Merge pull request #10 from rhthomas/blk-format

Consistent formatting with Black

Author: Amogh Singhal

BST_nodes_in_range.py

@@ -1,25 +1,30 @@
-# Problem: Find the no. of nodes in a BST 
+# Problem: Find the no. of nodes in a BST
 # that lies in a given range
 
+
 class Node:
-  def __init__(self, data):
-    self.data = data
-    self.left = None
-    self.right = None
+    def __init__(self, data):
+        self.data = data
+        self.left = None
+        self.right = None
+
 
 def nodesWithinRange(root, range):
-  low, high = range
-  if root is None:
-    return 0
-  elif root.data <= high and root.data >= low:
-    return 1 + nodesWithinRange(root.left, range) + nodesWithinRange(root.right, range)
-  elif root.data == high or root.data == low:
-    return 1
-  elif root.data > high:
-    return nodesWithinRange(root.left, range)
-  elif root.data < low:
-    return nodesWithinRange(root.right, range)
-    
+    low, high = range
+    if root is None:
+        return 0
+    elif root.data <= high and root.data >= low:
+        return (
+            1 + nodesWithinRange(root.left, range) + nodesWithinRange(root.right, range)
+        )
+    elif root.data == high or root.data == low:
+        return 1
+    elif root.data > high:
+        return nodesWithinRange(root.left, range)
+    elif root.data < low:
+        return nodesWithinRange(root.right, range)
+
+
 node = Node(10)
 node.left = Node(5)
 node.left.left = Node(1)

README.md

@@ -1,12 +1,15 @@
-# Python-Interview-Problems-for-Practice [![star this repo](http://githubbadges.com/star.svg?user=devAmoghS&repo=Python-Interview-Problems-for-Practice)](http://github.com/ddavison/github-badges) [![fork this repo](http://githubbadges.com/fork.svg?user=devAmoghS&repo=Python-Interview-Problems-for-Practice)](http://github.com/ddavison/github-badges/fork)
+# Python-Interview-Problems-for-Practice
+
+[![star this repo](http://githubbadges.com/star.svg?user=devAmoghS&repo=Python-Interview-Problems-for-Practice)](http://github.com/ddavison/github-badges) [![fork this repo](http://githubbadges.com/fork.svg?user=devAmoghS&repo=Python-Interview-Problems-for-Practice)](http://github.com/ddavison/github-badges/fork) [![Code style: black](https://img.shields.io/badge/code%20style-black-000000.svg)](https://github.com/python/black)
+
 ![alt text](https://images.unsplash.com/photo-1494178270175-e96de2971df9?ixlib=rb-1.2.1&auto=format&fit=crop&w=972&q=80)
 40+ Common code and interview problems solved in Python (it's growing...)
 
 ### The core idea is not to utilize built-in functions or library and giving it a more logic-based approach, so that it can be language-friendly and not end up being another repository of "Python Tips and Tricks" .
 
 ## How good is the code ?
 * It is well tested
-* It passes style checks (PEP8 compliant) [TODO]
+* Consistent formatting (using [Black](https://github.com/python/black))
 * It can compile in its current state (and there are relatively no issues)
 
 ## How much support is available?

binary_search_recursive.py

@@ -3,18 +3,20 @@
 # binary search. If the element is not found
 # it returns -1
 
+
 def binarySearch(lst, key, l, r):
-  if r>=l:
-    mid = l + (r-l) // 2
-    # use print(l, mid, r) to view the process
-    if lst[mid] == key:
-      return mid
-    elif lst[mid] < key:
-      return binarySearch(lst, key, mid+1, r)
-    elif lst[mid] > key:
-      return binarySearch(lst, key, l, mid-1)
-  else:
-    return -1
-    
+    if r >= l:
+        mid = l + (r - l) // 2
+        # use print(l, mid, r) to view the process
+        if lst[mid] == key:
+            return mid
+        elif lst[mid] < key:
+            return binarySearch(lst, key, mid + 1, r)
+        elif lst[mid] > key:
+            return binarySearch(lst, key, l, mid - 1)
+    else:
+        return -1
+
+
 arr = [int(i) for i in range(101)]
-print(binarySearch(arr, 67, 0, len(arr)-1))
+print(binarySearch(arr, 67, 0, len(arr) - 1))

