Python drawer = {}

for sock in ar:
    drawer[sock] = drawer.get(sock, 0) + 1

single_pair = sum(1 if x % 2 != 0 else 0 for x in drawer.values())
matching_pair = int((len(ar) - single_pair)/2)

return matching_pair

Here is my c++ solution , you can find the video explanation here : https://youtu.be/HPIhFXx_DVM

int sockMerchant(int n, vector<int> ar) {
    map<int, int>mp;
    int result = 0;
    for(int i = 0; i < ar.size(); i++){
        if(mp[ar[i]]){
            result ++;
            mp.erase(ar[i]);
            continue;
        }
        mp[ar[i]]++;
    }
    return result;
}

public static int sockMerchant(int n, List<Integer> ar) {
    int[] count = new int[101];
    int pairs = 0;

    for(int i = 0; i < ar.size(); i++) {
        count[ar.get(i)]++;
    }

    for(int i = 1; i < 101; i++) {
        pairs += count[i]/2;
    }

    return pairs;
}****

As a Linq lover, I would just do this

public static int sockMerchant(int n, List<int> ar)
{
    var ret = ar.GroupBy(x => x).Select(x => new {
        item = x.Key,
        freq = x.Count() / 2
    }).Sum(x => x.freq);
    return ret;
}`

Easily Understandable Approach

public static int sockMerchant(int n, List<Integer> ar) {
    // Write your code here
    
    Set<Integer> set = new HashSet<>();
    int c=0;

    for(Integer i:ar){
        if(set.contains(i)){
            set.remove(i);
            c++;
        }
        else{
            set.add(i);
        }
    }

    return c;
    }