Interview Questions: Elementary Symbol Tables

Author: proton

week4/elementary-symbol-tables/interview-questions/task1/TASK.md

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+### Java autoboxing and equals()
+
+Consider two double values `a` and `b` and their corresponding `Double` values `x` and `y`.
+
+- Find values such that `(a==b)` is `true` but `x.equals(y)` is `false`.
+- Find values such that `(a==b)` is `false` but `x.equals(y)` is `true`.

week4/elementary-symbol-tables/interview-questions/task2/TASK.md

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+### Check if a binary tree is a BST
+
+Given a binary tree where each Node contains a key, determine whether it is a binary search tree. Use extra space proportional to the height of the tree.

week4/elementary-symbol-tables/interview-questions/task2/solution.rb

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+class TreeNode
+  attr_accessor :value, :left, :right
+
+  def initialize(value = 0, left = nil, right = nil)
+    @value = value
+    @left = left
+    @right = right
+  end
+
+  def to_a
+    arr = []
+    arr += left.to_a if left
+    arr << value
+    arr += right.to_a if right
+    arr
+  end
+end
+
+def build_tree(array)
+  root = TreeNode.new(array[0])
+  queue = [root]
+  i = 1
+
+  while i < array.size
+    current = queue.shift
+
+    if array[i]
+      current.left = TreeNode.new(array[i])
+      queue << current.left
+    end
+    i += 1
+
+    if array[i]
+      current.right = TreeNode.new(array[i])
+      queue << current.right
+    end
+    i += 1
+  end
+
+  root
+end
+
+# Given a binary tree where each Node contains a key, determine whether it is a binary search tree. Use extra space proportional to the height of the tree.
+
+bsts = []
+bsts << build_tree([4, 2, 6, 1, 3, 5, 7])
+bsts << build_tree([1, nil, 2])
+bsts << build_tree([3, 2, nil, 1])
+bsts << build_tree([5, 2, 8, 1, 3, nil, 9, nil, nil, nil, 4])
+
+non_bsts = []
+non_bsts << build_tree([3, 4, 1, nil, 2])
+non_bsts << build_tree([1, 2, 3, 4, 5, 6, 7])
+non_bsts << build_tree([4, 2, 6, 1, 3, 7, 5])
+
+def is_bst?(root)
+  if root.left
+    return false if root.left.value > root.value
+    return false unless is_bst?(root.left)
+  end
+  if root.right
+    return false if root.right.value < root.value
+    return false unless is_bst?(root.right)
+  end
+  true
+end
+
+bsts.each do |tree|
+  fail unless is_bst?(tree)
+end
+
+non_bsts.each do |tree|
+  fail if is_bst?(tree)
+end