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add divide and conquer changes
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+196
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‎implementations/Data Structure - Graphs.ipynb‎

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{
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"cells": [
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{
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"# Data Structure - Graphs"
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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": 4,
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"metadata": {},
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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 [2, 3]\n",
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"2 [4, 5]\n",
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"4 [8, 9]\n",
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"8 []\n",
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"9 []\n",
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"5 [10]\n",
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"10 []\n",
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"3 [6, 7]\n",
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"6 [10]\n",
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"7 [11]\n",
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"11 []\n"
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]
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}
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],
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"source": [
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"from typing import Union, List\n",
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"\n",
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"class GraphNode:\n",
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" def __init__(self, value: int = None, children: Union[List, None] = None):\n",
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" self.children = children if children else []\n",
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" self.value = value\n",
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"\n",
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"\n",
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"nodes = {val: GraphNode(value=val) for val in range(1,12)}\n",
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"nodes[1].children += [nodes[2], nodes[3]]\n",
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"nodes[2].children += [nodes[4], nodes[5]]\n",
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"nodes[3].children += [nodes[6], nodes[7]]\n",
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"nodes[4].children += [nodes[8], nodes[9]]\n",
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"nodes[5].children += [nodes[10]]\n",
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"nodes[6].children += [nodes[10]]\n",
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"nodes[7].children += [nodes[11]]\n",
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"root = nodes[1]\n",
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"\n",
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"def create_adj_list(root: GraphNode, adj_list: dict):\n",
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" # Assume cycles may exist\n",
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" if root in adj_list: return \n",
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" adj_list[root] = set()\n",
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" \n",
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" for child in root.children:\n",
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" # set allows for checking if an edge exists in O(1) \n",
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" adj_list[root].add(child) \n",
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" create_adj_list(child, adj_list) if child not in adj_list else None\n",
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"\n",
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"adj_list = {}\n",
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"create_adj_list(root, adj_list)\n",
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"for node in adj_list:\n",
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" print(node.value, [child.value for child in node.children])"
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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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"source": []
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}
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],
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"metadata": {
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"kernelspec": {
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"display_name": "Python 3",
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"language": "python",
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"name": "python3"
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},
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"language_info": {
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"codemirror_mode": {
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"name": "ipython",
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"version": 3
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},
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"file_extension": ".py",
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"mimetype": "text/x-python",
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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": "3.8.5"
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}
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},
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"nbformat": 4,
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"nbformat_minor": 4
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}

‎implementations/Technique - Divide and Conquer.ipynb‎

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" assert actual == expected, f\"{matrix,val}: {expected} != {actual}\""
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]
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},
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{
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"## [City Skyline]()\n",
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"\n",
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"In: array of buildings (tuples); buildings[i] = [left_i, right_i, height_i]\n",
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"Out: the \"skyline\"; a list of \"key points\" (sorted by x coordinate). Each key point is left endpoint of a horizontal segment in the skyline. The last key point for a segment should have y==0. Any ground needs to be part of the contouring. There should be no consecutive heights in the skyline. \n",
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"\n",
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"Constraints:\n",
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" - 1 to 10000 buildings\n",
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" - Building left/right can be between 0 and int max; left is always left of right. \n",
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" - Height is 1 up to int max\n",
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" - Buildings are sorted by left i in nondecreasing order\n",
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" \n",
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"Edge cases\n",
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" - 1 building; should return [(building_left, height), (building_right, 0)]\n",
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" - Every building overlaps\n",
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" - 1 building completely covers another\n",
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"\n",
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"Do we need to look at every building?\n",
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" - Yes; suppose a case where every building is blocked out by the last one. \n",
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"\n",
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"Overlapping intervals? \n",
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"\n",
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"Skyline with \n",
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"\n",
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"We can probably D&C this:\n",
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" - A skyline can be split into two and merged afterwards\n",
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" - D&C on list of buildings\n",
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" - base case: one building - key points listed above\n",
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" - merge step: can we merge key points?\n",
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" - How do we merge without having to n^2 compare all key points \n",
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" - Mergesort merge starting from leftmost? \n",
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"\n",
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"What if we merge to \"multiheight buildings\"? \n",
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"\n",
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"Every building can start as two key points: (l, h), (r,0). We can convert them to these and then merge them. \n",
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"\n",
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"Critical points are merged in 3s. For points l,m,r where l.x <= m.x <= r.x:\n",
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"- if l.h > m.h\n",
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"\n",
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"Two separate things here (might be able to do them concurrently):\n",
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"- Merge buildings -> convert list of (l,r,h) to [(x,h),(x,h)....]\n",
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" - nlogn \n",
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" - starts with leftmost l and height; ends with rightmost r and 0. \n",
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" - We can do this in nlogn time by binary searching the list (or bisecting) \n",
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" - in the heights list, if l.h > r.h, area is obscured. if l.h < r.h, area open. Never l.h = r.h; not valid. \n",
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"- Get key points\n",
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" - Merged buildings have only one way of being drawn with critical points \n",
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" \n",
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" \n",
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"How do we merge two buildings? \n",
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"\n",
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"If we merge a single building with one that has multiple heights, we potentially have to compare the single height to every height in the building (O(n)).\n",
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" - If we use D&C, we do this log(n) times. \n",
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" \n",
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"D&C approach:\n",
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" - Divide our list of buildings in half\n",
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" - Base case: one building - return (l,h),(r,0)\n",
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" - combine: each recursive case returns a list of possible key points, which then need to be merged. \n",
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" - Do we need to store references to the buildings after returning key points? \n",
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" - No: we only remove points that were already there when adding new points to the list \n",
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" \n",
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"How do we combine (l1, h1), (r1, 0), (l2, h2), (r2, 0)?\n",
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"- lefts and rights either are separate, adjacent, overlapping, equal, or covering\n",
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" - separate: l1 < r1 < l2 < r2\n",
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" - Always: keep all\n",
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" - adjacent: r1 == l2\n",
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" - h1 == h2: (l1, h1), (r2, 0)\n",
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" - h1 != h2: (l1, h1), (r2, 0) (r1, h2)\n",
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" - overlapping: l1 <= l2 <= r1 <= r2 and not (l1 == l2 and r1==r2)\n",
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" - h1 == h2: (l1, h1), (r2, 0)\n",
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" - h1 != h2: above, and (r1,h2) if left is taller, else (l2, h2)\n",
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" - equal: l1 == l2 and r1 == r2\n",
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" - always: (l1, h1), (r2, 0)\n",
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" - covering: l1 < l2 < r2 < r1 or l2 < l1 < r1 < r2\n",
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" - eq height or inner smaller: (outer left, h) (outer right, 0)\n",
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" - diff height and inner greater: (outer l, outer h), (inner l, inner h), (inner r, inner h), (outer r, 0) \n",
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" \n",
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"How do we combine two n-len lists of points?\n",
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" - sorted insert of both, then reduce (might need to keep track of which points came from 1 vs 2, all points in 1 are fine as is)\n",
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" - for any three points l,r,m:\n",
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" - if l.h == m.h or m.h == r.h\n",
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" - if l/m or m/r came from same list, done\n",
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" - if l.h < m.h < r.h, done\n",
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"\n",
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"Can we do mergesort style merge for linear time?"
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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,

