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# Program to check if all characters have even frequency

• Last Updated : 14 Dec, 2022

Given a string S consisting only of lowercase letters check if the string has all characters appearing even times.

Examples:

Input : abaccaba
Output : Yes
Explanation: ‘a’ occurs four times, ‘b’ occurs twice, ‘c’ occurs twice and the other letters occur zero times.

Input:  hthth
Output : No

Approach: We will go through the string and count the occurrence of all characters after that we will check if the occurrences are even or not if there is any odd frequency character then we immediately print No.

## C++

 `// C++ implementation of the above approach``#include ``using` `namespace` `std;` `bool` `check(string s)``{``    ` `    ``// creating a frequency array``    ``int` `freq = {0};``    ` `    ``// Finding length of s``    ``int` `n = s.length();``    ``for` `(``int` `i = 0; i < s.length(); i++)``    ` `    ``// counting frequency of all characters``        ``freq[s[i] - 97]++;``    ` `    ``// checking if any odd frequency``    ``// is there or not``    ``for` `(``int` `i = 0; i < 26; i++)``        ``if` `(freq[i] % 2 == 1)``        ``return` `false``;``    ``return` `true``;``}` `// Driver Code``int` `main()``{``    ``string s = ``"abaccaba"``;``    ``check(s) ? cout << ``"Yes"` `<< endl :``               ``cout << ``"No"` `<< endl;``    ``return` `0;``}` `// This code is contributed by``// sanjeev2552`

## Java

 `// Java implementation of the above approach``class` `GFG``{``    ``static` `boolean` `check(String s)``    ``{` `        ``// creating a frequency array``        ``int``[] freq = ``new` `int``[``26``];` `        ``// Finding length of s``        ``int` `n = s.length();` `        ``// counting frequency of all characters``        ``for` `(``int` `i = ``0``; i < s.length(); i++)``        ``{``            ``freq[(s.charAt(i)) - ``97``] += ``1``;``        ``}` `        ``// checking if any odd frequency``        ``// is there or not``        ``for` `(``int` `i = ``0``; i < freq.length; i++)``        ``{``            ``if` `(freq[i] % ``2` `== ``1``)``            ``{``                ``return` `false``;``            ``}``        ``}``        ``return` `true``;``    ``}` `    ``// Driver Code``    ``public` `static` `void` `main(String[] args)``    ``{``        ``String s = ``"abaccaba"``;``        ``if` `(check(s))``        ``{``            ``System.out.println(``"Yes"``);``        ``}``        ``else``        ``{``            ``System.out.println(``"No"``);``        ``}``    ``}``}` `// This code is contributed by Rajput-Ji`

## Python3

 `# Python implementation of the above approach``def` `check(s):` `    ``# creating a frequency array``    ``freq ``=``[``0``]``*``26` `    ``# Finding length of s``    ``n ``=` `len``(s)` `    ``for` `i ``in` `range``(n):` `        ``# counting frequency of all characters``        ``freq[``ord``(s[i])``-``97``]``+``=` `1` `    ``for` `i ``in` `range``(``26``):` `        ``# checking if any odd frequency``        ``# is there or not``        ``if` `(freq[i]``%` `2` `=``=` `1``):``            ``return` `False``    ``return` `True` `# Driver code``s ``=``"abaccaba"``if``(check(s)):``    ``print``(``"Yes"``)``else``:``    ``print``(``"No"``)`

## C#

 `// C# implementation of the approach``using` `System;``    ` `class` `GFG``{``    ``static` `Boolean check(String s)``    ``{` `        ``// creating a frequency array``        ``int``[] freq = ``new` `int``;` `        ``// Finding length of s``        ``int` `n = s.Length;` `        ``// counting frequency of all characters``        ``for` `(``int` `i = 0; i < s.Length; i++)``        ``{``            ``freq[(s[i]) - 97] += 1;``        ``}` `        ``// checking if any odd frequency``        ``// is there or not``        ``for` `(``int` `i = 0; i < freq.Length; i++)``        ``{``            ``if` `(freq[i] % 2 == 1)``            ``{``                ``return` `false``;``            ``}``        ``}``        ``return` `true``;``    ``}` `    ``// Driver Code``    ``public` `static` `void` `Main(String[] args)``    ``{``        ``String s = ``"abaccaba"``;``        ``if` `(check(s))``        ``{``            ``Console.WriteLine(``"Yes"``);``        ``}``        ``else``        ``{``            ``Console.WriteLine(``"No"``);``        ``}``    ``}``}` `// This code is contributed by PrinciRaj1992`

## Javascript

 ``

Output

```Yes
```

Time Complexity: O(n), where n is the length of the given string.
Auxiliary Space: O(26) ⇒ O(1), no extra space is required, so it is a constant.

### Method #2:Using built-in python functions.

Approach: We will scan the string and count the occurrence of all characters using the built-in Counter() function after that we traverse the counter list and check if the occurrences are even or not if there is any odd frequency character then we immediately print No.

Note: This method is applicable for all types of characters

## Python3

 `# Python implementation for``# the above approach` `# importing Counter function``from` `collections ``import` `Counter` `# Function to check if all``# elements occur even times``def` `checkString(s):``  ` `    ``# Counting the frequency of all``    ``# character using Counter function``    ``frequency ``=` `Counter(s)``    ` `    ``# Traversing frequency``    ``for` `i ``in` `frequency:``      ` `        ``# Checking if any element``        ``# has odd count``        ``if` `(frequency[i] ``%` `2` `=``=` `1``):``            ``return` `False``    ``return` `True`  `# Driver code``s ``=` `"geeksforgeeksfor"``if``(checkString(s)):``    ``print``(``"Yes"``)``else``:``    ``print``(``"No"``)``    ` `# This code is contributed by vikkycirus`

Output

```Yes
```

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