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‎01_slides/1_motivation_big_o.ipynb

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"Here are seven Big-O run times that you’ll encounter frequently, sorted\n",
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"from fasted to slowest.\n",
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"\n",
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"![](attachment:images/big_o_viz.jpg)\n",
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"![](./images/big_o_viz.jpg)\n",
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"\n",
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"- $O(1)$, known as *constant time*. Ex: addition, division\n",
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"\n",

‎01_slides/2_ds_search_sort.ipynb

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"- You need to traverse the list once for each item in the list, so the\n",
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" Big-O is $O(n^2)$.\n",
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"\n",
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"![](attachment:images/insertion_sort.png)\n",
229+
"![](./images/insertion_sort.png)\n",
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"\n",
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"## Live Coding:\n",
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"\n",

‎01_slides/3_recursion.ipynb

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"- Suppose you are looking for a key in a box, but the box contains\n",
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" more boxes!\n",
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"\n",
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"![](attachment:images/box_recursion.png)\n",
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"![](./images/box_recursion.png)\n",
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"\n",
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"- 2 minutes: write down the steps of the algorithm you would take to\n",
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" search for the key\n",
@@ -224,15 +224,15 @@
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"\n",
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"## Recursion and the Call Stack\n",
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"\n",
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"![](attachment:images/rec_call_1.png)\n",
227+
"![](./images/rec_call_1.png)\n",
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"\n",
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"## Recursion and the Call Stack\n",
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"\n",
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"![](attachment:images/rec_call_2.png)\n",
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"![](./images/rec_call_2.png)\n",
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"\n",
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"## Recursion and the Call Stack\n",
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"\n",
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"![](attachment:images/rec_call_3.png)\n",
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"![](./images/rec_call_3.png)\n",
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"\n",
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"## Multiple Recursive Calls: Fibonacci Sequence\n",
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"\n",
@@ -489,7 +489,7 @@
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"- Python’s sort function uses a hybrid of merge and insertion sort,\n",
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" both of which you’ve learned!\n",
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"\n",
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"![](attachment:images/merge_sort.png)\n",
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"![](./images/merge_sort.png)\n",
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"\n",
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"## Big-O of Merge Sort\n",
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"\n",

‎01_slides/4_recursive_ds.ipynb

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"- The *height* of a tree is the longest path from its root to its\n",
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" leaves. The height of this tree is 4\n",
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"\n",
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"![](attachment:images/tree.png)\n",
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"![](./images/tree.png)\n",
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"\n",
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"## Anatomy of a Tree\n",
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"\n",
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"- The *ancestors* of a value are its parent, the parent of its parent,\n",
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" etc.\n",
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"\n",
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"![](attachment:images/tree.png)\n",
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"![](./images/tree.png)\n",
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"\n",
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"## Tree Traversal Methods\n",
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"\n",
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"\n",
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"**Result: 4 2 5 1 6 3**\n",
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"\n",
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"![](attachment:images/tree_num.png)\n",
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"![](./images/tree_num.png)\n",
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"\n",
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"## DFS: Inorder Traversal Code\n",
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"\n",
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"\n",
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"Preorder traversal is used to create a copy of the tree\n",
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"\n",
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"![](attachment:images/tree_num.png)\n",
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"![](./images/tree_num.png)\n",
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"\n",
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"## DFS: Postorder Traversal\n",
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"\n",
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"\n",
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"Preorder traversal is used to delete subtrees. (why?)\n",
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"\n",
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"![](attachment:images/tree_num.png)\n",
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"![](./images/tree_num.png)\n",
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"\n",
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"## BFS\n",
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"\n",
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"\n",
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"**Result: 1 2 3 4 5 6**\n",
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"\n",
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"![](attachment:images/tree_num.png)\n",
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"![](./images/tree_num.png)\n",
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"\n",
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"# Binary Search Trees\n",
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"\n",
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"\n",
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" - So if the tree was balanced, then it would be $O(\\text{log}n)$\n",
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"\n",
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"![](attachment:images/tree_unbal.png)\n",
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"![](./images/tree_unbal.png)\n",
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"\n",
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"## BST Efficiency\n",
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"\n",
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"\n",
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"Given a BST, insert a new node in this BST.\n",
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"\n",
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"![](attachment:images/insertion.png)\n"
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"![](./images/insertion.png)\n"
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]
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},
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{
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"\n",
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" - A node connected to another is a *neighbor*\n",
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"\n",
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"![](attachment:images/graph_anat.png)\n",
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"![](./images/graph_anat.png)\n",
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"\n",
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"## Types of Graphs\n",
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"\n",
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"There are two questions we ask about graphs: Is there a path from node A\n",
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"to B? What is the shortest path from node A to B? BFS answers both!\n",
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"\n",
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"![](attachment:images/graph_weight.png)\n",
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"![](./images/graph_weight.png)\n",
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"\n",
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"## BFS of Graphs\n",
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"\n",
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"\n",
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"Queue: \\[ , , , , \\]\n",
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"\n",
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"![](attachment:images/graph_bfs.png)\n",
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"![](./images/graph_bfs.png)\n",
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"\n",
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"## BFS Example\n",
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"\n",
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"\n",
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"Queue: \\[1, , , , \\]\n",
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"\n",
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"![](attachment:images/graph_bfs.png)\n",
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"![](./images/graph_bfs.png)\n",
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"\n",
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"## BFS Example\n",
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"\n",
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"\n",
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"Queue: \\[3, 6, , , \\]\n",
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"\n",
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"![](attachment:images/graph_bfs.png)\n",
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"![](./images/graph_bfs.png)\n",
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"\n",
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"## BFS Example\n",
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"\n",
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"\n",
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"Queue: \\[6, 10, , \\]\n",
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"\n",
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"![](attachment:images/graph_bfs.png)\n",
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"![](./images/graph_bfs.png)\n",
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"\n",
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"## BFS Example\n",
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"\n",
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"\n",
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"Queue: \\[10, 7, , \\]\n",
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"\n",
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"![](attachment:images/graph_bfs.png)\n",
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"![](./images/graph_bfs.png)\n",
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"\n",
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"## BFS Example\n",
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"\n",
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"\n",
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"Queue: \\[7, , , , \\]\n",
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"\n",
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"![](attachment:images/graph_bfs.png)\n",
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"![](./images/graph_bfs.png)\n",
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"\n",
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"## BFS Example\n",
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"\n",
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"\n",
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"Queue: \\[ , , , , \\]\n",
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"\n",
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"![](attachment:images/graph_bfs.png)\n",
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"![](./images/graph_bfs.png)\n",
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"\n",
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"## Time and Space Complexity of BFS\n",
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"\n",