‎implementations/Technique - Ordered Stacks.ipynb‎

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{
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"cells": [
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{
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"# Technique - Stacks \n"
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]
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},
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{
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"- Can we use recursion any time we use stacks?\n",
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"- Can we use stacks to preserve ordering?"
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]
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},
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{
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"## [Daily Temperatures](https://leetcode.com/problems/daily-temperatures/)"
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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": 9,
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"metadata": {},
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"outputs": [],
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"source": [
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"from typing import List\n",
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"\n",
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"class Solution:\n",
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" def dailyTemperatures(self, T: List[int]) -> List[int]:\n",
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" solution = []\n",
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" stack = []\n",
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" for i in range(len(T)-1, -1, -1):\n",
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" curr_temp = T[i]\n",
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" while stack and stack[-1][0] <= curr_temp: stack.pop()\n",
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" solution.append(0 if not stack else stack[-1][1]-i)\n",
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" stack.append((curr_temp,i))\n",
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" return solution[::-1]\n",
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"\n",
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"s = Solution()\n",
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"cases = [\n",
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" ([73, 74, 75, 71, 69, 72, 76, 73], [1, 1, 4, 2, 1, 1, 0, 0]),\n",
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" ([89,62,70,58,47,47,46,76,100,70], [8,1,5,4,3,2,1,1,0,0])\n",
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"]\n",
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"for temps, expected in cases:\n",
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" actual = s.dailyTemperatures(temps)\n",
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" assert actual == expected, f\"{temps}: {expected} != {actual}\""
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]
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},
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{
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"## [Basic Calculator](https://leetcode.com/problems/basic-calculator/)"
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]
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}
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],
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"metadata": {
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"kernelspec": {
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"display_name": "Python 3",
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"language": "python",
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"name": "python3"
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},
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"language_info": {
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"codemirror_mode": {
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"name": "ipython",
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"version": 3
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},
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"file_extension": ".py",
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"mimetype": "text/x-python",
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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": "3.8.5"
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}
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},
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"nbformat": 4,
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"nbformat_minor": 4
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}

‎implementations/Technique - Topological Sorting.ipynb‎

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"When in doubt:\n",
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"- Assume nodes may have unbounded children\n",
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"- Assume cycles may exist\n",
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"- Don't modify the input graph; generate an adjacency list or collection of nodes and modify that.\n",
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"- Rule out the possibility of unreachable nodes before proceeding.\n",
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"- If performance isn't totally critical, don't modify the input graph; generate an adjacency list or collection of nodes and modify that.\n",
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"- Rule out the possibility of unreachable nodes (either make sure you have a reference to every node, or that all nodes will be reachable from whatever subset you're given).\n",
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"\n",
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"\n",
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"The basic DFS implementation for topological ordering requires two functions:\n",
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" print(s.findOrder(n_courses, prereqs))"
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]
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},
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{
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"## Example: [Graph Valid Tree](https://leetcode.com/problems/graph-valid-tree/)"
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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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"source": []
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},
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{
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"cell_type": "markdown",
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"metadata": {},

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