check_anagrams.py

@@ -4,28 +4,33 @@
 # anagrams of each other
 
 import string
+
 letters = string.ascii_lowercase
-CHARACTER_HASH = dict(zip(letters, [0]*len(letters)))
+CHARACTER_HASH = dict(zip(letters, [0] * len(letters)))
+
 
 def mapLettersToHash(text_a):
-  for char in text_a:
-    if char in CHARACTER_HASH.keys():
-      CHARACTER_HASH[char]+=1
+    for char in text_a:
+        if char in CHARACTER_HASH.keys():
+            CHARACTER_HASH[char] += 1
+
 
 def computeCommonLetters(text_b):
-  common_letters = 0    
-  for char in text_b:
-    if CHARACTER_HASH[char] > 0:
-      common_letters+=1
-  return common_letters
+    common_letters = 0
+    for char in text_b:
+        if CHARACTER_HASH[char] > 0:
+            common_letters += 1
+    return common_letters
+
 
 def computeUncommonLetters(text_a, text_b, common_letters):
-  return abs(len(text_a)+len(text_b)-(2*common_letters))
+    return abs(len(text_a) + len(text_b) - (2 * common_letters))
+
 
 text_1 = "hello"
 text_2 = "billion"
 
 mapLettersToHash(text_1)
 common = computeCommonLetters(text_2)
 result = computeUncommonLetters(text_1, text_2, common)
-print(result)    
+print(result)

check_semiprime.py

@@ -5,28 +5,30 @@
 # Problem: Find the numbers which are semiprimes,
 # within a given range. For e.g. 1 to 100.
 
+
 def isSemiprime(num):
-  # start with the smallest prime
-  prime = 2
-  # initialize counter to 0
-  count = 0
-  # Design of while loop:
-  # 1. if count exceeds 2, it is not a semiprime, e.g. 30 = 2*3*5
-  # 2. when the number becomes 1, we have found the second prime
-  while(count<3 and num!=1):
-    # if the number is divisible by current prime,
-    # increment count, else move to new prime
-    if(not(num % prime)):
-      num = num / prime
-      count=count+1
-    else:
-      prime = prime + 1
-  # if count is two, given number is a semiprime
-  return count == 2
-  
-for i in range(1,100):
-  if(isSemiprime(i)):
-    print(i, end=' ')
-    
-# Result: 4 6 9 10 14 15 21 22 25 26 33 34 35 38 39 46 49 
-# 51 55 57 58 62 65 69 74 77 82 85 86 87 91 93 94 95      
+    # start with the smallest prime
+    prime = 2
+    # initialize counter to 0
+    count = 0
+    # Design of while loop:
+    # 1. if count exceeds 2, it is not a semiprime, e.g. 30 = 2*3*5
+    # 2. when the number becomes 1, we have found the second prime
+    while count < 3 and num != 1:
+        # if the number is divisible by current prime,
+        # increment count, else move to new prime
+        if not (num % prime):
+            num = num / prime
+            count = count + 1
+        else:
+            prime = prime + 1
+    # if count is two, given number is a semiprime
+    return count == 2
+
+
+for i in range(1, 100):
+    if isSemiprime(i):
+        print(i, end=" ")
+
+# Result: 4 6 9 10 14 15 21 22 25 26 33 34 35 38 39 46 49
+# 51 55 57 58 62 65 69 74 77 82 85 86 87 91 93 94 95

dfs_bfs.py

@@ -3,6 +3,7 @@
 # 1. Depth First Search (DFS)
 # 2. Breadth First Search (BFS)
 
+
 def dfs_1(graph, start):
     visited, stack = set(), [start]
     while stack:
@@ -12,6 +13,7 @@ def dfs_1(graph, start):
             stack.extend(graph[vertex] - visited)
     return visited
 
+
 def dfs_2(graph, start, visited=None):
     if visited is None:
         visited = set()
@@ -20,6 +22,7 @@ def dfs_2(graph, start, visited=None):
         dfs_2(graph, next, visited)
     return visited
 
+
 def bfs(graph, start):
     visited, queue = set(), [start]
     while queue:
@@ -29,8 +32,10 @@ def bfs(graph, start):
             queue.extend(graph[vertex] - visited)
     return visited
 
+
 # bfs(graph, 'A') # {'B', 'C', 'A', 'F', 'D', 'E'}
 
+
 def dfs_paths(graph, start, goal):
     stack = [(start, [start])]
     while stack:
@@ -41,12 +46,15 @@ def dfs_paths(graph, start, goal):
             else:
                 stack.append((next, path + [next]))
 
