Jethro's Braindump

Data Structures and Algorithms


Access Modifier

C: Class; P: Package; SC: Subclass; W: World

Modifier C P SC W
public Y Y Y Y
protected Y Y Y N
none Y Y N N
private Y N N N


Implement Comparable<T>:

int compareTo(T o)


  • If two objects are equal, hashCode must return the same result
  • hashCode must return the same result hen invoked on the same object more than once
int hashCode() {}
boolean equals(Object o) {}

Common Issues

Array out of bounds; divide by zero; infinite loop; exception thrown but not caught; access member variable in static method

Algorithmic Analysis

Amortized Cost
Algo has amortized cost \(T(n)\) if ∀ k ∈ \mathbb{Z}, cost of k operations is \(\leq kT(n)\)

Master’s Theorem

\(T(n) = aT(\frac{n}{b}) + f(n)\) where a ≥ 1, b > 1

  • \(f(n) \in O(n^c)\text{, } c < \log_ba \ \text{ then } T(n) \in \Theta(n^{\log_ba})\)

  • \(f(n) \in O(n^c\log^kn)\text{, } c = \log_ba \ \text{ then } T(n) \in \Theta(n^c\log^{k+1}n)\)

  • \(f(n) \in O(n^c), c > \log_ba \text{ and } \exists k \ \text{ st } af(\frac{n}{b})\le kf(n) \text{ then } T(n) \in \Theta(n^{\log_ba})\)

  • Common Ones

-   \\(T(n) = T(n/2) + \Theta(1) = O(\log n)\\) (Binary Search)
-   \\(T(n) = 2T(n/2) + \Theta(1) = O(n)\\) (Binary Tree Traversal)
-   \\(T(n) = 2T(n/2) + \Theta(\log n) = O(n)\\) (Optimal Sorted Matrix
-   \\(T(n) = 2T(n/2) + O(n) = O(n \log n)\\) (Merge Sort)

Abstract Data Types

Insert(i); Draw()
Stack (LIFO)
empty(); peek(); pop(); push(i)
Queue (FIFO)
add(); offer(i); peek(); poll(); remove()
double-ended queue


  • Binary Search \(O(\log n)\)
int binarySearch(int[] arr, int key) {
    int start = 0, end = arr.length - 1;
    int found = -1;
    while(start <= end) {
        int mid = start + (end - start)/2;
        if(arr[mid] < key) {
            start = mid + 1;
        } else if(arr[mid] > key) {
            end = mid - 1;
        } else {
            found = mid;
            // if we want first instance
            end = mid - 1;
            // if we want last instance
            // start = mid + 1;
    return found;
One sided Binary Search
Suppose one side is bounded, eg [1, ∞). Use the sequence [1,2,4,8,16…, \(2^k\)…] If it works for \(2^k\), then search on [\(2^{k-1}, 2^k\)]
Peak Finding
A[j] in array A is peak if (i) A[j] > A[j-1] (ii) A[j] > A[j+1]. If only one item in array, vacously true
1D Peak Finding \(O(\log n)\)
if a[n/2] < a[n/2-1] look at 1..n/2-1
else if a[n/2] < a[n/2+1] look at n/2+1..n
else return a[n/2]
2D Peak Finding \(O(m + n)\)
find max in border + cross O(m+n)
if max is peak return
else go into quadrant with higher number


Bubble Sort
Stable, In-place, W&A \(O(n^2)\), B \(O(n)\), S \(O(1)\); Invariant : At iteration i , the sub-array A[1 .. i] is sorted and any element in A[i + 1 .. A . size] is greater or equal to any element in A[1 .. i]
Selection Sort
In-Place, Unstable; find minimum element and swap. W,A,B \(O(n^2)\), S \(O(n/1)\); Invariant: a[0…i-1] is sorted all entries in a[i..n-1] are larger than or equal to the entries in a[0..i-1]
Insertion Sort
In-place, Stable; W \(O(n^2)\), B \(O(n \log n)\), S \(O(n)\); Invariant: The subarray a[i] consists of the original elements in sorted order.
Merge Sort
Stable, In Place; W/B \(O(n\log n)\), S \(O(n)\)
Quick Sort
In-place, Unstable; W \(O(n^2)\), A/B \(O(n\log n)\) S \(O(\log n)\)

