New local drive

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Author: msucorey

.byebug_history

@@ -0,0 +1,23 @@
+exit
+s
+line
+s
+key_length
+s
+set_header
+s
+header
+n
+s
+header
+s
+n
+exit
+_n
+_x
+s
+exit
+n
+s
+x
+s

README.md

@@ -1,2 +1,2 @@
-# codeval
-CodeEval stuff
+# codeval
+CodeEval stuff

bitpositions.rb

@@ -1,26 +1,26 @@
-=begin
-BIT POSITIONS
-CHALLENGE DESCRIPTION:
-Given a number n and two integers p1,p2 determine if the bits in position p1
-and p2 are the same or not. Positions p1 and p2 are 1 based.
-INPUT SAMPLE:
-The first argument will be a path to a filename containing a comma separated
-list of 3 integers, one list per line. E.g.
-86,2,3
-125,1,2
-OUTPUT SAMPLE:
-Print to stdout, 'true'(lowercase) if the bits are the same, else 'false'(lowercase). E.g.
-true
-false
-=end
-
-lines = File.readlines(ARGV[0])
-
-lines.each do |line|
-  num_strs = line.chomp.split(",")
-  n = num_strs.first
-  p1, p2 = num_strs[1..2]
-  n_bin_as_string = n.to_i.to_s(2)
-  n_bin_as_string.reverse!
-  puts n_bin_as_string[p1.to_i - 1] == n_bin_as_string[p2.to_i - 1]
-end
+=begin
+BIT POSITIONS
+CHALLENGE DESCRIPTION:
+Given a number n and two integers p1,p2 determine if the bits in position p1
+and p2 are the same or not. Positions p1 and p2 are 1 based.
+INPUT SAMPLE:
+The first argument will be a path to a filename containing a comma separated
+list of 3 integers, one list per line. E.g.
+86,2,3
+125,1,2
+OUTPUT SAMPLE:
+Print to stdout, 'true'(lowercase) if the bits are the same, else 'false'(lowercase). E.g.
+true
+false
+=end
+
+lines = File.readlines(ARGV[0])
+
+lines.each do |line|
+  num_strs = line.chomp.split(",")
+  n = num_strs.first
+  p1, p2 = num_strs[1..2]
+  n_bin_as_string = n.to_i.to_s(2)
+  n_bin_as_string.reverse!
+  puts n_bin_as_string[p1.to_i - 1] == n_bin_as_string[p2.to_i - 1]
+end

bptest.txt

@@ -1,2 +1,2 @@
-86,2,3
-125,1,2
+86,2,3
+125,1,2

cdtest.txt

@@ -1,3 +1,3 @@
-2 0 6 3 1 6 3 1 6 3 1
-3 4 8 0 11 9 7 2 5 6 10 1 49 49 49 49
-1 2 3 1 2 3 1 2 3
+2 0 6 3 1 6 3 1 6 3 1
+3 4 8 0 11 9 7 2 5 6 10 1 49 49 49 49
+1 2 3 1 2 3 1 2 3

