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

Sales by Match

Sort 4937 Discussions, By:

  • lukes712
     + 93 comments

    '''

        Set<Integer> colors = new HashSet<>();
    int pairs = 0;

    for (int i = 0; i < n; i++) {
        if (!colors.contains(c[i])) {
            colors.add(c[i]);
        } else {
            pairs++;
            colors.remove(c[i]);
        }
    }

    System.out.println(pairs);

    '''

    • Time Complexity: O(n)
    • Space Complexity: O(n)

  • tushar2693
     + 68 comments
    int count = 0;
    for(int i=0; i<n; i++){
        if(c[i]!=0){
             for(int j=i+1; j<n; j++){
                if(c[i]==c[j]){
                    count++;
                    c[j]=0;
                    break;
                    }
                }           
        }
    }
    printf("%d", count);

    Solution in C.. O(N^2).. Can it be optimized in C?? Please help..

  • kasyap_msk
     + 27 comments

    My oneliner in python:

    return sum([ar.count(i)//2 for i in set(ar)])

  • BeastT23
     + 28 comments

    Fun with Python :)

    from collections import Counter
input()
socks, pairs = Counter(map(int,input().strip().split())), 0
for s in socks: pairs += socks[s]//2
print(pairs)

  • simonas_dabrega
     + 19 comments

    Javascript solution that worked for me:

    function sockMerchant(n, ar) {
var res = 0;
    ar.sort();
    for(let i=0; i<n;i++){
        if(ar[i] == ar[i+1]){
            i++;
                  res++;
           }
    }
return res;
}
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