Initial commit

Author: isaccanedo

README.md

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+"""
+Problem:
+
+Given a list of numbers, return whether any two sums to k. For example, given
+[10, 15, 3, 7] and k of 17, return true since 10 + 7 is 17.
+
+Bonus: Can you do this in one pass?
+"""
+
+from typing import List
+
+
+def check_target_sum(arr: List[int], target: int) -> bool:
+    # using hash list to store the previously seen values to get access to them in O(1)
+    previous = set()
+    for elem in arr:
+        if (target - elem) in previous:
+            return True
+        previous.add(elem)
+    return False
+
+
+if __name__ == "__main__":
+    print(check_target_sum([], 17))
+    print(check_target_sum([10, 15, 3, 7], 17))
+    print(check_target_sum([10, 15, 3, 4], 17))
+
+
+"""
+SPECS:
+
+TIME COMPLEXITY: O(n)
+SPACE COMPLEXITY: O(n)
+"""

Solutions/001.py

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+"""
+Problem:
+
+Given an array of integers, return a new array such that each element at index i of the
+new array is the product of all the numbers in the original array except the one at i.
+
+For example, if our input was [1, 2, 3, 4, 5], the expected output would be
+[120, 60, 40, 30, 24]. If our input was [3, 2, 1], the expected output would be
+[2, 3, 6].
+
+Follow-up: what if you can't use division?
+"""
+
+from typing import List
+
+
+def product_of_arr_except_ith_elem(arr: List[int]) -> int:
+    length = len(arr)
+    result = [1 for _ in range(length)]
+    # multiplying all the elements on the left of the ith element in the 1st pass
+    # and all the elements on the right of the ith element in the 2nd pass
+    prod = 1
+    for i in range(length):
+        result[i] *= prod
+        prod *= arr[i]
+    prod = 1
+    for i in range(length - 1, -1, -1):
+        result[i] *= prod
+        prod *= arr[i]
+    return result
+
+
+if __name__ == "__main__":
+    print(product_of_arr_except_ith_elem([1, 2, 3, 4, 5]))
+    print(product_of_arr_except_ith_elem([3, 2, 1]))
+
+
+"""
+SPECS:
+
+TIME COMPLEXITY: O(n)
+SPACE COMPLEXITY: O(n)
+"""

Solutions/002.py

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+"""
+Problem:
+
+Write a program to serialize a tree into a string and deserialize a string into a tree.
+"""
+
+from DataStructures.Queue import Queue
+from DataStructures.Tree import Node, BinaryTree
+
+
+def serialize_helper(node: Node) -> str:
+    # helper function to serialize a binary tree (uses prefix traversal)
+    # data is padded with single quotes (') and comma (,) is used as a delimiter
+    if node.right is None and node.left is None:
+        return f"'{node.val}','None','None'"
+    elif node.left is not None and node.right is None:
+        return f"'{node.val}',{serialize_helper(node.left)},'None'"
+    elif node.left is None and node.right is not None:
+        return f"'{node.val}','None',{serialize_helper(node.right)}"
+    elif node.left is not None and node.right is not None:
+        return (
+            f"'{node.val}',"
+            + f"{serialize_helper(node.left)},"
+            + f"{serialize_helper(node.right)}"
+        )
+
+
+def serialize(tree: BinaryTree) -> str:
+    return serialize_helper(tree.root)
+
+
+def deserialize_helper(node: Node, queue: Queue) -> Node:
+    # helper function to deserialize a string into a Binary Tree
+    # data is a queue containing the data as a prefix notation can be easily decoded
+    # using a queue
+    left = queue.dequeue().strip("'")
+    if left != "None":
+        # if the left child exists, its added to the tree
+        node.left = Node(left)
+        node.left = deserialize_helper(node.left, queue)
+
+    right = queue.dequeue().strip("'")
+    if right != "None":
+        # if the right child exists, its added to the tree
+        node.right = Node(right)
+        node.right = deserialize_helper(node.right, queue)
+    return node
+
+
+def deserialize(string: str) -> BinaryTree:
+    # the string needs to have the same format as the binary tree serialization
+    # eg: data is padded with single quotes (') and comma (,) is used as a delimiter
+    data = string.split(",")
+    queue = Queue()
+    for node in data:
+        queue.enqueue(node)
+    tree = BinaryTree()
+    tree.root = Node(queue.dequeue().strip("'"))
+    deserialize_helper(tree.root, queue)
+    return tree
+
+
+if __name__ == "__main__":
+    tree = BinaryTree()
+
+    tree.root = Node("root")
+
+    tree.root.left = Node("left")
+    tree.root.right = Node("right")
+
+    tree.root.left.left = Node("left.left")
+
+    print(serialize(tree))
+
+    generated_tree = deserialize(
+        "'root','left','left.left','None','None','None','right','None','None'"
+    )
+
+    print(serialize(generated_tree))
+
+
+"""
+SPECS:
+
+SERIALIZE: (n = Number of Nodes)
+TIME COMPLEXITY: O(n) 
+SPACE COMPLEXITY: O(n)
+
+DESERIALIZE: (n = Number of Characters in the String)
+TIME COMPLEXITY: O(n) 
+SPACE COMPLEXITY: O(n)
+"""

