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  1. Sales by Match
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Sales by Match

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recency

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340 Discussions

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  • ahirshson
     + 0 comments

    Java solution using a Integer, Boolean map. Idea is to negate whatever bool is found in the map for each value. If it isn't already in the map put false. Increment a counter when something is made true. // Write your code here HashMap map = new HashMap<>(); int pairs = 0; System.out.println(map); for(Integer i : ar){ map.put(i, !map.getOrDefault(i,true)); if(map.get(i) == true){pairs++;} } S }

    }

  • EhsanAsh
     + 0 comments

    in Js:

    function sockMerchant(n, ar) {
    const pairsOfColors = ar.reduce((obj, num) => {
        obj[num] = (obj[num] || 0) + 1;
        return obj;
    }, {});
    let pairs = [];
    for (let key in pairsOfColors) {
        if (!(pairsOfColors[key] <= 1) && pairsOfColors[key] % 2 === 0) {
            pairs.push(pairsOfColors[key] / 2);
        } else if (!(pairsOfColors[key] <= 1) && pairsOfColors[key] % 2 === 1) {
            pairs.push((pairsOfColors[key] - 1) / 2);
        } else {
            pairs.push(0);
        }
    }
  return pairs.reduce((acc, num) => (acc + num), 0);
}

  • hasaladarshana2
     + 0 comments

    C#

            var cc = ar.GroupBy(x => x)
                       .Select(g => new { Count = g.Count(), Number = g.Key })
                       .ToList();

        var pairs = cc.Where(x => x.Count >= 2).Sum(x => x.Count / 2);
        return pairs;

  • ruan_amaral_lem1
     + 0 comments

    My asnwer C#

     int pairs = 0;
        ar.Sort();

        for (int i = 0; i < n - 1; i++)
        {
            if (ar[i] == ar[i + 1])
            {
                pairs++;
                i++;
            }
        }

        return pairs;

  • hank_ring
     + 0 comments
    def sockMerchant(n, ar):
    return sum([math.floor(ar.count(s) / 2) for s in set(ar)])