Geometric Algorithms

Jarvis March \(O(hn)\)

  1. Find somewhere to start, e.g. y-min coordinate
  2. Add point with maximum angle from horizon \(O(n)\)
  3. Keep adding points with maximum angle from previous

Line Intersection Algorithm \(O(n\log n)\)

  1. Divide into two equal size sets (along vertical line)
  2. Recursively find convex hulls (base case 3 points)
  3. Merge convex hulls
    1. Find upper tangent lines
      1. while \((u,v,w)\) clockwise, decrement \(v\)
      2. while \((v,w,z)\) clockwise, increment \(w\)
    2. Find lower tangent lines
      1. while \((w,v,u)\) clockwise, increment \(v\)
      2. while \((z,v,u)\) clockwise, decrement \(w\)

Quick Hull \(O(n \log n)\)

  1. Choose pivot, construct two subproblems, delete interior points
  2. recurse on subproblems


Binary Trees (height h)

\(h(v) = max\left(h(v.left), h(v.right)\right) + 1\)

  • BST: left ST < key < right ST
  • traversal \(O(n)\) IN:LSR, PRE:SLR, POST:LRS
  • insert, search, findMax, findMin: \(O(h)\)
  • successor \(O(h)\):
    • if hasRightChild, smallest node in right sub-tree
    • else, first parent node that is left child (parent of node is successor)
  • delete \(O(h)\): switch numChild
    • 0: remove v
    • 1: remove v, connect child(v) to parent(v)
    • 2: swap with successor(v), remove(v)

AVL Trees (height \(h = \log n\))

  • Property: Every node is height-balanced
  • \(\lvert v.left.height - v.right.height \rvert \le 1\)
  • insert \(O(\log n)\):
    • insert key in BST
    • walk up, perform max 2 rotations if out-of-balance
  • delete(v): (\(\log n\) rotations)
    • If v has 2 children, swap with successor
    • delete v, and reconnect children
    • for every ancestor of deleted node
      • rotate if out-of-balance
  • Splay Trees: Rotate nodes that are accessed to root. consider using where operations are non-random.

Augmented Trees

  • Rank Tree (Order Statistics)
-   store weight of tree in each node:
-   \\(w(v) = w(v.left) + w(v.right) + 1\\)
-   select(k) \\(O(\log n)\\): finds node with rank \\(k\\)


rank = left.weight + 1;
  if (k == rank)
    return v
  else if (k < rank)
  else return

-   rank(v) \\(O(\log n)\\): computes rank of node v


rank = v.left.weight + 1
  while (v != null) do
    if v is left child do nothing
    if v is right child,
       rank += v.parent.left.weight + 1
    v = v.parent
  • Interval Trees
-   Each node is an interval \\((m, n), m \le n\\)
-   Sort by \\(m\\), augment node with maximum \\(n\\) of children in each node
-   search(x) \\(O(\log n)\\):


if x in c
  return c
else if c has no left child
  search in right subtree
else if x > max endpoint in c.left
  search in right subtree
else search in left subtree

-   findAll(x) \\(O(k \log n)\\) for k overlapping intervals


store it somewhere else
remove interval
repeat until no intervals found
  • Orthogonal Range Searching
-    1D

    1.  use a binary tree search tree
    2.  store all points in the leaves of the tree, internal nodes store
        only copies
    3.  each internal node v stores the max of any leaf in the left subtree
    4.  Query Time: \\(O(k + \log n)\\)
    5.  Building Tree: \\(O(n \log n)\\)

-    k-dim Tree

    1.  each node in the x-tree has a set of points in its subtree
    2.  store the y-tree at each x-node containing all points
    3.  Query Time: \\(O(k + \log^d n)\\)
    4.  Building Tree: \\(O(n \log^{d-1}n)\\)
    5.  Space: \\(O(n \log^{d-1}n)\\)
  • Custom Augmentations
-   **Average height of people taller**: augment nodes to include the
    count of the number of nodes in that sub-tree, along with the sum
    of the heights of all the people in that sub-tree. To return the
    desired average, first search for the name in the hash table; assume
    it is at node v; then find the sum of the heights of: the right-child
    of v, and if w is on the path from v to the root and v is in w’s
    left-subtree, then w’s right-subtree and w.

