Backtracking

### Remarks

• The backtracking algorithm solves the optimization problem by incrementally builds out candidate solutions and backtracks a candidate as soon as we can determine it will not leads to optimal.
• If we represent the partial solutions as a tree, backtracking is doing depth-first search in this tree.
• Since it explores all of the tree in worst case, its upper bounded to have exponential time complexity. It usually need some constraints to prune the tree.
• Very often we would pass both the candidate solution as well as the state of partial solution to the recursive call to speed up checks.

### Template

To find one solution

def backtrack(candidate, state):
if criterion(candidate, state): return candidate
for choice in choices:
if not constraints(candidate, state, choice): continue
result = backtrack(expand(candidate, state, choice))
if result: return result
return

solution = backtrack(init_candidate, init_state)


To collect all possible solutions

solutions = set()
def backtrack(candidate, state):
for choice in choices:
if not constraints(candidate, state, choice): continue
backtrack(expand(candidate, state, choice))

backtrack(init_candidate, init_state)


Binary Search

### Remarks

• The binary search algorithm solves the problem of finding a target within a sorted sequence.
• The algorithm iteratively updating the lower and upper bound of the search space where the target is sure to reside. The update is done by checking the middle element of search space.
• The algorithm runs in O(logN) time worst case.
• We can apply binary search on the domain of a monotonic function.
• The given sequence as a whole may not be ordered, but if we know certain subsequence is ordered, we can apply binary search on the subsequence.

### Usage

1. To find a value in array:
def binary_search(arr, t):
l, h = 0, len(arr) - 1
while l <= h:
m = (h + l) // 2  # use l + (h - l) // 2 for non python
if arr[m] == t: return m
elif arr[m] > t: h = m - 1
else: l = m + 1
return -1


In case the target value occurrs more than once, this binary search code will find one, but not guarantee to be either leftmost or rightmost.

1. To find the left most occurrence of the value:
def binary_search_left(arr, t):
l, h = 0, len(arr)
while l < h:
m = (h + l) // 2
if arr[m] < t: l = m + 1
else: h = m
return l


In the case the target value does not exist in the array, the index returned by this code points to the first element that is larger than the target value. Thus it is the location where we can insert the target value while maintain the sorted order of the sequence.

That is to say, if binary_search_left(arr, t) returns i, then arr[:i] < t and arr[i:] >= t.

1. To find the right most occurrence of the value:
def binary_search_right(arr, t):
l, h = 0, len(arr)
while l < h:
m = (h + l) // 2
if arr[m] <= t: l = m + 1
else: h = m
return l - 1


In the case the target value does not exist in the array, the index returned by this code points to the last element that is smaller than the target value. Thus insert the target value at this location will violate the sorted order.

That is to say, if binary_search_right(arr, t) returns i, then arr[:i+1] <= t and arr[i+1:] > t.

Sample Questions

Binary Tree Traversals

### Remarks ### DFS traversals

• Recurrsive implementation of the tree dfs traversals are pretty straightforwardly easy.
• With a hashmap, the iterative implementation can be very easy, but it need extra O(n) space for the hash.
• Otherwise, the iterative version of the three dfs traversals can be quite different and nasty.
def inorder(node):
if node.left: inorder(node.left)
print(node.val)
if node.right: inorder(node.right)

def postorder(node):
if node.left: postorder(node.left)
if node.right: postorder(node.right)
print(node.val)

def preorder(node):
print(node.val)
if node.left: preorder(node.left)
if node.right: preorder(node.right)

def tree_dfs_iterative(root):
stack = [root]
visited = set()
while stack:
node = stack.pop()
if not node: continue
if node in visited:
print(node.val)
else:
# stack.extend([node.right, node, node.left])  # inorder
# stack.extend([node, node.right, node.left])  # postorder
stack.extend([node.right, node.left, node])  # preorder

def inorder_iterative(root):
node, stack = root, []
while stack or node:
while node:
stack.append(node)
node = node.left
node = stack.pop()
print(node.val)
node = node.right

def preorder_iterative(root):
stack = [root]
while stack:
node = stack.pop()
if not node: continue
print(node.val)
stack.append(node.right)
stack.append(node.left)

