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adamvm Avatar
Name:Unknown
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Member Since:Jan 2019
Last Seen:Jun 2025
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    • Conversations (11)
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      • WarpedReality
      • commented on "Switcheroo" cpp solution
      • 3 years ago

      yeah it is inefficient (and unreadable). Just wanted to make it weird

    • Custom User Avatar
      • 4eckah78
      • commented on "Maximum Product " cpp solution
      • 3 years ago

      thanks a lot! :)

    • Custom User Avatar
      • adamvm
      • commented on "Switcheroo" cpp solution
      • 3 years ago

      why transform all characters even if they are in final form ('c')? No sense

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      • adamvm
      • commented on "Maximum Product " cpp solution
      • 3 years ago

      Instead custom lambda, you can use predefined std::multiplies()

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      • adamvm
      • commented on "Fake Binary" cpp solution
      • 3 years ago

      any benefit from using auto && in this case?

    • Custom User Avatar
      • adamvm
      • commented on "Count by X" cpp solution
      • 3 years ago

      with iteration from 1 to n you can save n-times on adding i+1

    • Custom User Avatar
      • adamvm
      • commented on "Valid Braces" cpp solution
      • 3 years ago

      std::map dict should be const, also as mentioned std::vector v is redundant versus map keys

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      • adamvm
      • commented on "Sum of two lowest positive integers" kata
      • 3 years ago

      There is missing test at least for C++ that includes {INT_MAX, INT_MAX, INT_MAX, INT_MAX}
      Its exotic case but possible and curretly code that overflows to -2 is passing

    • Custom User Avatar
      • adamvm
      • commented on "Take a Ten Minutes Walk" cpp solution
      • 3 years ago

      4 times only if e != w - short circuit of logical phrases in c++, but I agree it is not efficient

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      • adamvm
      • commented on "The dropWhile Function" cpp solution
      • 3 years ago

      I suggest to exchange order on logic expression pred(*itr) && itr != arr.end().
      In your variant program first will evaluate pred(*itr) that can be already == arr.end(), what means that you will calculate value for out of container data. In reversed version first will be evaluated that end of container was reached - it will give false and rest of AND conditions will be not evaluated due to so called shortcircuit of operator

    • Custom User Avatar
      • adamvm
      • commented on "Number of Divisions" cpp solution
      • 3 years ago

      What if n is negative?

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