01. Odd Even Problem

The problem can be found at the following link: Question Link

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++)

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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