36_binary_flip.rb

# Task
# You are given a binary string (a string consisting of only '1' and '0'). 
# The only operation that can be performed on it is a Flip operation.

# It flips any binary character ( '0' to '1' and vice versa) 
# and all characters to the right of it.

# For example, applying the Flip operation to the 4th character of string "1001010"
# produces the "1000101" string, since all characters from the 4th to the 7th are flipped.

# Your task is to find the minimum number of flips required to convert
# the binary string to string consisting of all '0'.

# Example
# For s = "0101", the output should be 3.

# It's possible to convert the string in three steps:

#  "0101" -> "0010"
#     ^^^
#  "0010" -> "0001"
#      ^^
#  "0001" -> "0000"


def flip(string, index)
  while index < string.size
    if string[index] == "1"
      string[index] = "0"
    else
      string[index] = "1"
    end
    index += 1
  end
end

def bin_flip(binary_string)
  flips_used = 0

  i = 0
  while i < binary_string.length
    if binary_string[i] == "1"
      flip(binary_string, i)
      flips_used += 1
    end
    i += 1
  end

  return flips_used
end