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  1. Sales by Match
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Sales by Match

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recency

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357 Discussions

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  • Stelios10
     + 0 comments

    JAVA

    we just need to go through the n elements once (if we focus on performance and maybe not so clean code). I ve used the boolean array for memory optimization too, which makes the code a bit less readable too, possibly:

    public static int sockMerchant(int n, List<Integer> ar) {
         boolean[] socks = new boolean[101];
         int sockPairs = 0;
         
         for (int x: ar) {
            if (!socks[x]) {
                socks[x] ^= true;
            } else {
                socks[x] ^= true;
                sockPairs++;
            }
         }
         
         return sockPairs;
    }

  • ayushsaxena1512
     + 0 comments
    int sockMerchant(int n, vector<int> ar) {
    int pair = 0;
    unordered_map<int, int> map;
    for(size_t i = 0; i < ar.size(); i++){
        map[ar[i]]++;
    }
    for(auto it: map){
        pair += (it.second / 2);
    }
    return pair;
}

  • outsider973
     + 0 comments
       public static int sockMerchant(int n, List<Integer> ar) {
    // Write your code here
        var set = new HashSet<>();
        var numberPairs = 0;
        
        for (int sock : ar) {
            if (!set.add(sock)) {
                set.remove(sock);
                numberPairs++;
            }
        }
        return numberPairs;

    }

  • s_hmnd
     + 0 comments
    #Sorting, and counting stocks with same color, and division is not necesary
def sockMerchant(n, ar):
    p = defaultdict(bool)
    c=0
    for num in ar:
        if not p[num]:
            p[num] = True
        else:
            c += 1
            p[num] = False
    return c

  • anarod_val25
     + 0 comments

    `

    int sockMerchant(int n, vector ar) { std::sort(ar.begin(),ar.end()); int pairs = 0;

    for (int i = 0; i < n - 1;) {
    if (ar[i] == ar[i + 1]) {
        pairs++;
        i += 2; 
    } else {
        i += 1; 
    }
}    
return pairs;

    }

    `