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

Sales by Match

  • air_mvp_jordan
     + 0 comments

    This is my solution using C#.

    public static int sockMerchant(int n, List<int> ar)
{
		Dictionary<int, int> itemCounts = new Dictionary<int, int>();
		int pairCount = 0;

		for (int i = 0; i < n; i++) {
				if (itemCounts.ContainsKey(ar[i])) {
						itemCounts[ar[i]]++;
				} else {
						itemCounts[ar[i]] = 1;
				}
		}

		foreach(var item in itemCounts) {
				pairCount += (item.Value / 2);
		}

		return pairCount;
}

    Solution for Python3

    ` def sockMerchant(n, ar): # Write your code here arCount = Counter(ar) pairCount = 0

    for item in arCount:
    pairCount += arCount[item] // 2

return pairCount

    `