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  1. Sales by Match
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Sales by Match

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228 Discussions

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  • pradymm
     + 0 comments
    public static int sockMerchant(int n, List<Integer> ar) {
// Write your code here
    int result = 0;
    Map<Integer, Integer> countMap = new HashMap<>();
    for(Integer i : ar){
        countMap.put(i, countMap.getOrDefault(i, 0) + 1);
    }
    for(Map.Entry<Integer, Integer> entry : countMap.entrySet()){
        result = result + entry.getValue() / 2;
    }
    return result;
}

  • harrydunn333
     + 0 comments
    def sockMerchant(n, ar):
    # Write your code here
    frequency = {}
    result = 0
    for i in ar:
        frequency[i] = frequency.get(i,0) + 1
    for value in frequency.values():
        result += value // 2
    return result

  • ivan_romero_dev1
     + 0 comments
    public static void main(String[] args) {
        try (var br = new BufferedReader(new InputStreamReader(System.in));
             var bw = new BufferedWriter(new OutputStreamWriter(System.out))) {
            int n = Integer.parseInt(br.readLine().trim());
            String[] socks = br.readLine().trim().split(" ");
            int[] sockCounts = new int[101];
            for (String sock : socks) {
                int color = Integer.parseInt(sock);
                sockCounts[color]++;
            }
            int pairs = 0;
            for (int count : sockCounts) {
                pairs += count >> 1;
            }
            bw.write(String.valueOf(pairs));
            bw.newLine();
            bw.flush();
        } catch (Exception e) {
            e.printStackTrace();
        }
    }

  • victor_reial
     + 0 comments

    Rust best solution

    If you’re looking for solutions to the 3-month preparation kit in either Python or Rust, you can find them below: my solutions 

    fn sales_by_match(n: i32, ar: &[i32]) -> i32 {
    //Time complexity: O(n)
    //Space complexity (ignoring input): O(n)
    let mut socks_hash = std::collections::HashMap::new();
    for sock in ar {
        match socks_hash.get(sock) {
            Some(v) => socks_hash.insert(sock, v + 1),
            None => socks_hash.insert(sock, 1),
        };
    }
    let mut total_pairs = 0;
    for values in socks_hash.values() {
        total_pairs += values / 2;
    }
    total_pairs
}

  • victor_reial
     + 0 comments

    Python best solution

    If you’re looking for solutions to the 3-month preparation kit in either Python or Rust, you can find them below: my solutions 

    def sales_by_match(n, ar):
    #Time complexity: O(n)
    #Space complexity (ignoring input): O(n)
    socks_dict = {}
    for sock in ar:
        if sock in socks_dict:
            socks_dict[sock] += 1
        else:
            socks_dict[sock] = 1

    total_pairs = 0
    for values in socks_dict.values():
        total_pairs += values // 2

    return total_pairs