# this version has to cut links, but reads more like the previous two order in code
def postorder_iterative(root):
node, stack = root, []
while stack or node:
if node:
stack.append(node)
node = node.left
elif stack[-1].right:
node = stack[-1].right
stack[-1].right = None
else:
print(stack.pop().val)

# the double push version works better, but harder to explain what's going on.
# basically, the first bush is for backtrack, the second push is to traversal.
def postorder_iterative(root):
stack = [root, root]
while stack:
node = stack.pop()
if stack and stack[-1] is node:
if node.right:
stack.extend([node.right] * 2)
if node.left:
stack.extend([node.left] * 2)
else:
print(node.val)


### BFS traversal

• BFS traversal with queue is quite straightforward.
def tree_bfs_iterative(root):
queue = deque([root])
while queue:
node = queue.popleft()
if not node: continue
print(node.val)
queue.extend([node.left, node.right])


Bit Manipulation

### 4 bit signed integer decimal to binary table

decimal binary decimal binary
-8 1000 7 0111
-7 1001 6 0110
-6 1010 5 0101
-5 1011 4 0100
-4 1100 3 0011
-3 1101 2 0010
-2 1110 1 0001
-1 1111 0 0000

### Operators

operator name what it does example sample property / usage
~ complement switch each 0 to 1 and each 1 to 0 ~0101 = 1010 ~x = -x - 1 and ~(x - 1) = -x
& and 1 iff both bits are 1 else 0 0110 & 0101 = 0100 x & - x to extract last bit, x & (x - 1) to clear last bit
| or 0 iff both bits are 0 else 1 0110 | 0101 = 0111 flags |= x to set the flag
^ xor (eXclusive OR) 1 iff only one bit is 1 else 0 0110 ^ 0101 = 0011 0 ^ x = x, x ^ x = 0 simple checksum reduce(operator.xor, words, 0)
« left shift shift bits to left 0011 « 2 = 1100 x << y is $$x \times 2^y$$
» right shift shift bits to right 0110 » 2 = 0001 x >> y is $$\lfloor{x \div 2^y}\rfloor$$

### Usages

1. Set a bit at a position
def set_bit(x, pos):

2. Clear a bit at a position
def clear_bit(x, pos):

3. Clear leading bits from position
def clear_leading_bits(x, pos):
mask = (1 << pos) - 1

4. Clear trailing bits after position
def clear_trailing_bits(x, pos):
mask = (1 << 31) - 1 << pos

5. Flip a bit at position
def flip_bit(x, pos):

6. Check a bit at position
def check_bit(x, pos):
return x >> pos & 1

7. Check number is even
def check_even(x):
return (x & 1) == 0

8. Check number is power of 2
def check_pow_2(x):
return x & (x - 1) == 0

9. Check two numbers are of opposite sign
def check_opposite_sign(x, y):
return x ^ y < 0

10. Swap integers the bit manipulation way
def swap(x, y):
x = x ^ y
y = x ^ y
x = x ^ y
return x, y


Sample Questions

### Remarks

• The breadth first search algorithm traversals tree / graph by exploring all the immediate child nodes / neighboring nodes first before moving to next depth level.
• BFS is good for shortest path findings.

### Template

Some dummy criterion and constraints.

def criterion(node, target_val=4):
return node and node.val == target_val

def constraints(node):
return node


Iterative implementation using queue.

def bfs(root):
queue = deque([root])

while queue:
node = queue.popleft()
if criterion(node): return node

for child in node.childrens:
if not constraints(child): continue
queue.append(child)

result = bfs(root)


Depth First Search

### Remarks

• The depth first search algorithm traversals tree / graph by exploring the node branch / edges as far as possible before backtrack and explore other alternatives.
• DFS can be used to solve problems like:
1. Reachability.
2. Path finding.
3. Topological sort.
4. Directed cycle detection.

