Skip to content
New issue

Have a question about this project? Sign up for a free GitHub account to open an issue and contact its maintainers and the community.

Sign up for GitHub

By clicking “Sign up for GitHub”, you agree to our terms of service and privacy statement. We’ll occasionally send you account related emails.

Already on GitHub? Sign in to your account

DSA Assignment 1 #26

Closed
wants to merge 1 commit into from
Closed

DSA Assignment 1 #26

Changes from all commits
Commits
Show all changes
1 commit
Select commit
File filter

Filter by extension

Filter by extension
Conversations
Failed to load comments.
Loading
Jump to
Jump to file
Failed to load files.
Loading
Diff view
Diff view
244 changes: 218 additions & 26 deletions 02_activities/assignments/assignment_1.ipynb
View file
Open in desktop
Original file line number Diff line number Diff line change
Expand Up @@ -21,11 +21,19 @@
},
{
"cell_type": "code",
"execution_count": null,
"execution_count": 202,
"metadata": {},
"outputs": [],
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"1\n"
]
}
],
"source": [
"print((hash('your first name') % 3) + 1)"
"print((hash('sarita') % 3) + 1)"
]
},
{
Expand Down Expand Up @@ -72,20 +80,113 @@
},
{
"cell_type": "code",
"execution_count": null,
"execution_count": 203,
"metadata": {},
"outputs": [],
"outputs": [
{
"ename": "IndentationError",
"evalue": "expected an indented block (2850872121.py, line 8)",
"output_type": "error",
"traceback": [
"\u001b[1;36m Cell \u001b[1;32mIn[203], line 8\u001b[1;36m\u001b[0m\n\u001b[1;33m # TODO\u001b[0m\n\u001b[1;37m ^\u001b[0m\n\u001b[1;31mIndentationError\u001b[0m\u001b[1;31m:\u001b[0m expected an indented block\n"
]
}
],
"source": [
"# Definition for a binary tree node.\n",
"# class TreeNode(object):\n",
"# def __init__(self, val = 0, left = None, right = None):\n",
"# self.val = val\n",
"# self.left = left\n",
"# self.right = right\n",
"def is_symmetric(root: TreeNode) -> int:\n",
"def is_duplicate(root: TreeNode) -> int:\n",
" # TODO"
]
},
{
"cell_type": "code",
"execution_count": null,
"metadata": {},
"outputs": [
{
"data": {
"text/plain": [
"2"
]
},
"execution_count": 195,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"#Example1\n",
"root = TreeNode(1)\n",
"root.left = TreeNode(2)\n",
"root.right = TreeNode(2)\n",
"root.left.left = TreeNode(3)\n",
"root.left.right = TreeNode(5)\n",
"root.right.left = TreeNode(6)\n",
"root.right.right = TreeNode(7)\n",
"# print_in_order(root)\n",
"is_duplicate(root)"
]
},
{
"cell_type": "code",
"execution_count": null,
"metadata": {},
"outputs": [
{
"data": {
"text/plain": [
"10"
]
},
"execution_count": 196,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"#Example2\n",
"root = TreeNode(1)\n",
"root.left = TreeNode(10)\n",
"root.right = TreeNode(2)\n",
"root.left.left = TreeNode(3)\n",
"root.left.right = TreeNode(10)\n",
"root.right.left = TreeNode(12)\n",
"root.right.right = TreeNode(12)\n",
"# print_in_order(root)\n",
"is_duplicate(root)"
]
},
{
"cell_type": "code",
"execution_count": null,
"metadata": {},
"outputs": [
{
"data": {
"text/plain": [
"-1"
]
},
"execution_count": 197,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"#Example3\n",
"root = TreeNode(10)\n",
"root.left = TreeNode(9)\n",
"root.right = TreeNode(7)\n",
"root.left.left = TreeNode(8)\n",
"# print_in_order(root)\n",
"is_duplicate(root)"
]
},
{
"cell_type": "markdown",
"metadata": {},
Expand Down Expand Up @@ -193,12 +294,13 @@
]
},
{
"cell_type": "code",
"execution_count": null,
"cell_type": "markdown",
"metadata": {},
"outputs": [],
"source": [
"# Your answer here"
"# Your answer here\n",
"We need to traverse a binary tree and find any duplicate node values. \n",
" In case of multiple duplicates, the duplicate values closest to the root(Would be the first found duplicate in BFS) should be returned.\n",
"So, essentially we need to perform the breadth-first serach and return the first found duplicate value"
]
},
{
Expand All @@ -213,9 +315,62 @@
"cell_type": "code",
"execution_count": null,
