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  1. Practice
  2. Algorithms
  3. Implementation
  4. Birthday Chocolate
  5. Discussions

Birthday Chocolate

  • milczarek_r + 4 comments

    Same approach, but with iterators instead.

    int solve(int n, vector < int > s, int d, int m){
    int ways = 0;
    for(auto it = s.cbegin(); it != s.cend(); ++it){
        if(d == std::accumulate(it, it + m, 0))
            ways++;
    }
    return ways;
}
    • viigihabe + 0 comments
      [deleted]
    • tpacker + 2 comments

      I believe this is O(n^2). O(n) is also possible.

      • gk_2000 + 1 comment

        The OP is O(n)

        • victorz + 1 comment

          It's O(nm).

          • kjkanishk211194 + 0 comments

            its o(n^2), coz in worst case senario, all the choclate blocks would add up to d.

      • bineykingsley36 + 1 comment

        I have implemented it in O(n).

        • kjkanishk211194 + 0 comments

          can you show your logic ???

    • Spleen + 0 comments

      Why call accumulate function every iteration when you add only one square ? It's not the same approach : you are O(nlog(n)), his solution is in O(n)

    • 24god10 + 0 comments

      can U explain me about 2 lines of your code?: for(auto it = s.cbegin(); it != s.cend(); ++it){ if(d == std::accumulate(it, it + m, 0)) I think a accumulate will sum element out of vector, I think it will be it != cend() - m more saver.