Hash Tables

  • n: #items, m: #buckets
  • Simple Uniform Hashing: Keys are equally likely to map to every bucket, and are mapped independently
    • \(load(ht) = \frac{n}{m}\)
    • \(E_\text{search} = 1 + \frac{n}{m}\)
    • Assume \(m=\Omega(n)\), \(E_\text{search} = O(1)\)

Hash Functions

  • Division
-   \\(h(k) = k \text{ mod } m\\), choose m prime
  • Multiplication
-   fix table size: \\(m=2^r\\), for some \\(r\\)
-   fix word size: \\(w\\), size of key in bits
-   fix odd constant \\(A\\), \\(A > 2^{w-1}\\)
-   \\(h(k) = (Ak) \text{ mod } 2^w >> (w - r)\\)
  • Rolling Hash
-   When key changes by single character


  • bucket stores linked list, containing (object, value)
  • Worst insert \(O(1 + cost(h))\)
  • Expected search = \(1 + \frac{n}{m} = O(1)\)
  • Worst search \(O(n)\)

Open Addressing

  • One item per slot, probe sequence of buckets until find only one
  • \(h(key, i) : U \mapsto {1..m}\), \(i\) is no. of collisions
  • search: keep probing until empty bucket, or exhausted entire table
  • delete: set key to tombstone value, so probe sequence still works
  • insert: on deleted cell, overwrite, else find next available slot
  • good hash function:
    1. \(h(key, i)\) enumerates all possible buckets
    2. Simple Uniform Hashing
  • Linear: \(h(k,i) = h(k) + i\), Clustering
  • Double: \(h(k,i) = f(k) + i \cdot g(k) \text{ mod } m\)
  • Insert, Search: \(\frac{1}{1-\alpha}\) where \(\alpha = \frac{n}{m} \le 1\)
  • good: saves space, rare mem alloc, better cache perf
  • bad: sensitive to hash, load

Cuckoo Hashing

  • Resolving hash collisions with worst-case constant lookup time
  • Lookup: inspection of just two locations in the hash table
  • Insertion: Insert into first table if empty; else kick out other key to second location.
  • If infinite loop, hash function is rebuilt in place

Table resizing

  • Scan old table \(O(m_1)\), create new table \(O(m_2)\), insert each element \(O(1)\), total \(O(m_1 + m_2 + n)\)
  • \(O(n)\) amor: if \(n == m\), \(m = 2m\), if \(n < \frac{m}{4}\), \(m = \frac{m}{2}\)

Fingerprint Hash Table (FHT)

  • Vector of 0/1 bits
  • no false negatives, but has false positives. \(P_{\text{no FP}} = \left(\frac{1}{e}\right)^{n/m}\)

Bloom Filter

  • use \(n\) hash functions. More space per item, but require \(n\) collisions for false positive.
  • \(P_{\text{coll}} = \left(1- e^{-kn/m}\right)^k\)
  • Two hash functions, \(h(k)\) and \(t(k)\), two tables \(T_1\) and \(T_2\)
  • insert: \(T_1[h(k)] = 1\), \(T_2[h(k)] = 1\)
  • search: if \(T_1[h(k)]\) and \(T_2[h(k)]\) both 1 return true


Type Space v,w any all
List \(O(V+E)\) slow fast fast
Mat \(O(V^2)\) fast slow slow
  • BFS/DFS do not explore all paths

