Up to #25

Author: r1cc4rdo

problems/16/README.md

@@ -3,14 +3,15 @@
 You run a sneaker website and want to record the last N order ids in a log. Implement a data structure to
 accomplish this, with the following API:
 
-record(order_id): adds the order_id to the log
-get_last(i): gets the ith last element from the log. i is guaranteed to be smaller than or equal to N.
+* record(order_id): adds the order_id to the log
+* get_last(i): gets the ith last element from the log. i is guaranteed to be smaller than or equal to N.
 
 You should be as efficient with time and space as possible.
+
 Example:
 
     >>> log = coding_problem_16(10)
-    >>> for id in xrange(20):
+    >>> for id in range(20):
     ...     log.record(id)
 
     >>> log.get_last(0)
@@ -26,4 +27,4 @@ Example:
     >>> log.get_last(1)
     [21]
     >>> log.get_last(3)
-    [19, 20, 21]
+    [19, 20, 21]

problems/16/problem_16.py

@@ -10,7 +10,7 @@ def coding_problem_16(length):
     Example:
 
     >>> log = coding_problem_16(10)
-    >>> for id in xrange(20):
+    >>> for id in range(20):
     ...     log.record(id)
 
     >>> log.get_last(0)

problems/16/solution_16.py

@@ -10,7 +10,7 @@ def coding_problem_16(length):
     Example:
 
     >>> log = coding_problem_16(10)
-    >>> for id in xrange(20):
+    >>> for id in range(20):
     ...     log.record(id)
 
     >>> log.get_last(0)

problems/17/README.md

@@ -1,23 +1,23 @@
-##Problem 17
+## Problem 17
 
 Suppose we represent our file system by a string in the following manner:
 The string "dir\n\tsubdir1\n\tsubdir2\n\t\tfile.ext" represents:
 
-dir
-subdir1
-subdir2
-file.ext
+    dir
+        subdir1
+        subdir2
+            file.ext
 
 The directory dir contains an empty sub-directory subdir1 and a sub-directory subdir2 containing a file file.ext.
 The string "dir\n\tsubdir1\n\t\tfile1.ext\n\t\tsubsubdir1\n\tsubdir2\n\t\tsubsubdir2\n\t\t\tfile2.ext" represents:
 
-dir
-subdir1
-file1.ext
-subsubdir1
-subdir2
-subsubdir2
-file2.ext
+    dir
+        subdir1
+            file1.ext
+            subsubdir1
+        subdir2
+            subsubdir2
+                file2.ext
 
 The directory dir contains two sub-directories subdir1 and subdir2. subdir1 contains a file file1.ext and an empty
 second-level sub-directory subsubdir1. subdir2 contains a second-level sub-directory subsubdir2 containing a
@@ -32,6 +32,7 @@ a file in the abstracted file system. If there is no file in the system, return
 
 The name of a file contains at least a period and an extension.
 The name of a directory or sub-directory will not contain a period.
+
 Examples:
 
     >>> coding_problem_17('file1.ext')
@@ -45,13 +46,13 @@ Examples:
 
     >>> coding_problem_17('dir\n\t\tfile1.ext')
     Traceback (most recent call last):
-...
-RuntimeError: Malformed path string: nesting more than one level at a time.
+    ...
+    RuntimeError: Malformed path string: nesting more than one level at a time.
 
     >>> coding_problem_17('dir\n\tfile1.ext\n\t\tchild_of_a_file.ext')
     Traceback (most recent call last):
-...
-RuntimeError: Malformed path string: a file cannot contain something else.
+    ...
+    RuntimeError: Malformed path string: a file cannot contain something else.
 
     >>> coding_problem_17('dir\n\tfile1.ext\n\tsubdir\n\t\tsubsubdir\n\t\t\ttsubsubsubdir')
     13
@@ -61,4 +62,4 @@ RuntimeError: Malformed path string: a file cannot contain something else.
 
     >>> coding_problem_17('dir\n\tsubdir1\n\t\tfile1.ext\n\t\tsubsubdir1\n\tsubdir2\n\t\tsubsubdir2\n\t\t\tfile2.ext' +
     ...                   '\ndir2\n\tsubdir1\n\tsubdir2\n\t\tsubsubdir1\n\t\t\tsubsubsubdir3\n\t\t\t\tfile3.ext')
-47
+    47

problems/18/README.md

@@ -1,9 +1,10 @@
-##Problem 18
+## Problem 18
 
 Given an array of integers and a number k, where 1 <= k <= length of the array, compute the maximum values of each
 sub-array of length k. Do this in O(n) time and O(k) space. You can modify the input array in-place and you do not
 need to store the results. You can simply print them out as you compute them.
+
 Example:
 