-graph = {'A': set(['B', 'C']),
-         'B': set(['A', 'D', 'E']),
-         'C': set(['A', 'F']),
-         'D': set(['B']),
-         'E': set(['B', 'F']),
-         'F': set(['C', 'E'])}
 
-result = list(dfs_paths(graph, 'A', 'F')) # [['A', 'C', 'F'], ['A', 'B', 'E', 'F']]
+graph = {
+    "A": set(["B", "C"]),
+    "B": set(["A", "D", "E"]),
+    "C": set(["A", "F"]),
+    "D": set(["B"]),
+    "E": set(["B", "F"]),
+    "F": set(["C", "E"]),
+}
+
+result = list(dfs_paths(graph, "A", "F"))  # [['A', 'C', 'F'], ['A', 'B', 'E', 'F']]
 print(result)

find_m_to_last_llist.py

@@ -5,20 +5,22 @@
 
 from linked_list_data_structure import LinkedList
 
+
 def findMToLast(l_list, m):
-  current = l_list.head
-  count = 0
+    current = l_list.head
+    count = 0
+
+    while current is not None and count < m:
+        count += 1
+        current = current.getNextNode()
 
-  while current is not None and count < m:
-    count+=1
-    current = current.getNextNode()
+    m_behind = l_list.head
+    while current.next_node is not None:
+        current = current.getNextNode()
+        m_behind = m_behind.getNextNode()
 
-  m_behind = l_list.head
-  while current.next_node is not None:
-    current = current.getNextNode()
-    m_behind = m_behind.getNextNode()
+    return m_behind
 
-  return m_behind
 
 linked_list = LinkedList()
 m_to_last = 3

find_pairs_sum_k.py

@@ -1,12 +1,14 @@
 # Given an array of numbers, find all the
 # pairs of numbers which sum upto `k`
 
+
 def find_pairs(num_array, k):
-  pairs_array = []
-  for num in num_array:
-    if (k-num) in num_array:
-      pairs_array.append((num, (k-num)))
-  return pairs_array
+    pairs_array = []
+    for num in num_array:
+        if (k - num) in num_array:
+            pairs_array.append((num, (k - num)))
+    return pairs_array
+
 
 result = find_pairs([0, 14, 0, 4, 7, 8, 3, 5, 7], 11)
 print(result)

find_products_pair_k.py

@@ -1,15 +1,15 @@
 def product_pair(arr, x):
-  arr_sorted = sorted(arr)
+    arr_sorted = sorted(arr)
 
-  for i in range(0, len(arr_sorted)):
-    sub_array = arr_sorted[i+1:]
-    if x // arr_sorted[i] in sub_array:
-      return True
+    for i in range(0, len(arr_sorted)):
+        sub_array = arr_sorted[i + 1 :]
+        if x // arr_sorted[i] in sub_array:
+            return True
 
-  return False
+    return False
 
 
-arr = [10, 20, 9, 40]  
+arr = [10, 20, 9, 40]
 x = 400
 
 res = product_pair(arr, x)

find_second_largest_in_binary_tree.py

@@ -1,11 +1,13 @@
 # Given a binary tree, find the second largest
 # node in it
 
+
 class Node:
-  def __init__(self, data):
-    self.data = data
-    self.left = None
-    self.right = None
+    def __init__(self, data):
+        self.data = data
+        self.left = None
+        self.right = None
+
 
 def find_largest(root):
     current = root
@@ -14,21 +16,27 @@ def find_largest(root):
             return current.right.data
         current = current.right
 
+
 def find_second_largest(root):
     if root is None or (root.left is None and root.right is None):
         raise ValueError("Tree must atleast have 2 nodes")
-        
+
     current = root
-    
+
     while current is not None:
-        if(current.left is not None and current.right is None):
+        if current.left is not None and current.right is None:
             return find_largest(current.left)
-        
-        if(current.right is not None and current.right.left is None and current.right.right is None):
+
+        if (
+            current.right is not None
+            and current.right.left is None
+            and current.right.right is None
+        ):
             return current.data
-            
+
         current = current.right
-    
+
+
 node = Node(10)
 node.left = Node(5)
 node.left.left = Node(1)
@@ -37,4 +45,4 @@ def find_second_largest(root):
 node.right.right = Node(100)
 
 result = find_second_largest(node)
-print(result) # prints 50
+print(result)  # prints 50