### Template

Iterative implementation using queue.

def dfs(root):
stack = deque([root])

while stack:
node = stack.pop()
if criterion(node): return node

for child in node.childrens:
if not constraints(child): continue
stack.append(child)

result = dfs(root)


Recursive implementation with a hash map memorizing visited nodes.

visited = set()

def constraints(node):
return node and node not in visited

def dfs(node):
if criterion(node): return node

for child in node.childrens:
if constraints(child):
result = dfs(child)
if result: return result

result = dfs(root)


Dynamic Programming

### Remarks

• DP is a method for solving optimization problem by breaking the problem into smaller subproblems.
• DP requires that the overall optimal solution depends on the optimal solution to its subproblems.
• We might be able to apply DP when:
1. The subproblems can be seen as smaller version of the original problem.
2. The optimal solution of a problem is a function of the optimal solutions to one or more of its subproblems.

Two methods using dynamic programming are top-down memoization and bottom up tabulation.

• Top-down memoization is to recurssivly call the method to find solutions to subproblems and cache the results along the way to avoid redundent calculations.
• Bottom up tabulation is to solving the smallest subproblem first and then filling up the N dimension optimal solution table.

Classic example for dynammic programming is the fibonacci number calculation.

def fib(n):
if n <= 1: return 1
return fib(n - 1) + fib(n - 2)


The recursive relationship here is $$f(n) = f(n - 1) + f(n - 2)$$. The above code would work but will be very slow when n increases, due to the redundent calculations in the recursive calls.

With the top down memoization approach, we save the solutions of the subproblems and reuse the reuslts when we don’t need to recalculate.

memo = {0: 1, 1: 1}
def fib_memo(n):
if n in memo: return memo[n]
memo[n] = fib_memo(n - 1) + fib_memo(n - 2)
return memo[n]


With the bottom up tabulation approach, we iterativly fill up the 1 dimensional table up untill we can answer the problem.

def fib_tab(n):
fibs = [1, 1]
for i in range(2, n + 1):
fibs.append(fibs[-1] + fibs[-2])
return fibs[n]


Sample Questions ### Remarks

• Linked List is a sequence of node objects where each node store pointers to (1) some other node in the sequence, (2) the data this node is associated with.
• Linked List can be used as the underline data structure for many common abstract data types like lists, stacks, queues, etc.
operation time
push at front O(1)
pop at front O(1)
push at back O(1)
pop at back O(N)
access via index O(N)
insert in middle O(N)
delete in middle O(N)
swap two nodes O(N)

### Common operations on singly linked list.

1. Node class for singly linked list. python class ListNode(object): def init(self, item=None): self.item = item self.next = None

def repr(self): return ‘ ‘ .join([super().repr()[:-1], ‘of’, str(self.item), ‘ >’])

nodes = list(map(ListNode, [‘Head’, ‘Shoulders’, ‘Knees’, ‘Toes’])) for prev, next_ in zip(nodes, nodes[1:]): prev.next = next_

head = nodes tail = nodes[-1] del nodes


python
n = 0
while trav:
if verbose: print(trav.item)
if not trav.next: tail = trav
trav = trav.next
n += 1
return tail, n


1. Push and pop node at front. python def push_front(head, node): node.next = head return node


4. Push and pop node at back.
python
def push_back(tail, node):
tail.next = node
return node

while trav.next is not tail: trav = trav.next
trav.next = None
tail = trav
return tail

node = ListNode('Shoes')
tail = push_back(tail, node)
assert tail == node and traversal(head) == 5

assert traversal(head) == 4 and tail.next == None

1. Insert and delete nodes in the middle. python def get(head, i): trav = head prev = None while trav and i: i -= 1 prev = trav trav = trav.next return prev, trav

def insert(head, i, node): prev, next_ = get(head, i) prev.next = node node.next = next_


6. Swap nodes
python
i, j = sorted([i, j])
prev_j, node_j = get(node_i, j-i)
if prev_i: prev_i.next = node_j
prev_j.next = node_i
temp = node_i.next
node_i.next = node_j.next
node_j.next = temp
return head if prev_i else node_j


def reverse(head):
prev = None
while trav:
# trav.next, prev, trav = prev, trav, trav.next
next_ = trav.next
trav.next = prev
prev = trav
trav = next_
return prev


