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assignment 1 completed #104

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4 changes: 2 additions & 2 deletions 01_materials/slides/2_ds_search_sort.ipynb
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Original file line number Diff line number Diff line change
Expand Up @@ -422,7 +422,7 @@
],
"metadata": {
"kernelspec": {
"display_name": "Python 3 (ipykernel)",
"display_name": "algos-env",
"language": "python",
"name": "python3"
},
Expand All @@ -436,7 +436,7 @@
"name": "python",
"nbconvert_exporter": "python",
"pygments_lexer": "ipython3",
"version": "3.11.5"
"version": "3.11.15"
}
},
"nbformat": 4,
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114 changes: 101 additions & 13 deletions 02_activities/assignments/assignment_1.ipynb
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Original file line number Diff line number Diff line change
Expand Up @@ -25,17 +25,25 @@
},
{
"cell_type": "code",
"execution_count": null,
"execution_count": 1,
"metadata": {},
"outputs": [],
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"2\n"
]
}
],
"source": [
"import hashlib\n",
"\n",
"def hash_to_range(input_string: str) -> int:\n",
" hash_object = hashlib.sha256(input_string.encode())\n",
" hash_int = int(hash_object.hexdigest(), 16)\n",
" return (hash_int % 3) + 1\n",
"input_string = \"your_first_name_here\"\n",
"input_string = \"liliana\"\n",
"result = hash_to_range(input_string)\n",
"print(result)\n"
]
Expand Down Expand Up @@ -112,13 +120,58 @@
},
{
"cell_type": "code",
"execution_count": 1,
"execution_count": 5,
"metadata": {},
"outputs": [],
"source": [
"def is_valid_brackets(s: str) -> bool:\n",
" # TODO\n",
" pass"
" #create an empty list to store valid strings\n",
" stack = [] \n",
" #define the valid pairs\n",
" pairs = {\")\": \"(\", \"}\": \"{\", \"]\": \"[\"} \n",
"\n",
"#create a for loop:\n",
" for char in s:\n",
" if char in \"({[\":\n",
" #append the first left characters\n",
" stack.append(char)\n",
" else:\n",
" if not stack:\n",
" return False\n",
" # POP = remove the last item from the list AND hand that item back to you \n",
" # ugh suffered on this one. \n",
" # the whole idea was to remove the items (i.e., pop) out if they match . \n",
" # Once the stack is empty, it means that everything matched up \n",
" top = stack.pop()\n",
" if top != pairs[char]:\n",
" return False\n",
" #stack empty = False, therefore not stack = True \n",
" return not stack "
]
},
{
"cell_type": "code",
"execution_count": null,
"metadata": {},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"True\n",
"False\n",
"True\n",
"False\n"
]
}
],
"source": [
"#sanity check?\n",
"\n",
"print(is_valid_brackets(\"([]{})\")) # true\n",
"print(is_valid_brackets(\"([)]\")) # false\n",
"print(is_valid_brackets(\"()[]{}\")) # true\n",
"print(is_valid_brackets(\"[{]}\")) # false"
]
},
{
Expand Down Expand Up @@ -169,7 +222,8 @@
"metadata": {},
"outputs": [],
"source": [
"# Your answer here"
"# In this problem we had to determine whether a pair of strings is valid, where every opening bracket had a matching closing bracket\n",
"#of the same type of the same order. "
]
},
{
Expand All @@ -185,7 +239,33 @@
"metadata": {},
"outputs": [],
"source": [
"# Your answer here"
"# ex 1) descriptiom: A string representing a sequence of food containers and their lids.\n",
"\n",
"#The possible characters are:\n",
"\n",
"#'B' = open Burger bun\n",
"#'b' = close Burger bun\n",
"#'T' = open Taco shell\n",
"#'t' = close Taco shell\n",
"#'P' = open Pizza box\n",
"#'p' = close Pizza box\n",
"\n",
"# input: \"BbTt\"\n",
"# ouput: True\n",
"\n",
"\n",
"#ex 2) description: brand packaging check!\n",
"\n",
"#each brand uses a matching pair:\n",
"#'N' = Nike box is sealed\n",
"#'n' = Nike box is unsealed\n",
"#'A' = Apple package is sealed\n",
"#'a' = Apple package is unsealed\n",
"#'S' = Samsung package is sealed\n",
"#'s' = Samsung package is unsealed\n",
"\n",
"# input: \"NsAa\"\n",
"# ouput: False\n"
]
},
{
Expand All @@ -202,7 +282,10 @@
"metadata": {},
"outputs": [],
"source": [
"# Your answer here"
"# For example 2 :\n",
"\n",
"#Time complexity: I've looped through each string and did a pop plus a dictionary look up. I think the time : O(n) since every character needs to be iterated\n",
"#Space complexity: 0(n) because in the worst case the string is all opening brackets. Every character gets pushed and none get popped, so the stack grows until it holds all n characters "
]
},
{
Expand All @@ -219,7 +302,7 @@
"metadata": {},
"outputs": [],
"source": [
"# Your answer here"
"# as soon as it finds a matching pair it drops ( last in first out). At the end, if the stack is empty , it means that everything was matched"
]
},
{
Expand All @@ -236,7 +319,12 @@
"metadata": {},
"outputs": [],
"source": [
"# Your answer here"
"# Time complexity is O(n): I make a single pass through the string, and each character costs constant time a push, or a pop plus a dictionary lookup. \n",
"# Since I must look at every character to decide validity, O(n) is the best possible. \n",
"# \n",
"# Space complexity is also O(n): the stack is the only structure that grows with input size,\n",
"# and in the worst case (a string of all opening brackets) every character gets pushed with nothing popped,\n",
"# so the stack reaches size n"
]
},
{
Expand Down Expand Up @@ -301,7 +389,7 @@
],
"metadata": {
"kernelspec": {
"display_name": "Python 3 (ipykernel)",
"display_name": "algos-env",
"language": "python",
"name": "python3"
},
Expand All @@ -315,7 +403,7 @@
"name": "python",
"nbconvert_exporter": "python",
"pygments_lexer": "ipython3",
"version": "3.11.5"
"version": "3.11.15"
}
},
"nbformat": 4,
Expand Down
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