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  1. Sales by Match
  2. Discussions

Sales by Match

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225 Discussions

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  • 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

  • shiaramoon
     + 0 comments
    #Python Solution
def sockMerchant(n, ar):
    pairs = 0
    colors = set(ar)
    
    for color in colors:
        pairs += math.floor(ar.count(color)/2)
    
    return pairs

  • nagapattinam_sa1
     + 0 comments

    this should work for C as the problem says: 1 <= ar[i] <=100

    int i, res = 0;
int map[101] = {0};

for(i = 0; i < n; i++) {
    map[ar[i]]++;
    if (map[ar[i]] == 2) {
        res++;
        map[ar[i]] = 0;
    }
}
return res;

  • f_klaudia
     + 0 comments

    My 3 Python solutions - two O(n) and one O(n^2)

    # O(n^2)
def sockMerchant(n, ar):
    cnt = 0
    seen = set()
    for num in ar:
        if num not in seen:
            seen.add(num)
            cnt += ar.count(num) // 2
    
    return cnt


# O(n)
def sockMerchant(n, ar):
    ar.sort()
    idx = 0
    cnt = 0
    while idx < len(ar) - 1:
        if ar[idx] == ar[idx + 1]:
            cnt += 1
            idx += 2
        else:
            idx += 1
    return cnt


# O(n)
def sockMerchant(n, ar):
    cnt = [0] * (max(ar) + 1)
    for num in ar:
        cnt[num] += 1
    return sum([num // 2 for num in cnt])