### Common Problems

1. Two Pointers
1. left, right - Palindrome
2. slow, fast - cycle, last k
2. Multiple Lists
1. Merge, Intersection
3. Search and Sorting
4. Implement Stack/Queue/Deque etc.
5. Doubly Linked List + Hash Table

Sample Questions

Priority Queue / Heap

### Remarks

• Priority Queue is a Queue, in which elements will be dequeued based on their priorites(other than timestamp).
• Elements on the queue should be comparable, otherwise we can’t deduce their ralative priorities.
• Priority Queue can be implemented with heap.
• Heap is a tree in which the parent is always larger(max heap) or smaller(min heap) than its children. (Heap property)
• We can implement priority queue using complete binary heap.

### Time complexity of a Binary Heap Queue

| operation | time | | —————- | —: | | construct | O(N) | | front | O(1) | | dequeue | O(logN) | | enqueue | O(logN) |

### Implementation

class HeapQueue(object):
from operator import lt, gt
comparators = {'min': lt, 'max': gt}

def __init__(self, data=None, comparator='min'):
self._heapq = [None] + list(data) if data else [None]
self._comparator = self.comparators.get(comparator, comparator)
self._n = len(self._heapq) - 1
if data: self.heapify()

def _compare(self, i, j):
return self._comparator(self._heapq[i], self._heapq[j])

def __len__(self):
return self._n

def is_empty(self):
return self._n == 0

def _swap(self, i, j):
self._heapq[i], self._heapq[j] = self._heapq[j], self._heapq[i]

def front(self):
assert not self.is_empty(), 'heapq is empty'
return self._heapq

def enqueue(self, x):
self._heapq.append(x)
self._n += 1
self.swim(self._n)

def dequeue(self):
assert not self.is_empty(), 'heapq is empty'
self._swap(1, self._n)
self._n -= 1
self.sink(1)
return self._heapq.pop()

def swim(self, k):
while k > 1 and self._compare(k, k >> 1):
self._swap(k, k >> 1); k >>= 1

def sink(self, k):
n = self._n
while k * 2 <= n:
j = k * 2
if j < n and self._compare(j + 1, j): j += 1
if self._compare(k, j): break
self._swap(k, j); k = j

def heapify(self):
for k in range(self._n >> 1, 0, -1):
self.sink(k)


Queue

### Remarks

• A queue is a abstract data type that is First-In-First-Out(FIFO).

### Implementation

An implementation of Queue using linked list.

class Queue(object):
class Node(object):
def __init__(self, item=None):
self.item = item
self.next = None

def __init__(self):
self.first = None
self.last = None
self._n = 0

@property
def is_empty(self):
return not self.first

def __len__(self):
return self.n

def peek(self):

def enqueue(self, item):
prev = self.last
self.last = self.Node(item)
if self.is_empty():
self.first = self.last
else:
prev.next = self.last
self._n += 1

def dequeue(self):
assert not self.is_empty(), "Queue is empty"
item = self.first.item
self.first = self.first.next
self._n -= 1
if self.is_empty():
self.last = None
return item


An implementation of Queue using fixed length array and pointer. Note the formula for pointer updates:

• Enqueue: (first + n) % capacity
• Dequeue: (first + 1) % capacity
class Queue(object):
def __init__(self, size):
self.size = size
self.queue = [None] * size
self.first = 0
self.n = 0

def enqueue(self, item):
assert not self.is_full(), 'Queue is full'
self.queue[(self.first + self.n) % self.size] = value
self.n += 1

def dequeue(self):
assert not self.is_empty(), 'Queue is empty'
item = self.queue[self.first]
self.queue[self.first] = None
self.first = (self.first + 1) % self.size
self.n -= 1
return item

def peek(self):
return self.queue[self.first]

def is_empty(self):
return self.n == 0

def is_full(self):
return self.n == self.size


Sample Questions

Sorting

Here are implementations of a few array sorting algorithms.

a = list('sortingexample')