"metadata": {},
"outputs": [],
"outputs": [
{
"data": {
"text/plain": [
"2"
]
},
"execution_count": 198,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"# Your answer here\n",
"# Example 1 Input: root = [10, 7, 2, 7, 3, 5, 2], output = 2\n",
"root = TreeNode(4)\n",
"root.left = TreeNode(7)\n",
"root.right = TreeNode(2)\n",
"root.left.left = TreeNode(11)\n",
"root.left.right = TreeNode(12)\n",
"root.right.left = TreeNode(5)\n",
"root.right.left.right = TreeNode(2)\n",
"root.right.left.left = TreeNode(14)\n",
"# print_in_order(root)\n",
"is_duplicate(root)"
]
},
{
"cell_type": "code",
"execution_count": null,
"metadata": {},
"outputs": [
{
"data": {
"text/plain": [
"10"
]
},
"execution_count": 199,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"# Your answer here"
"# Example 2 Input: root = [1, 5, 7, 2, 2, 3, 7] Output =10\n",
"root = TreeNode(10)\n",
"root.left = TreeNode(7)\n",
"root.right = TreeNode(2)\n",
"root.left.left = TreeNode(1)\n",
"root.left.right = TreeNode(3)\n",
"root.right.left = TreeNode(5)\n",
"root.right.right = TreeNode(18)\n",
"root.right.left.left = TreeNode(12)\n",
"root.right.right.left = TreeNode(10)\n",
"# print_in_order(root)\n",
"is_duplicate(root)"
]
},
{
Expand All @@ -232,7 +387,30 @@
"metadata": {},
"outputs": [],
"source": [
"# Your answer here"
"# Your answer here\n",
"# Definition for a binary tree node.\n",
"class TreeNode(object):\n",
" def __init__(self, val = 0, left = None, right = None):\n",
" self.val = val\n",
" self.left = left\n",
" self.right = right\n",
"\n",
"def is_duplicate(root: TreeNode) -> int:\n",
" if not root:\n",
" return -1\n",
" queue = deque([root])\n",
" seen_values = set() \n",
" while queue:\n",
" node = queue.popleft()\n",
" if node.val in seen_values:\n",
" # if value is seen already return as duplicate \n",
" return node.val\n",
" seen_values.add(node.val)\n",
" if node.left:\n",
" queue.append(node.left)\n",
" if node.right:\n",
" queue.append(node.right)\n",
" return -1"
]
},
{
Expand All @@ -244,12 +422,12 @@
]
},
{
"cell_type": "code",
"execution_count": null,
"cell_type": "markdown",
"metadata": {},
"outputs": [],
"source": [
"# Your answer here"
"# Your answer here\n",
"While traversing the nodes, we maintain a set of unique values that we come across. If we encounter a duplicate, we return that value. In case it is a new value, we add it to the set.\n",
"The deque function helps us to create a queue which we then access from the left, to process the node values sequentially."
]
},
{
Expand All @@ -261,12 +439,13 @@
]
},
{
"cell_type": "code",
"execution_count": null,
"cell_type": "markdown",
"metadata": {},
"outputs": [],
"source": [
"# Your answer here"
"# Your answer here\n",
"Time complexity - In our solution code, each node is visited once , creating a time complexity of O(n)\n",
"\n",
"Space Complexity - The worst case for a highly skewed tree would have a space complexity of O(n)"
]
},
{
Expand All @@ -278,12 +457,25 @@
]
},
{
"cell_type": "code",
"execution_count": null,
"cell_type": "markdown",
"metadata": {},
"outputs": [],
"source": [
"# Your answer here"
"# Your answer here\n",
"I was thinking of using a recursive function\n",
"\n",
"The function would need two parameters, the node value and the seen list to keep track of the already visited values\n",
"\n",
"Under a function is_duplicate, do below:\n",
"\n",
"create an empty set to store unique values\n",
"\n",
"Evaluate the first node\n",
"if the node value is encountered the first time, add it to the set\n",
"otherwise, when encountering a value that already exists in set, return the value as a duplicate(and exit the function)\n",
"\n",
"continue evaluating the left and right branches and call the is_duplicate function recursively with the left node value and the set in it's current state\n",
"\n",
"\n"
]
},
{
Expand Down Expand Up @@ -345,7 +537,7 @@
"name": "python",
"nbconvert_exporter": "python",
"pygments_lexer": "ipython3",
"version": "3.11.7"
"version": "3.9.15"
}
},
"nbformat": 4,
Expand Down
Loading