‎02_assignments/assignment_1.ipynb

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"\n",
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" ### Example 1\n",
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"\n",
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" ![](images/q1_ex1.png \"Example 1\")\n",
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" ![](./images/q1_ex1.png)\n",
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"\n",
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" Input: `root = [1, 2, 2, 3, 5, 6, 7]` *What traversal method is this?*\n",
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"\n",
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" Output: 2\n",
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"\n",
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" ### Example 2\n",
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"\n",
54-
" ![](images/q1_ex2.png \"Example 2\")\n",
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" ![](./images/q1_ex2.png)\n",
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"\n",
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" Input: `root = [1, 10, 2, 3, 10, 12, 12]`\n",
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"\n",
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" Output: 10\n",
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"\n",
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" ### Example 3\n",
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"\n",
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" ![](images/q1_ex3.png \"Example 3\")\n",
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" ![](./images/q1_ex3.png)\n",
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"\n",
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" Input: `root = [10, 9, 8, 7]`\n",
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"\n",
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"\n",
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" ### Example 1\n",
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"\n",
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" ![](images/q1_ex1.png \"Example 1\")\n",
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" ![](./images/q1_ex1.png)\n",
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"\n",
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" Input: `root = [1, 2, 2, 3, 5, 6, 7]` *What traversal method is this?*\n",
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"\n",
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" Output: [[1, 2, 3], [1, 2, 5], [1, 2, 6], [1, 2, 7]]\n",
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"\n",
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" ### Example 2\n",
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"\n",
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" ![](images/q1_ex3.png \"Example 2\")\n",
112+
" ![](./images/q1_ex3.png)\n",
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"\n",
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" Input: `root = [10, 9, 8, 7]`\n",
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"\n",

‎05_cohort_three/additional_resources/lessons/5_optimization.ipynb

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"- For cell Guitar 1, a guitar will fit there. It will also fit in cell Guitar 2, 3, 4\n",
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"- Sounds redundant, but let's keep going\n",
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"\n",
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"![](images/dynamic.png)\n",
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"\n",
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"## Stereo Row\n",
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"\n",
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"- In the second row, we can steal the stereo or the guitar.\n",
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"- At 1 kg, you can only steal the guitar, same as for every other cell until Stereo 4, at which point you can steal the stereo and only the stereo.\n",
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"\n",
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"![](images/dynamic.png)\n",
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"\n",
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"## Laptop Row\n",
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"\n",
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"- Now we can steal all 3 items\n",
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"- In the first two columns, we still can only steal the guitar. But in Laptop 3, we can steal the laptop\n",
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"- Laptop 4 is the interesting step. We could steal only the stereo, or the laptop and something else for 1 kg. What is that 1 kg item?\n",
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"- According to the above row, the max value for 1 kg is the guitar!\n",
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"\n",
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"![](images/dynamic.png)\n",
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"\n",
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"## Solution\n",
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"\n",
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"- If we stole the guitar and laptop, the total value is 3500, which is greater than just stealing the stereo\n",
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"- Thus, we should steal guitar and laptop\n",
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"\n",
238-
"![](images/dynamic.png)\n",
239-
"\n",
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"## Formula for each cell\n",
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"\n",
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"- We skipped some very trivial steps in calculating cells aside from the last one\n",

‎05_cohort_three/additional_resources/lessons/6_slow_code.ipynb

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"- Memoization can also avoid the maximum recursion depth error because\n",
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" the call stack is smaller\n",
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"\n",
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"![](attachment:./images/memo.png)\n",
72-
"\n",
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"## Memoization Python\n",
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"\n",
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"[1] From Bhargava chapter 8"

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