  • BFS \(O(V+E)\)


  while Q is not empty:
    current = Q.dequeue()
    for each node n adj to current
      if n not visited
        n.parent = current
  • DFS \(O(V+E)\)
-   Same as BFS, but use stack instead of queue
  • Topological Sort (DAG)
-   Post-order DFS
-   Kahn's Algorithm (first append all nodes with no incoming edges to
    result set, remove edges connected to these nodes and repeat,
    also O(V+E))


  • Bellman-Ford \(O(EV)\)
-   \\(O(V^3)\\) if using Adj Matrix


do V number of times
  for (Edge e : graph)

-   can terminate early if no improvement
-   can detect negative cycle: perform V times, then perform once more,
    if have changes it has negative cycle
-   if all weights are the same, use BFS
  • Dijkstra \(O(E\log V)\)
[Interesting article on how Dijkstra's algorithm is everywhere](

-   Doesn't work with negative edge weights
-   can terminate once end is found


add start to PQ
dist[i] = INF for all i
dist[start] = 0
while PQ not empty
  w = pq.dequeue()
  for each edge e connected to w
    if edge is improvement
      update pq[w] O(logn)
      update dist[w]
  • DAG
-   Toposort, relax in order
-   SSSP on DAG: run topo sort, and relax edges in that order in \\(O(V+E)\\)
-   Single Source Longest Path problem is easy on DAG: multiply edge
    weights by -1 and run SSSP


  • implements priority queue, is a complete binary tree

  • priority of parent > priority of child

  • insert: create new leaf, bubbleUp

  • decreaseKey: update priority, bubbleDown

  • delete: swap with leaf, delete, and then bubble

  • store in array:

    • \(left(x) = 2x + 1\)
    • \(right(x) = 2x + 2\)
    • \(parent(x) = \lfloor(x-1)/2\rfloor\)
  • Heap Sort

1.  Heapify (insert n items)  O(n log n)
2.  Extract from heap n times (O(n log n))

3.  **Improvement**: recursively join 2 heaps and bubble root down (base
    case single node) O(n)
4.  O(n log n) worst case, deterministic, in-place
  • UFDS (weighted)
-   union(p,q) \\(O(\log n)\\)
    -   find parent of p and q
    -   make root of smaller tree root of larger tree
-   find(k) \\(O(\log n)\\)
    -   search up the tree, return the root
    -   (PC): update all traversed nodes parent to root

-   WU with PC, union and find \\(O(\alpha(m,n))\\)


  • acyclic subset of edges that connects all nodes, and has minimum weight

  • Properties

1.  Cutting edge in MST results in 2 MSTs
2.  **Cycle Poperty**: \\(\forall\\) cycle, max weight edge is not in MST
3.  **Cut Property**: \\(\forall\\) partitions, min weight edge
    across cut is in MST
  • Prim’s \(O(E \log V)\)
-   Uses cycle property


T = {start}
enqueue start's edges in PQ
while PQ not empty
  e = PQ.dequeue()
  if (vertex v linked with e not in T)
    T = T U {v, e}
    ignore edge
  • Kruskal’s \(O(E\log V)\)
-   Uses UFDS
-   It is possible that some edge in the first \\(V-1\\) edges will form a
    cycle with pre-existing MST solution


Sort E edges by increasing weight
T = {}
for (i = 0; i < edgeList.length; i++)
  if adding e = edgelist[i] does
  not form a cycle
    add e to T
    else ignore e
  • Boruvka’s \(O(E\log V)\)
T = { one-vertex trees }
While T has more than one component:
 For each component C of T:
   Begin with an empty set of edges S
   For each vertex v in C:
     Find the cheapest edge from v
     to a vertex outside of C, and
     add it to S
   Add the cheapest edge in S to T
 Combine trees connected by edges
  • Variants
1.  Same weight: BFS/DFS \\(O(E)\\)
2.  Edges have weight \\(1..k\\):
    -   Kruskal's
        -   Bucket sort Edges \\(O(E)\\)
        -   Union/check \\(O(\alpha (V))\\)
        -   Total cost: \\(O(\alpha(V)E)\\)
    -   Prim's
        -   Use array of size k as PQ, each slot holds linked
            list of nodes
    -   insert/remove nodes \\(O(V)\\)
    -   decreaseKey \\(O(E)\\)
3.  Directed MST
    -   \\(\forall\\) node except root, add minimum incoming edge \\(O(E)\\)
4.  MaxST
    -   negate all weights, run MST algo
  • MST Problems
-    How do I add an edge (A,B) of weight k into graph G and find MST quickly?