cycledetection.rb

@@ -1,46 +1,46 @@
-=begin
-INPUT SAMPLE:
-Your program should accept as its first argument a path to a filename containing a sequence of numbers (space delimited). The file can have multiple such lines. E.g
-2 0 6 3 1 6 3 1 6 3 1
-3 4 8 0 11 9 7 2 5 6 10 1 49 49 49 49
-1 2 3 1 2 3 1 2 3
-OUTPUT SAMPLE:
-Print to stdout the first cycle you find in each sequence. Ensure that there are no trailing empty spaces on each line you print. E.g.
-6 3 1
-49
-1 2 3
-The cycle detection problem is explained more widely on wiki
-Constrains:
-The elements of the sequence are integers in range [0, 99]
-The length of the sequence is in range [0, 50]
-=end
-
-lines = File.readlines(ARGV[0])
-
-lines.each do |line|
-  numbers = line.split(" ").map(&:to_i)
-  found_cycle = false
-  cycle_length = 1
-  while (!found_cycle) && (cycle_length < numbers.length)
-    tortoise_index = 0
-    hare_index = tortoise_index + cycle_length
-    while (hare_index < numbers.length - cycle_length - 1)
-      if (numbers[tortoise_index] == numbers[hare_index]) && !found_cycle
-        found_cycle = true
-        cycle = numbers[tortoise_index..(hare_index - 1)]
-        cycle_start_index = tortoise_index
-        while cycle_start_index < (numbers.length - cycle_length)
-          found_cycle = false if numbers[cycle_start_index] != numbers[cycle_start_index + cycle_length]
-          found_cycle = false if numbers[cycle_start_index..(cycle_start_index + cycle_length - 1)] != cycle
-          cycle_start_index += cycle_length
-        end
-        puts cycle.map(&:to_s).join(" ") if found_cycle
-      end
-      tortoise_index +=1
-      hare_index += 1
-    end
-    cycle_length +=1
-  end
-end
-
-__END__
+=begin
+INPUT SAMPLE:
+Your program should accept as its first argument a path to a filename containing a sequence of numbers (space delimited). The file can have multiple such lines. E.g
+2 0 6 3 1 6 3 1 6 3 1
+3 4 8 0 11 9 7 2 5 6 10 1 49 49 49 49
+1 2 3 1 2 3 1 2 3
+OUTPUT SAMPLE:
+Print to stdout the first cycle you find in each sequence. Ensure that there are no trailing empty spaces on each line you print. E.g.
+6 3 1
+49
+1 2 3
+The cycle detection problem is explained more widely on wiki
+Constrains:
+The elements of the sequence are integers in range [0, 99]
+The length of the sequence is in range [0, 50]
+=end
+
+lines = File.readlines(ARGV[0])
+
+lines.each do |line|
+  numbers = line.split(" ").map(&:to_i)
+  found_cycle = false
+  cycle_length = 1
+  while (!found_cycle) && (cycle_length < numbers.length)
+    tortoise_index = 0
+    hare_index = tortoise_index + cycle_length
+    while (hare_index < numbers.length - cycle_length - 1)
+      if (numbers[tortoise_index] == numbers[hare_index]) && !found_cycle
+        found_cycle = true
+        cycle = numbers[tortoise_index..(hare_index - 1)]
+        cycle_start_index = tortoise_index
+        while cycle_start_index < (numbers.length - cycle_length)
+          found_cycle = false if numbers[cycle_start_index] != numbers[cycle_start_index + cycle_length]
+          found_cycle = false if numbers[cycle_start_index..(cycle_start_index + cycle_length - 1)] != cycle
+          cycle_start_index += cycle_length
+        end
+        puts cycle.map(&:to_s).join(" ") if found_cycle
+      end
+      tortoise_index +=1
+      hare_index += 1
+    end
+    cycle_length +=1
+  end
+end
+
+__END__

dectobin.rb

@@ -1,25 +1,25 @@
-=begin
-DECIMAL TO BINARY
-CHALLENGE DESCRIPTION:
-You are given a decimal (base 10) number, print its binary representation.
-INPUT SAMPLE:
-Your program should accept as its first argument a path to a filename
-containing decimal numbers greater or equal to 0, one per line.
-Ignore all empty lines.
-For example:
-2
-10
-67
-OUTPUT SAMPLE:
-Print the binary representation, one per line.
-For example:
-10
-1010
-1000011
-=end
-
-lines = File.readlines(ARGV[0])
-
-lines.each do |line|
-  puts line.chomp.to_i.to_s(2)
-end
+=begin
+DECIMAL TO BINARY
+CHALLENGE DESCRIPTION:
+You are given a decimal (base 10) number, print its binary representation.
+INPUT SAMPLE:
+Your program should accept as its first argument a path to a filename
+containing decimal numbers greater or equal to 0, one per line.
+Ignore all empty lines.
+For example:
+2
+10
+67
+OUTPUT SAMPLE:
+Print the binary representation, one per line.
+For example:
+10
+1010
+1000011
+=end
+
+lines = File.readlines(ARGV[0])
+
+lines.each do |line|
+  puts line.chomp.to_i.to_s(2)
+end

disemvowel.js

@@ -1,9 +1,9 @@
-function disemvowel(str) {
-  var output = "";
-  for (var n = 0;n < str.length; n++) {
-    if (!"aeiou".includes(str.charAt(n))) {
-      output += str.charAt(n);
-    }
-  }
-  return output;
-}
+function disemvowel(str) {
+  var output = "";
+  for (var n = 0;n < str.length; n++) {
+    if (!"aeiou".includes(str.charAt(n))) {
+      output += str.charAt(n);
+    }
+  }
+  return output;
+}