     >>> coding_problem_18([10, 5, 2, 7, 8, 7], 3)
-    [10, 7, 8, 8]
+    [10, 7, 8, 8]

problems/18/solution_18.py

@@ -8,7 +8,7 @@ def coding_problem_18(arr, k):
     >>> coding_problem_18([10, 5, 2, 7, 8, 7], 3)
     [10, 7, 8, 8]
     """
-    for cnt in xrange(k - 1):
+    for _ in range(k - 1):
         arr = [max(value, other) for value, other in zip(arr[:-1], arr[1:])]
 
     return arr

problems/19/README.md

@@ -1,5 +1,13 @@
-##Problem 19
+## Problem 19
 
 A builder is looking to build a row of N houses that can be of K different colors. He has a goal of minimizing cost
 while ensuring that no two neighboring houses are of the same color. Given an N by K matrix where the nth row and
-kth column represents the cost to build the nth house with kth color, return the minimum cost.
+kth column represents the cost to build the nth house with kth color, return the minimum cost.
+
+Example:
+
+    >>> house_size = [1, 2, 4, 2]  # size of the house
+    >>> color_cost = [1, 2, 4]  # cost of paint for each color
+    >>> house_costs = [[cc * hs for cc in color_cost] for hs in house_size]  # 4 houses, 3 colors
+    >>> coding_problem_19(house_costs)  # [1,2,4,2] . [1,2,1,2] == 1*1 + 2*2 + 1*4 + 2*2 == 13
+    13

problems/19/problem_19.py

@@ -1,8 +1,15 @@
-def coding_problem_19(costs):
+def coding_problem_19(house_costs):
     """
     A builder is looking to build a row of N houses that can be of K different colors. He has a goal of minimizing cost
     while ensuring that no two neighboring houses are of the same color. Given an N by K matrix where the nth row and
     kth column represents the cost to build the nth house with kth color, return the minimum cost.
+    Example:
+
+    >>> house_size = [1, 2, 4, 2]  # size of the house
+    >>> color_cost = [1, 2, 4]  # cost of paint for each color
+    >>> house_costs = [[cc * hs for cc in color_cost] for hs in house_size]  # 4 houses, 3 colors
+    >>> coding_problem_19(house_costs)  # [1,2,4,2] . [1,2,1,2] == 1*1 + 2*2 + 1*4 + 2*2 == 13
+    13
     """
     pass
 

problems/19/solution_19.py

@@ -1,15 +1,25 @@
-def coding_problem_19(costs):
+def coding_problem_19(house_costs):
     """
     A builder is looking to build a row of N houses that can be of K different colors. He has a goal of minimizing cost
     while ensuring that no two neighboring houses are of the same color. Given an N by K matrix where the nth row and
     kth column represents the cost to build the nth house with kth color, return the minimum cost.
+    Example:
+
+    >>> house_size = [1, 2, 4, 2]  # size of the house
+    >>> color_cost = [1, 2, 4]  # cost of paint for each color
+    >>> house_costs = [[cc * hs for cc in color_cost] for hs in house_size]  # 4 houses, 3 colors
+    >>> coding_problem_19(house_costs)  # [1,2,4,2] . [1,2,1,2] == 1*1 + 2*2 + 1*4 + 2*2 == 13
+    13
+
+    Notes: this is dynamic programming in disguise. We assign a color to each house in order, and keep
+    track of the minimum cost associated with each choice of color. These costs are the minimum over all
+    possible permutation in which the last added house is of a particular color.
     """
-    best_cost = [0] * len(costs[0])
-    for cost in costs:  # add a house at a time
-        for index in xrange(len(cost)):  # best cost is the one for that color plus min cost between every other color
-            best_cost[index] = cost[index] + min(best_cost[:index] + best_cost[index + 1:])
+    best_costs = [0] * len(house_costs[0])
+    for color_costs in house_costs:  # color_costs are those paid to paint last added house with a certain color
+        best_costs = [color_costs[i] + min(best_costs[:i] + best_costs[i + 1:]) for i in range(len(best_costs))]
 
-    return min(best_cost)
+    return min(best_costs)
 
 
 if __name__ == '__main__':

problems/20/README.md

@@ -1,33 +1,30 @@
-##Problem 20
+## Problem 20
 
 Given two singly linked lists that intersect at some point, find the intersecting node.
 Do this in O(M + N) time (where M and N are the lengths of the lists) and constant space.
 For example, given A = 3 -> 7 -> 8 -> 10 -> 1 and B = 99 -> 1 -> 8 -> 10, return the node with value 8.
+
 Example:
 
     >>> class LinkedListNode(object):
     ...
-...     def __init__(self, value, child=None):
-...         self.value = value
-...         self.next = child
-...
-...     def add(self, value):
-...         return LinkedListNode(value, self)
-...
-...     @classmethod
-...     def len(cls, node):
-...         count = 0
-...         while node:
-...             node = node.next
-...             count += 1
-...         return count
+    ...     def __init__(self, value, child=None):
+    ...         self.value = value
+    ...         self.next = child
+    ...
+    ...     def add(self, value):
+    ...         return LinkedListNode(value, self)
+    ...
+    ...     @classmethod
+    ...     def len(cls, node):
+    ...         count = 0
+    ...         while node:
+    ...             node = node.next
+    ...             count += 1
+    ...         return count
 
