Brute force solution

  • 0

    Hi All,

    I can't understand why DFS would work here, would someone please explain it ?

    I would like to see the brute force solution, so that I can learn from it.

  • 1

    I haven't figured this out yet, but I think one brute force solution is to generate all permutations of the + and - signs, and that will require 2^n permutations.

    For example, for a list with 3 numbers, you have 2^3 possible options. And then you loop through each one to see if it produces the target sum. Each option is just the binary representation of that number, from 0 to 2^n

    0 ---
    1 --+
    2 -+-
    3 -++
    4 +--
    5 +-+
    6 ++-
    7 +++

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