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DSA Assignment 1
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‎02_activities/assignments/assignment_1.ipynb

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@@ -21,11 +21,19 @@
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},
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{
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"cell_type": "code",
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"execution_count": null,
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"execution_count": 202,
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"metadata": {},
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"outputs": [],
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"outputs": [
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{
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"name": "stdout",
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"output_type": "stream",
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"text": [
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"1\n"
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]
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}
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],
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"source": [
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"print((hash('your first name') % 3) + 1)"
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"print((hash('sarita') % 3) + 1)"
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]
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},
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{
@@ -72,20 +80,113 @@
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},
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{
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"cell_type": "code",
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"execution_count": null,
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"execution_count": 203,
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"metadata": {},
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"outputs": [],
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"outputs": [
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{
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"ename": "IndentationError",
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"evalue": "expected an indented block (2850872121.py, line 8)",
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"output_type": "error",
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"traceback": [
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"\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"
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]
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}
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],
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"source": [
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"# Definition for a binary tree node.\n",
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"# class TreeNode(object):\n",
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"# def __init__(self, val = 0, left = None, right = None):\n",
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"# self.val = val\n",
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"# self.left = left\n",
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"# self.right = right\n",
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"def is_symmetric(root: TreeNode) -> int:\n",
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"def is_duplicate(root: TreeNode) -> int:\n",
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" # TODO"
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]
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},
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{
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"cell_type": "code",
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"execution_count": null,
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"metadata": {},
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"outputs": [
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{
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"data": {
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"text/plain": [
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"2"
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]
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},
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"execution_count": 195,
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"metadata": {},
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"output_type": "execute_result"
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}
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],
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"source": [
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"#Example1\n",
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"root = TreeNode(1)\n",
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"root.left = TreeNode(2)\n",
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"root.right = TreeNode(2)\n",
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"root.left.left = TreeNode(3)\n",
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"root.left.right = TreeNode(5)\n",
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"root.right.left = TreeNode(6)\n",
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"root.right.right = TreeNode(7)\n",
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"# print_in_order(root)\n",
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"is_duplicate(root)"
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]
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},
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{
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"cell_type": "code",
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"execution_count": null,
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"metadata": {},
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"outputs": [
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{
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"data": {
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"text/plain": [
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"10"
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]
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},
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"execution_count": 196,
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"metadata": {},
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"output_type": "execute_result"
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}
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],
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"source": [
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"#Example2\n",
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"root = TreeNode(1)\n",
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"root.left = TreeNode(10)\n",
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"root.right = TreeNode(2)\n",
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"root.left.left = TreeNode(3)\n",
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"root.left.right = TreeNode(10)\n",
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"root.right.left = TreeNode(12)\n",
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"root.right.right = TreeNode(12)\n",
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"# print_in_order(root)\n",
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"is_duplicate(root)"
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]
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},
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{
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"cell_type": "code",
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"execution_count": null,
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"metadata": {},
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"outputs": [
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{
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"data": {
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"text/plain": [
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"-1"
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]
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},
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"execution_count": 197,
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"metadata": {},
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"output_type": "execute_result"
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}
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],
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"source": [
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"#Example3\n",
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"root = TreeNode(10)\n",
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"root.left = TreeNode(9)\n",
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"root.right = TreeNode(7)\n",
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"root.left.left = TreeNode(8)\n",
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"# print_in_order(root)\n",
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"is_duplicate(root)"
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]
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},
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{
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"cell_type": "markdown",
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"metadata": {},
@@ -193,12 +294,13 @@
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]
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},
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{
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"cell_type": "code",
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"execution_count": null,
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"cell_type": "markdown",
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"metadata": {},
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"outputs": [],
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"source": [
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"# Your answer here"
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"# Your answer here\n",
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"We need to traverse a binary tree and find any duplicate node values. \n",
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" In case of multiple duplicates, the duplicate values closest to the root(Would be the first found duplicate in BFS) should be returned.\n",
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"So, essentially we need to perform the breadth-first serach and return the first found duplicate value"
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]
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},
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{
@@ -213,9 +315,62 @@
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"cell_type": "code",
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"execution_count": null,
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"metadata": {},
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"outputs": [],
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"outputs": [
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{
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"data": {
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"text/plain": [