### 1. Selection Sort

• Iteratively put the ith smallest element at ith location from the left.
• We maintain a sorted section from the left.
• In each iteration, we expand the sorted section by swap the smallest element from the unsorted section into the location.
def selection_sort(a):
n = len(a)
for i in range(n):
s = i
for j in range(i + 1, n):
if a[j] < a[s]:
s = j
a[i], a[s] = a[s], a[i]


### 2. Bubble Sort

• Iteratively put the ith largest element to ith location from the end.
• We maintain a sorted seciton from the right.
• In each iteration, we compare adjacent pairs of elements and swap larger values to the right until we reach the sorted part.
def bubble_sort(a):
n = len(a)
for i in range(1, n):
for j in range(n - i):
if a[j] > a[j + 1]:
a[j + 1], a[j] = a[j], a[j + 1]


### 3. Insertion Sort

• Iteratively insert new element into the sorted part.
• We maintain a sorted section from the left.
• In each iteration, we sink one new element from the right to the correct location in the sorted part.
def insertion_sort(a):
n = len(a)
for i in range(n):
for j in range(i, 0, -1):
if a[j] < a[j - 1]:
a[j - 1], a[j] = a[j], a[j - 1]


### 4. Merge Sort

• Recursively divide the array into two halves, sort both(recursively) and then merge the two sorted array.
• It follows the Divide-and-Conquer paradigm.
# A quick and dirty non-inplace version
def merge_sort(a):
def merge(left, right):
merged = []
while left and right:
merged.append(left.pop() if left[-1] > right[-1] else right.pop())
merged.extend(left + right)
return merged[::-1]

if len(a) <= 1: return a
mid = len(a) // 2
left = merge_sort(a[:mid])
right = merge_sort(a[mid:])
return merge(left, right)

# A in place version
def merge_sort(a):
temp = a.copy()

def merge(low, mid, high):
for i in range(low, high + 1):
temp[i] = a[i]

i, j = low, mid + 1
for k in range(low, high + 1):
if i > mid:
a[k] = temp[j]
j += 1
elif j > high:
a[k] = temp[i]
i += 1
elif temp[j] < temp[i]:
a[k] = temp[j]
j += 1
else:
a[k] = temp[i]
i += 1

# bottom-up version
# n = len(a)
# step_size = 1
# while step_size < n:
#     for low in range(0, n, 2 * step_size):
#         mid = low + step_size - 1
#         high = min(n - 1, low + 2 * step_size - 1)
#         merge(low, mid, high)
#     step_size *= 2

# # top-down version
def _merge_sort(low, high):
if low >= high: return
mid = (low + high) // 2
_merge_sort(low, mid)
_merge_sort(mid + 1, high)
merge(low, mid, high)

_merge_sort(0, len(a) - 1)


### 5. Quick Sort

• Pick a random pivot item, put it in the right place by moving items smaller than it to its left and items larger than it to its right.
• Use the pivot item(since its in the right location now), to cut the problem in two halves and recursively sort the whole thing.
import random

# an inplace implementation, with random pivot item instead of shuffle and alwasy pick leftmost as pivot.
def quick_sort(a):
def partition(l, h):
if l >= h: return
p = random.randint(l, h)
a[l], a[p] = a[p], a[l]
pivot = a[l]
i, j = l, h
while i < j:
while i < h and a[i] <= pivot: i += 1
while j > l and a[j] >= pivot: j -= 1
if j > i and a[i] > a[j]: a[i], a[j] = a[j], a[i]
a[j], a[l] = a[l], a[j]
return j

def sort(l, h):
if h > l:
j = partition(l, h)
sort(l, j - 1)
sort(j + 1, h)

sort(0, len(a) - 1)


### 6. Heap Sort

• Heapify the array then poping item one at a time.
• See Heap section for how heap works.

Stack

### Remarks

• A stack is a collection that is Last-In-First-Out(LIFO).

