> For the complete documentation index, see [llms.txt](https://gl01.gitbook.io/gfg-editorials/llms.txt). Markdown versions of documentation pages are available by appending `.md` to page URLs; this page is available as [Markdown](https://gl01.gitbook.io/gfg-editorials/2024/06-2024-june/01-odd-even-problem.md).

# 01. Odd Even Problem

The problem can be found at the following link: [Question Link](https://www.geeksforgeeks.org/problems/help-nobita0532/1)

## My Approach

* Initialize a vector `c` of size 26 to count the frequency of each character in the string `s`.
* Traverse through the string `s` and update the count of each character in the vector `c`.
* Initialize a counter `cnt` to zero to keep track of characters that meet the odd/even criteria.
* Iterate through the vector `c`, checking if the frequency of each character is odd or even based on the character's position in the alphabet (i.e., position % 2).
* Increment the counter `cnt` for each character that meets the condition where the frequency's odd/even status matches the character's position % 2.
* Return "ODD" if `cnt` is odd; otherwise, return "EVEN".

## Time and Auxiliary Space Complexity

* **Time Complexity**: `(O(n))`, where `( n )` is the length of the string `s`, since we are traversing the string and then iterating through a fixed-size array of length 26.
* **Auxiliary Space Complexity**: `( O(1))`, since the vector `c` has a fixed size of 26 regardless of the input size.

## Code (C++)

```cpp
class Solution {
public:
    string oddEven(string s) {
        vector<int> c(26, 0);
        for(int i = 0; i < s.size(); ++i)
            ++c[s[i] - 'a'];
            
        int cnt = 0;
        for(int i = 0; i < 26; ++i)
            if(c[i] && c[i] % 2 == (i + 1) % 2)
                ++cnt;
                
        return cnt % 2 ? "ODD" : "EVEN";
    }
};
```

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