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

Sales by Match

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6060 Discussions

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  • abythomas300
     + 0 comments

    Initial solution in JS. I will post an optimized solution later.

    function sockMerchant(n, ar){
  
    let pairCount = 0;
    
    while(ar.length>1) {
      for(let i = 1; i < ar.length; i++) {
        if(ar[0] === ar[i]) {
          pairCount += 1
          ar.splice(0, 1)
          ar.splice(i-1, 1)
          i = 0
        }
      }
      ar.splice(0, 1)
    }
    return pairCount

}

  • rharikrishnan871
     + 0 comments

    Easy to read code, concise and efficient

    def sockMerchant(n, ar): Sum = 0 for i in list(set(ar)): Sum += ar.count(i)//2 return Sum `

  • mohamedkhallaou1
     + 0 comments
     public static int sockMerchant(int n, List<Integer> ar) {
    // Write your code here
        Map<Integer,Integer> colorCount =new HashMap<>();
        for(int sock : ar ){
            colorCount.put(sock, colorCount.getOrDefault(sock,0)+1);
        }
        
        Integer pairs = 0;
        for(Integer c : colorCount.values()){
            pairs+=c/2;
            

        }
        return pairs;

    }

}

  • carlos_pancioni
     + 1 comment

    Java using Collections frequency utility:

    public static int sockMerchant(int n, List<Integer> ar) {
    Set<Integer> unique = new HashSet<>(ar);

    int pairs = 0;
    for(Integer i : unique){
        Integer frequency = Collections.frequency(ar, i);
        pairs = pairs + (frequency / 2);
    }
    return pairs;
}

  • wmgarciaporcel
     + 0 comments

    This is my solution in JS:

    function sockMerchant(n, ar) {
        
    let pairCounter=0;
    
    while(ar.length>1){
        let color= ar.shift();
        for(let i=0; i<ar.length; i++){
            if(color===ar[i]) {
             pairCounter++;
             ar.splice(i,1);
             break;   
            }
        }
    }
    return pairCounter;
}