     >>> common_tail = LinkedListNode(1).add(10).add(8)
     >>> list_a = LinkedListNode(7, common_tail).add(3)
     >>> list_b = LinkedListNode(1, common_tail).add(99).add(14)
     >>> coding_problem_20(list_a, list_b)
     8
-
-Note: the problem statement above is ambiguous and misleading. I think B list was originally supposed to end
-up with a -> 1. This is how most problems of this type I googled are formulated. If this is not the case, this
-is akin to finding the longest common list between the lists, which I believe cannot be solved in O(M+N).

problems/20/problem_20.py

@@ -27,9 +27,5 @@ def coding_problem_20(list_a, list_b):
     >>> list_b = LinkedListNode(1, common_tail).add(99).add(14)
     >>> coding_problem_20(list_a, list_b)
     8
-
-    Note: the problem statement above is ambiguous and misleading. I think B list was originally supposed to end
-    up with a -> 1. This is how most problems of this type I googled are formulated. If this is not the case, this
-    is akin to finding the longest common list between the lists, which I believe cannot be solved in O(M+N).
     """
-    pass
+    pass

problems/20/solution_20.py

@@ -37,10 +37,10 @@ def coding_problem_20(list_a, list_b):
     if len_b > len_a:
         list_a, list_b = list_b, list_a
 
-    for advance in xrange(abs(len_a - len_b)):
+    for advance in range(abs(len_a - len_b)):
         list_a = list_a.next
 
-    for check in xrange(len_b):
+    for check in range(len_b):
         if list_a is list_b:
             return list_a.value
 
@@ -53,4 +53,4 @@ def coding_problem_20(list_a, list_b):
 if __name__ == '__main__':
 
     import doctest
-    doctest.testmod(verbose=True)
+    doctest.testmod(verbose=True)

problems/21/README.md

@@ -1,7 +1,8 @@
-##Problem 21
+## Problem 21
 
 Given an array of time intervals (start, end) for classroom lectures (possibly overlapping),
 find the minimum number of rooms required.
+
 Example:
 
     >>> coding_problem_21([(30, 75), (0, 50), (60, 150)])

problems/22/README.md

@@ -1,8 +1,9 @@
-##Problem 22
+## Problem 22
 
 Given a dictionary of words and a string made up of those words (no spaces), return the original sentence in a
 list. If there is more than one possible reconstruction, return any of them. If there is no possible
 reconstruction, then return null.
+
 Examples:
 
     >>> coding_problem_22(['Riccardo', 'Brigittie', 'and', 'lollipop'], 'RiccardoandBrigittie')
@@ -12,4 +13,4 @@ Examples:
     ['the', 'quick', 'brown', 'fox']
 
     >>> coding_problem_22(['bed', 'bath', 'bedbath', 'and', 'beyond'], 'bedbathandbeyond')
-    ['bed', 'bath', 'and', 'beyond']
+    ['bed', 'bath', 'and', 'beyond']

problems/23/README.md

@@ -1,15 +1,18 @@
-##Problem 23
+## Problem 23
 
 You are given an M by N matrix consisting of booleans that represents a board. Each True boolean represents a wall.
 Each False boolean represents a tile you can walk on. Given this matrix, a start coordinate, and an end coordinate,
 return the minimum number of steps required to reach the end coordinate from the start. If there is no possible
 path, then return null. You can move up, left, down, and right. You cannot move through walls. You cannot wrap
 around the edges of the board.
+
 Examples:
 
     >>> map = [[False, False, False, False], [True, True, False, True],
     ...        [False, False, False, False], [False, False, False, False]]
     >>> coding_problem_23(matrix, (3, 0), (0, 0))
     7
+    
     >>> map[1][2] = True  # close off path
-    >>> coding_problem_23(matrix, (3, 0), (0, 0))  # None
+    >>> coding_problem_23(matrix, (3, 0), (0, 0))  # returns None
+    

problems/23/problem_23.py

@@ -11,8 +11,10 @@ def coding_problem_23(matrix, start, end):
     ...        [False, False, False, False], [False, False, False, False]]
     >>> coding_problem_23(matrix, (3, 0), (0, 0))
     7
+
     >>> map[1][2] = True  # close off path
     >>> coding_problem_23(matrix, (3, 0), (0, 0))  # None
+
     """
     pass
 

problems/23/solution_23.py

@@ -11,14 +11,16 @@ def coding_problem_23(matrix, start, end):
     ...        [False, False, False, False], [False, False, False, False]]
     >>> coding_problem_23(matrix, (3, 0), (0, 0))
     7
+
     >>> map[1][2] = True  # close off path
     >>> coding_problem_23(matrix, (3, 0), (0, 0))  # None
+
     """
     coords = [(index_r, index_c) for index_r, row in enumerate(matrix)
               for index_c, element in enumerate(row) if not element]
 
     current_distance = 0
-    distances = [[None for col in xrange(len(matrix[0]))] for row in xrange(len(matrix))]
+    distances = [[None for col in range(len(matrix[0]))] for row in range(len(matrix))]
     distances[start[0]][start[1]] = 0
     while True: