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  1. Ice Cream Parlor
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Ice Cream Parlor

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73 Discussions

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  • hank_ring
     + 0 comments
    def icecreamParlor(m, arr):
    return [[i + 1, len(arr) - arr[::-1].index(m - v)] for i, v in enumerate(arr) if m - v in arr[i+1:]][0]

  • trip2ken
     + 0 comments

    Java 8 - Used an Index Sort to order the values. One time through the list and to find the correct pair.

        public static List<Integer> icecreamParlor(int m, List<Integer> arr) {
        List<Integer> rtn = new ArrayList<>();
        
        // Index Sort
        int[] indices = IntStream.range(0, arr.size())
                .boxed()
                .sorted(Comparator.comparingInt(i -> arr.get(i)))
                .mapToInt(Integer::intValue)
                .toArray();
        
        // Close in on the solution
        int l = 0;
        int h = arr.size()-1;
        while(l<h && arr.get(indices[l]) + arr.get(indices[h]) != m) {
            int vh = arr.get(indices[h]);
            int vl = arr.get(indices[l]);
            
            if (vh >= m || (vh + vl) > m)
                --h;
            else 
                ++l;           
        }
        
        // sort result into asscending order.
        rtn.add(Math.min( indices[l], indices[h] )+1); 
        rtn.add(Math.max( indices[l], indices[h] )+1); 
        
        return rtn ;
    }

  • ghoultarek
     + 0 comments

    Compact Python solution:

    def icecreamParlor(m, arr):

for i_1,a in enumerate(arr):

    for i_2,b in enumerate(arr):

        if i_2>i_1:
            if m-arr[i_1]==arr[i_2]:
                return [i_1+1,i_2+1]

  • yashdeora98294
     + 0 comments

    Here is hackerrank ice cream parlor problem solution in Python, Java, C++, C and javascript

  • dLuxHu
     + 0 comments

    A more elegant O(n) solution, similar to cb_kalz's solution (Scala):

      def icecreamParlor(m: Int, arr: Array[Int]): Array[Int] = {
    arr.zip(1 to Int.MaxValue).foldLeft((None: Option[Array[Int]], Map.empty[Int, Int])) {
      case ((result @ Some(_), acc), _) => (result, acc)
      case ((None, acc), (elem1, i)) =>
        val elem2 = m - elem1
        if (acc.contains(elem2)) (Some(Array(acc(elem2), i).sorted), acc)
        else (None, acc + (elem1 -> i))
    }._1.get
  }