doublesquares.rb

@@ -1,49 +1,49 @@
-=begin
-DOUBLE SQUARES
-CHALLENGE DESCRIPTION:
-Credits: This challenge appeared in the Facebook Hacker Cup 2011.
-A double-square number is an integer X which can be expressed as the sum of
-two perfect squares. For example, 10 is a double-square because 10 = 3^2 +
-1^2. Your task in this problem is, given X, determine the number of ways in
-which it can be written as the sum of two squares.
-For example, 10 can only be written as 3^2 + 1^2 (we don't count 1^2 + 3^2 as
-being different). On the other hand, 25 can be written as 5^2 + 0^2 or as 4^2
-+ 3^2.
-NOTE: Do NOT attempt a brute force approach. It will not work. The following
-constraints hold:
-0 <= X <= 2147483647
-1 <= N <= 100
-INPUT SAMPLE:
-Your program should accept as its first argument a path to a filename. You
-should first read an integer N, the number of test cases. The next N lines
-will contain N values of X.
-5
-10
-25
-3
-0
-1
-OUTPUT SAMPLE:
-E.g.
-1
-2
-0
-1
-1
-=end
-
-lines = File.readlines(ARGV[0])
-
-N = lines[0].chomp.to_i
-
-lines[1..N].each do |line|
-  x = line.chomp.to_i
-  count = 0
-  (0..Math.sqrt(x).to_i).each do |n|
-    root = Math.sqrt(x - n**2)
-    count +=1 if root == root.to_i.to_f
-  end
-  count /= 2
-  count +=1 if x == 0
-  puts count
-end
+=begin
+DOUBLE SQUARES
+CHALLENGE DESCRIPTION:
+Credits: This challenge appeared in the Facebook Hacker Cup 2011.
+A double-square number is an integer X which can be expressed as the sum of
+two perfect squares. For example, 10 is a double-square because 10 = 3^2 +
+1^2. Your task in this problem is, given X, determine the number of ways in
+which it can be written as the sum of two squares.
+For example, 10 can only be written as 3^2 + 1^2 (we don't count 1^2 + 3^2 as
+being different). On the other hand, 25 can be written as 5^2 + 0^2 or as 4^2
++ 3^2.
+NOTE: Do NOT attempt a brute force approach. It will not work. The following
+constraints hold:
+0 <= X <= 2147483647
+1 <= N <= 100
+INPUT SAMPLE:
+Your program should accept as its first argument a path to a filename. You
+should first read an integer N, the number of test cases. The next N lines
+will contain N values of X.
+5
+10
+25
+3
+0
+1
+OUTPUT SAMPLE:
+E.g.
+1
+2
+0
+1
+1
+=end
+
+lines = File.readlines(ARGV[0])
+
+N = lines[0].chomp.to_i
+
+lines[1..N].each do |line|
+  x = line.chomp.to_i
+  count = 0
+  (0..Math.sqrt(x).to_i).each do |n|
+    root = Math.sqrt(x - n**2)
+    count +=1 if root == root.to_i.to_f
+  end
+  count /= 2
+  count +=1 if x == 0
+  puts count
+end

dstest.txt

@@ -1,6 +1,6 @@
-5
-10
-25
-3
-0
-1
+5
+10
+25
+3
+0
+1

dtbtest.txt

@@ -1,3 +1,3 @@
-2
-10
-67
+2
+10
+67

endianness.rb

@@ -1,9 +1,9 @@
-=begin
-Print to stdout the endianness, wheather it is LittleEndian or BigEndian.
-=end
-
-if [1].pack("I") == [1].pack("N")
-  puts "BigEndian"
-else
-  puts "LittleEndian"
-end
+=begin
+Print to stdout the endianness, wheather it is LittleEndian or BigEndian.
+=end
+
+if [1].pack("I") == [1].pack("N")
+  puts "BigEndian"
+else
+  puts "LittleEndian"
+end

filesize.rb

@@ -1 +1 @@
-puts File.size(ARGV[0])
+puts File.size(ARGV[0])