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"2"
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]
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},
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"execution_count": 198,
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"metadata": {},
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"output_type": "execute_result"
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}
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],
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"source": [
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"# Your answer here\n",
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"# Example 1 Input: root = [10, 7, 2, 7, 3, 5, 2], output = 2\n",
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"root = TreeNode(4)\n",
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"root.left = TreeNode(7)\n",
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"root.right = TreeNode(2)\n",
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"root.left.left = TreeNode(11)\n",
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"root.left.right = TreeNode(12)\n",
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"root.right.left = TreeNode(5)\n",
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"root.right.left.right = TreeNode(2)\n",
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"root.right.left.left = TreeNode(14)\n",
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"# print_in_order(root)\n",
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"is_duplicate(root)"
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]
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},
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{
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"cell_type": "code",
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"execution_count": null,
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"metadata": {},
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"outputs": [
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{
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"data": {
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"text/plain": [
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"10"
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]
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},
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"execution_count": 199,
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"metadata": {},
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"output_type": "execute_result"
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}
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],
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"source": [
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"# Your answer here"
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"# Example 2 Input: root = [1, 5, 7, 2, 2, 3, 7] Output =10\n",
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"root = TreeNode(10)\n",
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"root.left = TreeNode(7)\n",
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"root.right = TreeNode(2)\n",
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"root.left.left = TreeNode(1)\n",
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"root.left.right = TreeNode(3)\n",
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"root.right.left = TreeNode(5)\n",
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"root.right.right = TreeNode(18)\n",
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"root.right.left.left = TreeNode(12)\n",
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"root.right.right.left = TreeNode(10)\n",
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"# print_in_order(root)\n",
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"is_duplicate(root)"
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]
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},
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{
@@ -232,7 +387,30 @@
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"metadata": {},
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"outputs": [],
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"source": [
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"# Your answer here"
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"# Your answer here\n",
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"# Definition for a binary tree node.\n",
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"class TreeNode(object):\n",
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" def __init__(self, val = 0, left = None, right = None):\n",
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" self.val = val\n",
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" self.left = left\n",
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" self.right = right\n",
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"\n",
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"def is_duplicate(root: TreeNode) -> int:\n",
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" if not root:\n",
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" return -1\n",
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" queue = deque([root])\n",
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" seen_values = set() \n",
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" while queue:\n",
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" node = queue.popleft()\n",
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" if node.val in seen_values:\n",
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" # if value is seen already return as duplicate \n",
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" return node.val\n",
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" seen_values.add(node.val)\n",
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" if node.left:\n",
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" queue.append(node.left)\n",
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" if node.right:\n",
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" queue.append(node.right)\n",
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" return -1"
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]
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},
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{
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]
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},
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{
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"cell_type": "code",
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"execution_count": null,
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"cell_type": "markdown",
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"metadata": {},
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"outputs": [],
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"source": [
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"# Your answer here"
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"# Your answer here\n",
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"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",
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"The deque function helps us to create a queue which we then access from the left, to process the node values sequentially."
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]
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},
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{
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]
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},
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{
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"cell_type": "code",
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"execution_count": null,
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"cell_type": "markdown",
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"metadata": {},
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"outputs": [],
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"source": [
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"# Your answer here"
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"# Your answer here\n",
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"Time complexity - In our solution code, each node is visited once , creating a time complexity of O(n)\n",
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"\n",
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"Space Complexity - The worst case for a highly skewed tree would have a space complexity of O(n)"
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]
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},
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{
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]
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},
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{
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"cell_type": "code",
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"execution_count": null,
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"cell_type": "markdown",
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"metadata": {},
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"outputs": [],
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"source": [
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"# Your answer here"
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"# Your answer here\n",
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"I was thinking of using a recursive function\n",
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"\n",
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"The function would need two parameters, the node value and the seen list to keep track of the already visited values\n",
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"\n",
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"Under a function is_duplicate, do below:\n",
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"\n",
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"create an empty set to store unique values\n",
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"\n",
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"Evaluate the first node\n",
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"if the node value is encountered the first time, add it to the set\n",
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"otherwise, when encountering a value that already exists in set, return the value as a duplicate(and exit the function)\n",
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"\n",
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"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",
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"\n",
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"\n"
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]
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},
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{
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"name": "python",
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"nbconvert_exporter": "python",
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"pygments_lexer": "ipython3",
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"version": "3.11.7"
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"version": "3.9.15"
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}
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},
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"nbformat": 4,

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