Added Set in .py

Set/Randomized_Set.py

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+#! /usr/bin/env python3
+
+"""
+Design a data structure that supports all following operations
+in average O(1) time.
+insert(val): Inserts an item val to the set if not already present.
+remove(val): Removes an item val from the set if present.
+random_element: Returns a random element from current set of elements.
+           Each element must have the same probability of being returned.
+"""
+
+import random
+
+
+class RandomizedSet():
+    """
+    idea: shoot
+    """
+
+    def __init__(self):
+        self.elements = []
+        self.index_map = {}  # element -> index
+
+    def insert(self, new_one):
+        if new_one in self.index_map:
+            return
+        self.index_map[new_one] = len(self.elements)
+        self.elements.append(new_one)
+
+    def remove(self, old_one):
+        if not old_one in self.index_map:
+            return
+        index = self.index_map[old_one]
+        last = self.elements.pop()
+        self.index_map.pop(old_one)
+        if index == len(self.elements):
+            return
+        self.elements[index] = last
+        self.index_map[last] = index
+
+    def random_element(self):
+        return random.choice(self.elements)
+
+
+def __test():
+    rset = RandomizedSet()
+    ground_truth = set()
+    n = 64
+
+    for i in range(n):
+        rset.insert(i)
+        ground_truth.add(i)
+
+    # Remove a half
+    for i in random.sample(range(n), n // 2):
+        rset.remove(i)
+        ground_truth.remove(i)
+
+    print(len(ground_truth), len(rset.elements), len(rset.index_map))
+    for i in ground_truth:
+        assert(i == rset.elements[rset.index_map[i]])
+
+    for i in range(n):
+        print(rset.random_element(), end=' ')
+    print()
+
+
+if __name__ == "__main__":
+    __test()

Set/Set_Covering.py

@@ -0,0 +1,108 @@
+from itertools import chain, combinations
+
+"""
+Universe *U* of n elements
+Collection of subsets of U:
+    S = S1,S2...,Sm
+    Where every substet Si has an associated cost.
+Find a minimum cost subcollection of S that covers all elements of U
+Example:
+    U = {1,2,3,4,5}
+    S = {S1,S2,S3}
+    S1 = {4,1,3},    Cost(S1) = 5
+    S2 = {2,5},      Cost(S2) = 10
+    S3 = {1,4,3,2},  Cost(S3) = 3
+    Output:
+        Set cover = {S2, S3}
+        Min Cost = 13
+"""
+
+
+def powerset(iterable):
+    """Calculate the powerset of any iterable.
+    For a range of integers up to the length of the given list,
+    make all possible combinations and chain them together as one object.
+    From https://docs.python.org/3/library/itertools.html#itertools-recipes
+    """
+    "list(powerset([1,2,3])) --> [(), (1,), (2,), (3,), (1,2), (1,3), (2,3), (1,2,3)]"
+    s = list(iterable)
+    return chain.from_iterable(combinations(s, r) for r in range(len(s) + 1))
+
+
+def optimal_set_cover(universe, subsets, costs):
+    """ Optimal algorithm - DONT USE ON BIG INPUTS - O(2^n) complexity!
+    Finds the minimum cost subcollection os S that covers all elements of U
+    Args:
+        universe (list): Universe of elements
+        subsets (dict): Subsets of U {S1:elements,S2:elements}
+        costs (dict): Costs of each subset in S - {S1:cost, S2:cost...}
+    """
+    pset = powerset(subsets.keys())
+    best_set = None
+    best_cost = float("inf")
+    for subset in pset:
+        covered = set()
+        cost = 0
+        for s in subset:
+            covered.update(subsets[s])
+            cost += costs[s]
+        if len(covered) == len(universe) and cost < best_cost:
+            best_set = subset
+            best_cost = cost
+    return best_set
+
+
+def greedy_set_cover(universe, subsets, costs):
+    """Approximate greedy algorithm for set-covering. Can be used on large
+    inputs - though not an optimal solution.
+    Args:
+        universe (list): Universe of elements
+        subsets (dict): Subsets of U {S1:elements,S2:elements}
+        costs (dict): Costs of each subset in S - {S1:cost, S2:cost...}
+    """
+    elements = set(e for s in subsets.keys() for e in subsets[s])
+    # elements don't cover universe -> invalid input for set cover
+    if elements != universe:
+        return None
+
+    # track elements of universe covered
+    covered = set()
+    cover_sets = []
+
+    while covered != universe:
+        min_cost_elem_ratio = float("inf")
+        min_set = None
+        # find set with minimum cost:elements_added ratio
+        for s, elements in subsets.items():
+            new_elements = len(elements - covered)
+            # set may have same elements as already covered -> new_elements = 0
+            # check to avoid division by 0 error
+            if new_elements != 0:
+                cost_elem_ratio = costs[s] / new_elements
+                if cost_elem_ratio < min_cost_elem_ratio:
+                    min_cost_elem_ratio = cost_elem_ratio
+                    min_set = s
+        cover_sets.append(min_set)
+        # union
+        covered |= subsets[min_set]
+    return cover_sets
+
+
+if __name__ == '__main__':
+    universe = {1, 2, 3, 4, 5}
+    subsets = {'S1': {4, 1, 3}, 'S2': {2, 5}, 'S3': {1, 4, 3, 2}}
+    costs = {'S1': 5, 'S2': 10, 'S3': 3}
+
+    optimal_cover = optimal_set_cover(universe, subsets, costs)
+    optimal_cost = sum(costs[s] for s in optimal_cover)
+
+    greedy_cover = greedy_set_cover(universe, subsets, costs)
+    greedy_cost = sum(costs[s] for s in greedy_cover)
+
+    print('Optimal Set Cover:')
+    print(optimal_cover)
+    print('Cost = %s' % optimal_cost)
+
+    print('Greedy Set Cover:')
+    print(greedy_cover)
+    print('Cost = %s' % greedy_cost)

Set/__init__.py

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+from .find_keyboard_row import *

Set/find_keyboard_row.py

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+"""
+Given a List of words, return the words that can be typed using letters of
+alphabet on only one row's of American keyboard.
+For example:
+Input: ["Hello", "Alaska", "Dad", "Peace"]
+Output: ["Alaska", "Dad"]
+Reference: https://leetcode.com/problems/keyboard-row/description/
+"""
+def find_keyboard_row(words):
+    """
+    :type words: List[str]
+    :rtype: List[str]
+    """
+    keyboard = [
+        set('qwertyuiop'),
+        set('asdfghjkl'),
+        set('zxcvbnm'),
+    ]
+    result = []
+    for word in words:
+        for key in keyboard:
+            if set(word.lower()).issubset(key):
+                result.append(word)
+    return result