ``// if the character has ` `        ``// even frequency then skip ` `        ``if` `(m[s[i]] % 2 == 0) ` `            ``continue``; ` ` `  `        ``// else concatenate the ` `        ``// character to the new string ` `        ``new_string += s[i]; ` `    ``} ` ` `  `    ``// display the modified string ` `    ``cout << new_string << endl; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``string s = ``"aabbbddeeecc"``; ` ` `  `    ``// remove the characters which ` `    ``// have even frequencies ` `    ``solve(s); ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java implementation of the approach ` `import` `java.util.*; ` ` `  `class` `GFG ` `{ ` `    ``// Function that removes the ` `    ``// characters which have even ` `    ``// frequencies in the string ` `    ``static` `void` `solve(String s) ` `    ``{ ` `        ``// create a map to store the ` `        ``// frequency of each character ` `        ``HashMap m = ``new` `HashMap<>(); ` `         `  `        ``for` `(``int` `i = ``0``; i < s.length(); i++)  ` `        ``{ ` `            ``if``(m.containsKey(s.charAt(i))) ` `                        ``m.put(s.charAt(i),  ` `                        ``m.get(s.charAt(i)) + ``1``);  ` `            ``else` `                ``m.put(s.charAt(i), ``1``); ` `        ``} ` `     `  `        ``// to store the new string ` `        ``String new_string = ``""``; ` `     `  `        ``// remove the characters which ` `        ``// have even frequencies ` `        ``for` `(``int` `i = ``0``; i < s.length(); i++) ` `        ``{ ` `     `  `            ``// if the character has ` `            ``// even frequency then skip ` `            ``if` `(m.get(s.charAt(i)) % ``2` `== ``0``) ` `                ``continue``; ` `     `  `            ``// else concatenate the ` `            ``// character to the new string ` `            ``new_string = new_string + s.charAt(i); ` `        ``} ` `     `  `        ``// display the modified string ` `        ``System.out.println(new_string); ` `    ``} ` `     `  `    ``// Driver code ` `    ``public` `static` `void` `main(String []args) ` `    ``{ ` `        ``String s = ``"aabbbddeeecc"``; ` `     `  `        ``// remove the characters which ` `        ``// have even frequencies ` `        ``solve(s); ` `    ``} ` `} ` ` `  `// This code is contributed by ihritik `

## Python3

 `# Python3 implementation of  ` `# above approach ` ` `  `# Function that removes the ` `# characters which have even ` `# frequencies in the string ` `def` `solve(s): ` `     `  `    ``# create a map to store the ` `    ``# frequency of each character ` `    ``m ``=` `dict``() ` `    ``for` `i ``in` `range``(``len``(s)): ` `        ``if` `s[i] ``in` `m: ` `            ``m[s[i]] ``=` `m[s[i]]``+``1` `        ``else``: ` `            ``m[s[i]] ``=` `1` `             `  `    ``# to store the new string ` `    ``new_string ``=` `"" ` `     `  `    ``# remove the characters which ` `    ``# have even frequencies ` `    ``for` `i ``in` `range``(``len``(s)): ` `         `  `        ``# if the character has  ` `        ``# even frequency then skip ` `        ``if` `m[s[i]]``%``2` `=``=` `0``: ` `            ``continue` `         `  `        ``# else concatenate the ` `        ``# character to the new string ` `        ``new_string ``=` `new_string``+``s[i] ` `         `  `    ``# display the modified string ` `    ``print``(new_string) ` `     `  `#Driver code ` `if` `__name__``=``=``'__main__'``: ` `    ``s ``=` `"aabbbddeeecc"` ` `  `# remove the characters which ` `# have even frequencies ` `    ``solve(s) ` ` `  `# this code is contributed by  ` `# Shashank_Sharma `

## C#

 `// C# implementation of the approach ` `using` `System; ` `using` `System.Collections.Generic;  ` ` `  `class` `GFG ` `{ ` `    ``// Function that removes the ` `    ``// characters which have even ` `    ``// frequencies in the string ` `    ``static` `void` `solve(String s) ` `    ``{ ` `        ``// create a map to store the ` `        ``// frequency of each character ` `        ``Dictionary<``char``, ``int``> m = ``new` `Dictionary<``char``, ``int``>(); ` `         `  `        ``for` `(``int` `i = 0; i < s.Length; i++)  ` `        ``{ ` `            ``if``(m.ContainsKey(s[i])) ` `            ``{ ` `                ``var` `val = m[s[i]]; ` `                ``m.Remove(s[i]); ` `                ``m.Add(s[i], val + 1);  ` `                 `  `            ``}          ` `            ``else` `                ``m.Add(s[i], 1); ` `        ``} ` `     `  `        ``// to store the new string ` `        ``String new_string = ``""``; ` `     `  `        ``// remove the characters which ` `        ``// have even frequencies ` `        ``for` `(``int` `i = 0; i < s.Length; i++) ` `        ``{ ` `     `  `            ``// if the character has ` `            ``// even frequency then skip ` `            ``if` `(m[s[i]] % 2 == 0) ` `                ``continue``; ` `     `  `            ``// else concatenate the ` `            ``// character to the new string ` `            ``new_string = new_string + s[i]; ` `        ``} ` `     `  `        ``// display the modified string ` `        ``Console.WriteLine(new_string); ` `    ``} ` `     `  `    ``// Driver code ` `    ``public` `static` `void` `Main(String []args) ` `    ``{ ` `        ``String s = ``"aabbbddeeecc"``; ` `     `  `        ``// remove the characters which ` `        ``// have even frequencies ` `        ``solve(s); ` `    ``} ` `} ` ` `  `// This code has been contributed by 29AjayKumar `

Output:

```bbbeee
```

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

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