Discussion on Sales by Match Challenge

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1 day ago+ 0 comments

Java

 public static int sockMerchant(int n, List<Integer> ar) {
        int pairs = 0;
        int size = ar.size();
        while(size-- > 0){
            int sock = ar.get(0);
            ar.remove(0);
	        if (ar.contains(sock)){
	            pairs++;
	            ar.remove(ar.indexOf(sock));
	            size--;
	            }
	        }
        return pairs;
}

2 days ago+ 0 comments

Python Solution

def sockMerchant(n, ar):
    n_pairs = 0
    pair_dict = Counter(ar)
    for value in pair_dict.values():
        n_pairs += (value//2)
    return n_pairs

6 days ago+ 0 comments

def sockMerchant(n, ar):
    sock_counts = {}  
    pairs = 0  
    
    for sock in ar:
        if sock in sock_counts:
            sock_counts[sock] += 1
        else:
            sock_counts[sock] = 1
    
    for count in sock_counts.values():
        pairs += count // 2  
    
    return pairs

1 week ago+ 0 comments

Typescript solution. Time O(n), Space O(k). k is the number of colors.

function sockMerchant(n: number, ar: number[]): number { if (n < 2) return 0;

const unpairedSocks = new Set<number>();
let pairs = 0;

for (const color of ar) {
    if (unpairedSocks.has(color)) {
        pairs++;
        unpairedSocks.delete(color);
    } else {
        unpairedSocks.add(color);
    }
}

return pairs;

}

1 week ago+ 0 comments

C++ Solution:

int sockMerchant(int n, vector<int> ar) {
    map<int, int> mp;
    int pair = 0;
    for(int color : ar){
        mp[color]++;
        if(mp[color]%2 == 0){
            pair++;
        }
    }
    return pair;
}

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