### Implementation

An implementation of Stack using linked list.

class Stack(object):
class Node(object):
def __init__(self, item=None):
self.item = item
self.next = None

def __init__(self):
self.node = None
self._n = 0

@property
def is_empty(self):
return not self.node

def __len__(self):
return self.n

def push(self, item):
prev = self.node
self.node = self.Node(item)
self.node.next = prev
self._n += 1

def pop(self):
assert not self.is_empty, "Stack is empty"
item = self.node.item
self.node = self.node.next
self._n -= 1
return item

def peek(self):
return self.node.item


Sweep Line

### Remarks

• It is more of a way of approaching problems than an algorithm.
• It generally means that We will process data in some kind of order, and once we processed one piece of data, we won’t go back anymore.
• Sweep a line across problem space, events of interest popsup, keep track of these events.
• Deal with the events at the line and be done with it.
• The information about past events can be summarized in various format to handle new events. Example can be past min/max etc for single reference, a monotone stack or a priority queue for multi past events.
• Given a problem, there might be more than one way to define what is a event, which can lead to different solutions.

### Common Problems

• Segment intersection. 1D
• Rectangle intersection. 2D

Topological Sort ### Remarks

• Topological sort gives a topological order of the nodes in a directed acyclic graph(DAG).
• A topological ordering implies: for each directed edge u -> v, u must comes before v in the ordering.
• It is a useful preprocessing on the graph, we can implement search for shortest/longest path easily based on this ordering.
• It also gives sort of a notion of levels/diameter of the connected components in the graph.

### Implementation

• One implementation of topological sort is to do DFS and put the postorder on to an array.
• The reverse of the postorder traversal is the topological order.
• We can also incorporate cycle detection into the DFS topological sort implementation. We can stop early when we found a cycle.
def topsort(nodes, edges):
n = len(nodes)
visited = set()
path = set() # for cycle detection
top_order = deque([])

def dfs(node):
''' Perform dfs traversal, Return True if cycle is found. '''
if node in visited: return False
if node in path: return True

for neighbor in edges[node]:
if dfs(neighbor): return True
top_order.appendleft(node)  # put postorder on to array
return False

for node in nodes:
if dfs(node): return []

nodes = set(range(7))
edges = {
0: [2, 5, 1],
1: [4, 6],
2: ,
3: ,
4: ,
5: [2,3],
6: []
}
print(topsort(nodes, edges))

• Another implementation for topological sort is BFS based, aka the Khan’s algorith.
• The BFS topological sort would require more graph preprocessing, we will also need the in degrees(or in edges) for each node.
• We can still detect cycle with BFS topological sort. But only after the procedure is done. When there are cycles in the graph, the ordering will be incomplete. Because, the in_degrees for the nodes in the cycle won’t be 0.
• With BFS topological sort, we can also test the uniqueness of the topological ordering. If the topological ordering is unique, the queue can only hold one node at any given time.
def topsort(queue, out_edges, in_degrees):
ordering = []
while queue:
node = queue.popleft()
ordering.append(node)
for next_ in out_edges[node]:
in_degrees[next_] -= 1
if not in_degrees[next_]:
queue.append(next_)

return ordering

in_degrees = {0:0, 1:1, 2:2, 3:2, 4:2, 5:1, 6:2}
queue = deque([node for node in in_degrees if not in_degrees[node]])
print(topsort(queue, edges, in_degrees))


Tries / Prefix Trees

### Remarks

• Trie(pronounce as try) or Prefix Tree is a N-way tree used to facilitate search of string valued keys.
• The node path corresponds to the characters in the key.
• We use a flag to indicate the end of key.
• Trie can be implemented with Tree Nodes or as hashmap of hashmaps.
• The basic operations like add, get or prefix search startswith all take $$O(k)$$ time where k = len(key)

### Implementation

Below we implement a Trie using python dictionaries.

from collections import defaultdict

class Trie(object):
def __init__(self):
_trie = lambda: defaultdict(_trie)
self._trie = _trie()
self.flag = '#'

trie = self._trie
for char in word:
trie = trie[char]
trie[self.flag] = trie.get(self.flag, 0) + 1

def get(self, word):
trie = self._trie
for char in word:
if char not in trie: return False
trie = trie[char]
return False if self.flag not in trie else trie[self.flag]

def contains(self, word):
return self.get(word) is not False

def startswith(self, prefix):
trie = self._trie
for char in prefix:
if char not in trie: return False
trie = trie[char]
return True


