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

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6049 Discussions

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  • dcruz14
     + 0 comments

    I want to recieve feedback about my solution in Javascript. How can I do it better?:

    function sockMerchant(n, ar) {
    ar.sort((a, b)=> a - b)
    let pairs=0;
    for(let i=0;i<ar.length;i++){
        if(ar[i]===ar[i+1]){
            pairs++;
            i++;
        }
    }
    return pairs;
}

  • zold_ithunter
     + 0 comments

    Java

    Map<Integer, Integer> check = new HashMap<>();
        AtomicInteger rs = new AtomicInteger(0);
        ar.forEach(it -> {
            if (check.containsKey(it)) {
                rs.getAndIncrement();
                check.remove(it);
            } else {
                check.put(it, it);
            }
        });
        return rs.get();

  • macellinoss035
     + 0 comments

    C# Solution

        public static int sockMerchant(int n, List<int> ar)
    {
        Dictionary<int, int> pairs = new Dictionary<int, int>();
        
        for (int i = 0; i < n; i++) {
            if (pairs.ContainsKey(ar[i])) pairs[ar[i]]++;
            else pairs[ar[i]] = 1;
        }
        
        int sum = 0;
        foreach (KeyValuePair<int, int> pair in pairs) {
            sum += pair.Value / 2;
        }
        
        return sum;
    }

    Grouping up each socks and then just divide each group by 2. In C#, when a mathematical operation would result in a floating-point number for integer, the decimal would be concatenated. So if Sock 1 has 5 socks, divided by 2 it would result in 2 pairs - which is correct. Probably could optimize the loop further but the idea was there.

  • asimparui2003
     + 0 comments

    def sockMerchant(n, ar): # Write your code here count=0 y= Counter(ar) for i in y.values(): t=int(i/2) count+=t * return count

  • asimparui2003
     + 0 comments

    def sockMerchant(n, ar): # Write your code here count=0 y= Counter(ar) for i in y.values(): t=int(i/2) count+=t * return count