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  1. Prepare
  2. Algorithms
  3. Implementation
  4. Sales by Match
  5. Discussions

Sales by Match

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6054 Discussions

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  • ai_13310411
     + 0 comments

    java 8 Collections.sort(ar); int count = 0; int temp = 0; for(int i = 0; i < n; i++){

            if(temp == ar.get(i)){
            temp = 0;
            continue;
        }
        temp = ar.get(i);
        if((i + 1) < n && temp == ar.get(i + 1)){
            count++;
        }
    }
    return count;

  • kepliu
     + 0 comments

    Using Java stream

        public static int sockMerchant(int n, List<Integer> ar){

        Map<Integer, Integer> map = ar.stream().map(i -> new AbstractMap.SimpleEntry<>(i, 1)).collect(Collectors.toMap(AbstractMap.SimpleEntry::getKey,
                AbstractMap.SimpleEntry::getValue,
                Integer::sum));
        return map.values().stream().mapToInt(integer -> integer / 2).sum();
    }

  • quierountacopic1
     + 0 comments

    in c#

    int pairs = 0;
        var colors = ar.Distinct();
        foreach(var a in colors){
            var pair = ar.Where(x => x == a).ToList();
            if(pair.Count() % 2 != 0) pair.Remove(a);
            pairs += pair.Count()/2;
        }
        return pairs;

  • just_nick
     + 0 comments

    Python

    def sockMerchant(n, ar):
    seen = set()
    pairs = 0
    for c in ar:
        if c not in seen:
            seen.add(c)
        else:
            pairs +=1
            seen.remove(c)
    return pairs

  • Mukultyagi8387
     + 0 comments

    public static int sockMerchant(int n, List ar) { int pairCount = 0; Dictionary DictionaryObj = new Dictionary();

        foreach (var item in ar)
    {
        if (DictionaryObj.ContainsKey(item))
        {
            DictionaryObj[item]++;
        }
        else
        {
            DictionaryObj[item] = 1;
        }
    }

    foreach (int count in DictionaryObj.Values)
    {
        pairCount += count / 2;
    }
    return pairCount;
}****