Lecture 12: Midterm #1 Review and AVL Trees

This lecture we talked about the midterm we just took as well as implementing the add method in an AVL tree


  • Neal Malani

Midterm 1

Problem 1 A

  • Mostly marked “best answer”, but not all “correct answers”
    • Property
      • f = O(g), g(n) < = h(n)
      • f = O(h)
      • For f = O(n2) means there exists N and C such that for n >= N, f(n) < = C * n2

Problem 1 B

  • foo = O(log n) bar = O(n) not O(log n) Empirical running time?
    • Do not know constant factors
      • 1000000 log n < 0.001 n for large n
    • O is only for worst-case, not necessary what typically happens

Problem 3 A

  • Stack: resize for push to increase capacity, pop to decrease
    • Push/pop O(n) is worst case
  • Amortized cost?
    • If pop and resize to N1 then next decrease of N/2 will be cost of Cn/2
    • Each pop pay 1/(N/2) fraction of CN/2 = O(1) so amortized cost is O(1)

Problem 3 B

  • What about pushes together w/ pops?
    • Push: size N → N + 1, capacity N → 2N
    • Pop: size N+1 → N, capacity 2N → N
    • Repeat 1 and 2

Problem 4

  • Multi S Set
    • Sequence of elements in an array w restrictions
      • Find multiplicity of elements in O(log n)
    • How to find # of x’s
      • Use binary search to go through the entire array to search for x
      • Once x is found, split up the left and the right halves in order to conduct binary search for the first and last values of when x appears
      • Once found the indices of those two values, subtract them from each other and add one in order to get the multiplicity

AVL Trees

  • Last time
    • T a binary search tree and node v is balanced if h(left child) - h(right child) <= 1
    • T is a AVL tree if all nodes are balanced
  • Claim: if T is an AVL tree w/ n nodes, then h(T) = O(log n)
    • Add, remove, find have running time O(log n)
  • Note: If T is AVL then add/remove could destroy AVL property
  • Goal: Supplement the add and remove operations efficiently maintain balance
  • Dealing with add
    • Add creates imbalance
      • Adding node w:
        • Z = first ancestor of w that is unbalanced
        • Y = child of z toward w
        • X = child of y toward w
      • Make x the new root and y its left and z its right child
      • Why would the result be a BST?
        • Relative order is maintained
    • What is the running time?
      • O(1) once z is located
      • Finding z efficiently
        • Each node stores height as variables
        • When add/remove/restructure we update heights
      • O(height) b/c only need to update ancestors