Solutions/003.py

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+"""
+Problem:
+
+Given an array of integers, find the first missing positive integer in linear time and
+constant space. In other words, find the lowest positive integer that does not exist in
+the array. The array can contain duplicates and negative numbers as well.
+
+For example, the input [3, 4, -1, 1] should give 2. The input [1, 2, 0] should give 3.
+
+You can modify the input array in-place.
+"""
+
+from typing import List
+
+
+def first_missing_positive_integer(arr: List[int]) -> int:
+    # placing the positive elements (< length) in their proper position
+    # proper position index = element - 1
+    # if after the palcement is complete, index of the 1st element not in its proper
+    # position is the answer
+    length = len(arr)
+    for i in range(length):
+        correctPos = arr[i] - 1
+        while 1 <= arr[i] <= length and arr[i] != arr[correctPos]:
+            arr[i], arr[correctPos] = arr[correctPos], arr[i]
+            correctPos = arr[i] - 1
+    # finding the first missing positive integer
+    for i in range(length):
+        if i + 1 != arr[i]:
+            return i + 1
+    return length + 1
+
+
+if __name__ == "__main__":
+    print(first_missing_positive_integer([3, 4, 2, 1]))
+    print(first_missing_positive_integer([3, 4, -1, 1]))
+    print(first_missing_positive_integer([1, 2, 5]))
+    print(first_missing_positive_integer([-1, -2]))
+
+
+"""
+SPECS:
+
+TIME COMPLEXITY: O(n)
+SPACE COMPLEXITY: O(1)
+
+NOTE: Even though there is a nested loop it is a O(n) algorithm as the cap on the
+      maximum iterations is 2 * n [Amortised analysis]
+"""

Solutions/004.py

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+"""
+Problem:
+
+cons(a, b) constructs a pair, and car(pair) and cdr(pair) returns the first and last
+element of that pair. For example, car(cons(3, 4)) returns 3, and cdr(cons(3, 4))
+returns 4.
+
+Given this implementation of cons:
+
+def cons(a, b):
+    def pair(f):
+        return f(a, b)
+    return pair
+"""
+
+from typing import Callable
+
+
+# given implementation of cons:
+def cons(a, b):
+    def pair(f):
+        return f(a, b)
+
+    return pair
+
+
+# car implementation
+def car(f: Callable) -> int:
+    z = lambda x, y: x
+    return f(z)
+
+
+# cdr implementation
+def cdr(f: Callable) -> int:
+    z = lambda x, y: y
+    return f(z)
+
+
+if __name__ == "__main__":
+    pair = cons(1, 3)
+
+    print(car(pair))
+    print(cdr(pair))
+
+
+"""
+SPECS:
+
+car:
+TIME COMPLEXITY: O(1) 
+SPACE COMPLEXITY: O(1)
+
+cdr:
+TIME COMPLEXITY: O(1) 
+SPACE COMPLEXITY: O(1)
+"""

Solutions/005.py

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+"""
+Problem:
+
+An XOR linked list is a more memory efficient doubly linked list. Instead of each node
+holding next and prev fields, it holds a field named both, which is an XOR of the next
+node and the previous node. Implement an XOR linked list; it has an add(element) which
+adds the element to the end, and a get(index) which returns the node at index.
+
+If using a language that has no pointers (such as Python), you can assume you have
+access to get_pointer and dereference_pointer functions that converts between nodes
+and memory addresses.
+"""
+
+
+# Solution copied from:
+# https://github.com/r1cc4rdo/daily_coding_problem/blob/master/problems/06
+
+
+"""
+An XOR linked list is a more memory efficient doubly linked list.
+Instead of each node holding next and prev fields, it holds a field named both, which
+is a XOR of the next node and the previous node. Implement a XOR linked list; it has an
+add(element) which adds the element to the end, and a get(index) which returns the node
+at index.
+
+NOTE: python does not have actual pointers (id() exists but it is not an actual pointer
+in all implementations). For this reason, we use a python list to simulate memory.
+Indexes are the addresses in memory. This has the unfortunate consequence that the
+travel logic needs to reside in the List class rather than the Node one.
+"""
+
+from typing import Tuple
+
+
+class XORLinkedListNode:
+    def __init__(self, val: int, prev: int, next: int) -> None:
+        self.val = val
+        self.both = prev ^ next
+
+    def next_node(self, prev_idx: int) -> int:
+        return self.both ^ prev_idx
+
+    def prev_node(self, next_idx: int) -> int:
+        return self.both ^ next_idx
+
+
+class XORLinkedList:
+    def __init__(self) -> None:
+        self.memory = [XORLinkedListNode(None, -1, -1)]
+
+    def head(self) -> Tuple[int, int, XORLinkedListNode]:
+        # head node index, prev node index, head node
+        return 0, -1, self.memory[0]
+
+    def add(self, val: int) -> None:
+        current_node_index, previous_node_index, current_node = self.head()
+        while True:
+            # walk down the list until the end is found
+            next_node_index = current_node.next_node(previous_node_index)
+            if next_node_index == -1:
+                # the end is reached
+                break
+            previous_node_index, current_node_index = (
+                current_node_index,
+                next_node_index,
+            )
+            current_node = self.memory[next_node_index]
+        # allocation
+        new_node_index = len(self.memory)
+        current_node.both = previous_node_index ^ new_node_index
+        self.memory.append(XORLinkedListNode(val, current_node_index, -1))
+
+    def get(self, index: int) -> int:
+        current_index, previous_index, current_node = self.head()
+        for _ in range(index + 1):
+            previous_index, current_index = (
+                current_index,
+                current_node.next_node(previous_index),
+            )
+            current_node = self.memory[current_index]
+        return current_node.val
+
+
+if __name__ == "__main__":
+    xor_linked_list = XORLinkedList()
+
+    xor_linked_list.add(1)
+    xor_linked_list.add(2)
+    xor_linked_list.add(3)
+    xor_linked_list.add(4)
+
+    print(xor_linked_list.get(0))
+    print(xor_linked_list.get(1))
+    print(xor_linked_list.get(2))
+    print(xor_linked_list.get(3))