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  1. Prepare
  2. Algorithms
  3. Strings
  4. Gemstones
  5. Discussions

Gemstones

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1047 Discussions

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  • geovanni_luna
     + 0 comments

    Here is a O(n+m) time and O(1) space solution in python.

    O(n+m) is the average of the worse case scenarion, where all 26 chars will be present in every string. This means in the worse case the time will be O(Ch*n).

    def gemstones(arr):
    res = set(arr[0])
    
    for i in range(1, len(arr)):
        res &= set(arr[i])
    
    return len(res)

  • fruggiero
     + 0 comments

    C# solution:

    public static int gemstones(List<string> arr)
    {
        HashSet<char> common = new HashSet<char>(arr[0]);
        for (int i = 1; i < arr.Count; i++)
        {
            common.IntersectWith(arr[i]);
        }
        return common.Count;
    }

  • afzal_saiyed931
     + 0 comments
    def gemstones(arr):
    # Write your code here
    common = set(arr[0])
    
    for i in range(1, len(arr)):
        common &= set(arr[i])
        
    return len(common)

  • afzal_saiyed931
     + 0 comments
    def gemstones(arr):
    # Write your code here
    gemstones = 0
    map1 = {}
    
    for item in arr:
        map2 = {}
        for i in item:
            if map1.get(i):
                if not map2.get(i):
                    map1[i] += 1
                    map2[i] = 1
                else:
                    continue
                if map1[i] == len(arr):
                    gemstones += 1
            else:
                map1[i] = 1
                map2[i] = 1
            
    return gemstones

  • yashdeora98294
     + 0 comments

    Here is problem solution in python java c++ c and javascript - https://programmingoneonone.com/hackerrank-gemstones-problem-solution.html