    -   Use cycle property; max edge in any cycle is not in MST
    -   only add (A,B) if k is not the max weight edge
    -   O(V + E) time to find max edge along A → B with DFS

-    Given an undirected graph with \\(K\\) power plants, find the minimum cost to connect all other sites.

    -   run Prim’s, use super source
    -   weight of new edges are zero
    -   this is a single MST

-    How do I make Kruskal run faster when sorting?

    -   Store edges in separate linked lists
    -   To process edges in increasing weight, process all edges in one
        linked list then the next
    -   Time: \\(O(E)\\) or \\(O(E\alpha(m, n))\\)
    -   Space: \\(O(E)\\), need to store all \\(E\\) edges

-    Minimum Bottleneck Spanning Tree (MBST)

    -   General idea: If I use some edge e that is not in the MST to replace
        some edge e’ in the MST, then my max. edge is max (max edge on
        original MST, e).
    -   Intuitively, my MST would then fulfill the condition of MBST.
    -   Note: Every MST is an MBST, but not every MBST is an MST

-    Find maximum distance between 2 vertices in MST

    -   Bruteforce: perform DFS starting from every single location since
        there is only one path from any node to another
    -   DFS: \\(O(V+E)\\), doing it \\(V\\) times, \\(O(V(V+E)) =O(V^2)\\) since \\(E =
          V - 1\\)
    -   Space: \\(O(V)\\), need to store all the edges in MST

Floyd-Warshall (APSP)

  • Shortest paths have optimal substructure
  • Shortest paths have overlapping subproblems
  • Idea: gradually allow usage of intermediate vertices
  • Invariant: At step k, shortest path via nodes 0 to k are correct
// precondition: A[i][j] contains weight
//  of edge (i,j) or inf if no edge
int[][] APSP(A) {
  // len = # vertices
  // clone A into S
  for(int k = 0; k < len; k++)
      for(int i = 0; i < len; i++)
          for(int j = 0; j < len; j++)
              S[i][j] =
                           S[i][k] + S[k][j]);
  return S;

Network Flow

k-edge connected
Source and target are k-edge connected if there are k edge disjoint paths(don’t share edges) from source to target.
Max flow
st-cut property with minimum capacity(outgoing from s, ignore incoming to s)
Min cut
Let \(S\) be the nodes reachable from the source in the residual graph. \(T\) = all other nodes, S → T is minimum cut
Augmenting Path
path in residual graph from s to t that has no 0 weight edges
  • Ford-Fulkerson
1.  Start with 0 flow
2.  While there exists augmenting path:
    -   find an augmenting path
    -   compute bottleneck (min edge)
    -   increase flow on the path by bottleneck capacity

Time Complexity:

-   DFS: \\(O(|F|E)\\)
-   BFS(Edmonds-Karp, shortest augmenting path): \\(O(VE^2)\\)
-   Dinitz: \\(O(V^2E)\\)

