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349 Discussions

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  • mcoroz96
     + 0 comments

    c++

    int sockMerchant(int n, vector<int> ar) {
    unordered_map<int, int> frequencyMap;
    
    int nPairs = 0;
    
    for(int i = 0; i < ar.size(); ++i){
        frequencyMap[ar[i]] += 1;
        if(frequencyMap[ar[i]] % 2 == 0){
            nPairs +=1;
        }
    }
    
    return nPairs;
}

  • trugamer00x
     + 0 comments

    python

    def sockMerchant(n, ar):
    d={i:ar.count(i) for i in set(ar)}
    res=0
    for i in d:
        if d[i] >= 2:
            res = res + d[i]//2
    return res

  • cooldilcool
     + 0 comments

    C# Code 

    public static int sockMerchant(int n, List<int> ar)
    {
        int pairCount = 0;
        
        var groupList = ar.Order()
            .GroupBy(e => e)
            .Select(e => new { SockId = e.Key, Count = e.Count() })
        .ToList();
        
        foreach(var sockGroup in groupList)
        {
            pairCount += sockGroup.Count / 2;
        }
        
        return pairCount;
    }

  • rtpardavila
     + 0 comments

    ` //// Python, using dict and flags:

    def sockMerchant(n, ar): no_pairs : int = 0 pair_nopair : dict[int, bool] = {}

    for sock in ar:
    if sock not in pair_nopair:
        pair_nopair[sock] = True
        continue
    if pair_nopair[sock]:
        no_pairs = no_pairs + 1
    pair_nopair[sock] = not pair_nopair[sock]

return no_pairs

    `

  • fagundes4181
     + 0 comments

    Kotlin Solution

    fun sockMerchant(n: Int, ar: Array<Int>): Int {
        var numberOfPairs = 0
        ar.groupBy { it }.forEach {
            numberOfPairs += it.value.size / 2
        }
        return numberOfPairs
    }