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Why binary search is O(log_2(n)):

  Every iteration, binary search halves the search space,
  until the search space has only one/zero items.

  Suppose the array size is n. To find out how many times
  we need to halve n by to get 1, we solve for x in the
  following:

      n 
    ----- = 1
     2^x

    n = 2^x

    x = log_2(n)

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Time complexity examples:

  O(n^2):
    double input size -> algo takes four times as long

    triple input size -> algo takes nine times as long

    T(n) = n^2

    T(3n) = (3n)^2
          = 9n^2
	  = 9T(n)

  O(n^3):
    double input size -> algo take eight times as long

    triple input size -> algo takes 27 times as long

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Exercise in appendix section:

  O(n^2) algorithm

  If input size is:
  - 1000 => 1.2 seconds
  - 2000 => 4.8 seconds
  - 10,000 => 120 seconds
  - 100,000 => 12,000 seconds => 200 minutes
  - 1,000,000 => 20,000 minutes