When there are more stuff the node needs to keep track of, we can implement Trie with actual TrieNode objects:

class Trie(object):
class TrieNode:
def __init__(self):
self.childrens = defaultdict(Trie.TrieNode)
self.terminal = False

def __init__(self):
self.root = self.TrieNode()

def add(self, word: str) -> None:
trie = self.root
for char in word:
trie = trie.childrens[char]
trie.terminal = True

def contains(self, word: str) -> bool:
trie = self.root
for char in word:
if char not in trie.childrens: return False
trie = trie.childrens[char]
return trie.terminal

def startswith(self, prefix: str) -> bool:
trie = self.root
for char in prefix:
if char not in trie.childrens: return False
trie = trie.childrens[char]
return True


Union Find / Disjoint Sets

### Remarks

• Union Find is a data structure keeps track of a set of elements that are partitioned into a number of disjoint subsets.
• It has two operations union(p, q) and find(p).
• the find/search finds the subset element p belongs to.
• and union/merge merges the subsets containing p and q.
• It memory usage is O(N), and each find and union operations are near O(1) in time.
• The algorithm can be used to find all connected components in a network.
• It is also used in kruskal’s algorithm to find the minimal spanning tree for a graph.

### Implementation

class UnionFind(object):
def __init__(self, n):
self.parents = list(range(n))
self.sizes =  * n

def find(self, i):
# while i != self.parents[i]:
#     # path compression, have i points to the cluster centroid
#     self.parents[i] = self.find(self.parents[i])
#     i = self.parents[i]
# return i
root = i
while self.parents[root] != root:
root = self.parents[root]
while self.parents[i] != root:
parent = self.parents[i]
self.parents[i] = root
i = parent
return root

def union(self, p, q):
root_p, root_q = map(self.find, (p, q))
if root_p == root_q: return
small, big = sorted([root_p, root_q], key=lambda x: self.sizes[x])
self.parents[small] = big
self.sizes[big] += self.sizes[small]


Use Union Find to find connected components in undirected graph.

connections = [[0, 1], [1, 2], [2, 3], [3, 4], [5, 6], [6, 8], [7, 9]]

uf = UnionFind(10)
for p, q in connections: uf.union(p, q)
num_components = len(set(uf.find(i) for i in range(10)))
print(num_components) • To find minimum spanning tree in the graph where there are weights on the edges.
1. Sort edges by edge weights ascendingly.
2. Iteratve over the edges unify nodes when two nodes don’t belong to same cluster.
3. Repeat 2 until no nodes or no edges.

### An Implementation without pre-allocation

class UnionFind(object):
def __init__(self):
self.parents = dict()
self.sizes = dict()
self.n_sets = 0

def __contains__(self, i):
return i in self.parents

def insert(self, i):
if self.__contains__(i): return
self.parents[i] = i
self.sizes = 1
self.n_sets += 1

def find(self, i):
while i != self.parents[i]:
self.parents[i] = self.find(self.parents[i])
i = self.parents[i]
return i

def union(self, p, q):
root_p, root_q = map(self.find, (p, q))
if root_p == root_q: return
small, big = sorted([root_p, root_q], key=lambda x: self.sizes[x])
self.parents[small] = big
self.sizes[big] += self.sizes[small]
self.n_sets += 1


Miscellaneous

### A collection of handy preprocessing code for common data.

• Get unique characters from a collection of strings.
functools.reduce(set.union, map(set, list_of_words))


Patterns

### 1. Sliding Window

For problems on Array or Linked List, where the task is to find a subarray/ a section of the linked list satisify certain criterion. General approach is to use two pointers to indicate the left and right of the window, move, expand and contract the window along the line.

1. Use some variables to summarize the window for simple questions like sum, avg.
2. Use a hashmap/hashset/orderedhashmap to keep track of counts of elements in the window for more complex problems.