3223. Minimum Length of String After Operations

📝 Problem Details

• Title: 3223. Minimum Length of String After Operations
• Link: https://leetcode.com/problems/minimum-length-of-string-after-operations/
• Difficulty: Medium
• Tags/Categories: {{tags/categories}}

💭What Were My Initial Thoughts?

- need at least 3 elements relating to a particular value in order to conduct a removal operation
- could create a hashmap, where the key is the char and the value is a list of indices where the chars are located 

🤔What Did I Struggle With?

~

💡 Explanation of Solution

Observation:
- if a character appears odd times --> 1 character remains
- if a character appears even times --> 2 characcters remain
- instead of tracking indices explicitly , we can just count occurences of each character 
- based on the above observation, we only need to process each character once 

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- **Create a frequency hashmap** to count occurrences of each character.
- **For each character**, determine how many will remain after applying the removal rule:
    - If `freq % 2 == 1`, keep `1` character.
    - If `freq % 2 == 0`, keep `2` characters.
- **Sum up the remaining counts** to get the final string length.

⌛ Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(n)

💻 Implementation of Solution

class Solution {
public:
    int minimumLength(string s) {
        unordered_map<char,int> map;
        int res = 0;
 
        for(char c : s) {
            map[c]++;
        }
 
        for(auto it = map.begin(); it != map.end(); it++) {
            if(it->second % 2 == 1) res++;
            else res += 2;
        }
        return res;
    }
};