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Assignment 1 #1

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ff2ae10
This commit includes the solution of assignment 1
Issy90 Oct 11, 2025
b574681
adjusting the comments
Issy90 Oct 11, 2025
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137 changes: 119 additions & 18 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": [
"1\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 = \"Isra\"\n",
"result = hash_to_range(input_string)\n",
"print(result)\n"
]
Expand Down Expand Up @@ -165,11 +173,12 @@
},
{
"cell_type": "code",
"execution_count": null,
"execution_count": 34,
"metadata": {},
"outputs": [],
"source": [
"# Your answer here"
"# The problem is to find the first number that is repeated when iterating through the list and return that number, \n",
"# if there was no duplicate, return -1"
]
},
{
Expand All @@ -181,11 +190,13 @@
},
{
"cell_type": "code",
"execution_count": null,
"execution_count": 15,
"metadata": {},
"outputs": [],
"source": [
"# Your answer here"
"# nums = [2,1,3,5,3,2], this should return 3\n",
"# nums = [5,6,7,8], this should return -1\n",
"# nums = [1,2,3,4,4,5,5,6,6,7], this should return 4"
]
},
{
Expand All @@ -198,11 +209,92 @@
},
{
"cell_type": "code",
"execution_count": null,
"execution_count": 26,
"metadata": {},
"outputs": [],
"outputs": [
{
"data": {
"text/plain": [
"3"
]
},
"execution_count": 26,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"from typing import List\n",
"\n",
"def first_duplicate(nums: List[int]) -> int:\n",
" found = set()\n",
" for num in nums: \n",
" if num in found:\n",
" return num\n",
" found.add(num)\n",
" return -1\n",
"\n",
"first_duplicate([2, 1, 3, 5, 3, 2]) "
]
},
{
"cell_type": "code",
"execution_count": 27,
"metadata": {},
"outputs": [
{
"data": {
"text/plain": [
"1"
]
},
"execution_count": 27,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"first_duplicate([3, 1, 4, 2, 5, 1, 6])"
]
},
{
"cell_type": "code",
"execution_count": 28,
"metadata": {},
"outputs": [
{
"data": {
"text/plain": [
"-1"
]
},
"execution_count": 28,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"first_duplicate([7, 8, 9, 10])"
]
},
{
"cell_type": "code",
"execution_count": 29,
"metadata": {},
"outputs": [
{
"data": {
"text/plain": [
"4"
]
},
"execution_count": 29,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"# Your answer here"
"first_duplicate([1,2,3,4,4,5,5,6,6,7])"
]
},
{
Expand All @@ -215,11 +307,15 @@
},
{
"cell_type": "code",
"execution_count": null,
"execution_count": 30,
"metadata": {},
"outputs": [],
"source": [
"# Your answer here"
"# In my code, I used a set to keep track of the the numbers that have already been found. \n",
"# As I iterate through the list, I check if the number was already in the set. \n",
"# If it was, I will return that number as it is the first duplicate. \n",
"# If it wasn't found, I will add it to the set and continue until the end of the list.\n",
"# If no duplicates were found, I return -1."
]
},
{
Expand All @@ -232,11 +328,14 @@
},
{
"cell_type": "code",
"execution_count": null,
"execution_count": 31,
"metadata": {},
"outputs": [],
"source": [
"# Your answer here"
"# In terms of time complexity, this solution is O(n) because we only iterate through it once. \n",
"# In terms of space complexity, this solution is O(n) in the worst case scenario, if there are no duplicates.\n",
"# In the best scenario, if the first two numbers are the same, the space complexity is O(1).\n",
"# This solution is effiencet in time complexity but not in space complexity. "
]
},
{
Expand All @@ -249,11 +348,13 @@
},
{
"cell_type": "code",
"execution_count": null,
"execution_count": 33,
"metadata": {},
"outputs": [],
"source": [
"# Your answer here"
"# An alterative solution would be to sort the list and then look for duplicates. \n",
"# This would add a time complexity because of the sorting step.\n",
"# making the time complexity O(n Log n) but we would need to store anything so the space complexity would be O(1)"
]
},
{
Expand Down Expand Up @@ -301,7 +402,7 @@
],
"metadata": {
"kernelspec": {
"display_name": "Python 3 (ipykernel)",
"display_name": "dsi_participant",
"language": "python",
"name": "python3"
},
Expand All @@ -315,7 +416,7 @@
"name": "python",
"nbconvert_exporter": "python",
"pygments_lexer": "ipython3",
"version": "3.11.5"
"version": "3.9.23"
}
},
"nbformat": 4,
Expand Down