Graph Algorithms on Trees

  • Check if connected graph is tree
Run DFS, stop when after traversing \\(V-1\\) edges, return true if all
nodes connected and no other used edge. False otherwise. \\(O(V)\\)
  • Min Vertex Cover
-   Idea: transform tree into DAG, run DP
-   only two possiblities for each vertex; taken or not


int MVC(int v, int flag) {
    int ans = 0;
    if (memo[v][flag] != -1)
        return memo[v][flag];
    else if (leaf[v]) //if v is leaf
        ans = flag;
    else if (flag == 0) {
        ans = 0;
        for(child : adjList[v]) {
            ans+= MVC(child, 1);
    else if (flag == 1) {
        for (child : adjList[v]) {
            ans += min(MVC(child,1),
  • SSSP
-   On a weighted tree, any graph traversal algorithm (eg. DFS, BFS) can
    obtain the shortest path to any vertice in \\(O(V)\\)
-   Weight of shortest path between two vertices is the sum of the
    weights of edges on the unique path
  • ASSP
-   Run SSSP on V vertices in total \\(O(V^2)\\), compared to \\(O(V^3)\\) FW algorithm
  • Diameter
-   Originally, run FW in \\(O(V^3)\\) and do an \\(O(V^2)\\) all-pairs check,
    to total \\(O(V^2)\\).
-   Now, only need 2 \\(O(V)\\) traversals: DFS/BFS from any vertex \\(s\\) to find
    the furthest vertex \\(x\\). Then do a DFS/BFS one more time from vertex
    \\(x\\) to find furthest vertex \\(y\\). Length of unique path along x to y
    is the diameter of the tree.

Graph Modelling Techniques

  1. minimum shortest path from many source to one destination: run SSSP treating destination as source.
  2. multiple sources to multiple destinations: consider super source and super sink, with edge weight 0, and run Dijkstra (if no negative edge weights), BF otherwise.
  3. Attempt to convert graph into a DAG and use DP techniques. Example: attaching a variable to a vertex that is monotonically decreasing
  4. Shortest path between X and Y that passes through node \(A\): Compute two shortest paths; X to A, A to Y, and join the paths.

Parallel Algorithms

Parallel Fibonacci

parallelFib(n) {
  if(n < 2) then
  return n;
  x = spawn parallelFib(n - 1);
  y = spawn parallelFib(n - 2);
  return x + y;
  • Critical Path: \(T_\infty\), Parallelism = \(T_1/T_\infty\)
  • \(T_\infty(n) = max(T_\infty(n - 1), T_\infty(n - 2)) + O(1) = O(n)\)
  • \(T_p> T_1/p\)
  • \(T_p > T_\infty\), cannot run slower on more processors
  • Goal: \(T_p = (T_1/p) + T_\infty\), \(T_1/p\) is the parallel part, \(T_\infty\) is the sequential part

Matrix Addition

Before: • Work analysis: \(T_1(n) = O(n^2)\) • critical path analysis: \(T_\infty(n) = O(n^2)\) After:

  if(n == 1)
    C[i,j] = A[i,j] + B[i,j];
    spawn pMatAdd(A,B,C,i,j,n/2);
    spawn pMatAdd(A,B,C,i,j + n/2,n/2);
    spawn pMatAdd(A,B,C,i + n/2,j,n/2);
    spawn pMatAdd(A,B,C,i + n/2,j + n/2,n/2);
  • Work Analysis: \(T_1(n) = 4T_1(n/2) + O(1) = O(n^2)\)
  • Critical Path Analysis: \(T_\infty(n) = T_\infty(n/2) + O(1) = O(\log n)\)

Parallelized Merge Sort \(O(\log^3n)\)

pMerge(A[1..k], B[1..m], C[1..n])
  if (m > k) then pMerge(B, A, C);
  else if (n==1) then C[1] = A[1];
  else if (k==1) and (m==1) then
    if (A[1] <= B[1]) then
      C[1] = A[1]; C[2] = B[1];
      C[1] = B[1]; C[2] = A[1];
    // binary search for j where
    // B[j] <= A[k/2] <= B[j+1]
    spawn pMerge(A[1..k/2],
    spawn pMerge(A[k/2+1..l],

pMergeSort(A, n)
  if (n=1) then return;
    X = spawn pMergeSort(A[1..n/2], n/2)
    Y = spawn pMergeSort(A[n/2+1, n], n/2)
    A = spawn pMerge(X, Y);

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