Basic Datastructures and Heaps.ipynb

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      "source": "# Problem Set \\# 2 (Basic Datastructures and Heaps)\n\nTopics covered:\n  - Basic data-structures\n  - Heap data-structures\n  - Using heaps and arrays to realize interesting functionality.",
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      "source": "## Problem 1 (Least-k Elements Datastructure)\n\n\nWe saw how min-heaps can efficiently allow us to query the least element in a heap (array). We would like to modify minheaps in this exercise to design a data structure to maintain the __least k__ elements for a  given $k \\geq 1$ with $$k = 1$$ being the minheap data-structure.\n\nOur design is to hold two arrays: \n  - (a) a sorted array `A` of $k$ elements that forms our least k elements; and \n  - (b) a minheap `H` with the remaining $n-k$ elements. \n\nOur data structure will itself be a pair of arrays `(A,H)` with the following property:\n - `H` must be a minheap\n - `A` must be sorted of size $k$.\n - Every element of `A` must be smaller than every element of `H`.\n\nThe key operations to implement in this assignment include: \n  - insert a new element into the data-structure\n  - delete an existing element from the data-structure.\n\n\nWe will first ask you to design the data structure and them implement it.\n\n### (A) Design Insertion  Algorithm\n\nSuppose we wish to insert a new element with key $j$ into this data structure. Describe the pseudocode. Your pseudocode must deal with two cases: when the inserted element $j$ would be one of the `least k` elements i.e, it belongs to the array `A`; or when the inserted element belongs to the heap `H`. How would you distinguish between the two cases? \n\n- You can assume that heap operations such as `insert(H, key)` and `delete(H, index)` are defined.\n- Assume that the heap is indexed as  `H[1]`,...,`H[n -k]` with `H[0]` being unused.\n- Assume $ n > k$, i.e, there are already more than $k$ elements in the data structure.\n\n\nWhat is the complexity of the insertion operation in the worst case in terms of $k, n$.\n\n__Unfortunately, we cannot grade your answer. We hope you will use this to design your datastructure on paper before attempting to code it up__",
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      "source": "### (B) Design Deletion Algorithm\n\nSuppose we wish to delete an index $j$ from the top-k array $A$. Design an algorithm to perform this deletion. Assume that the heap is not empty, in which case you can assume that the deletion fails. \n\n\n\n- You can assume that heap operations such as `insert(H, key)` and `delete(H, index)` are defined.\n- Assume that the heap is indexed as  `H[1]`,...,`H[n -k]` with `H[0]` being unused.\n- Assume $ n > k$, i.e, there are already more than $k$ elements in the data structure.\n\nWhat is the complexity of the insertion operation in the worst case in terms of $k, n$.\n\n__Unfortunately, we cannot grade your answer. We hope you will use this to design your datastructure on paper before attempting to code it up__",
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      "source": "## (C) Program your solution by completing the code below\n\nNote that although your algorithm design above assume that your are inserting and deleting from cases where $n \\geq k$, the data structure implementation below must handle $n < k$ as well. We have provided implementations for that portion to help you out.",
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      "source": "# First let us complete a minheap data structure.\n# Please complete missing parts below.\n\nclass MinHeap:\n    def __init__(self):\n        self.H = [None]\n \n    def size(self):\n        return len(self.H)-1\n    \n    def __repr__(self):\n        return str(self.H[1:])\n        \n    def satisfies_assertions(self):\n        for i in range(2, len(self.H)):\n            assert self.H[i] >= self.H[i//2],  f'Min heap property fails at position {i//2}, parent elt: {self.H[i//2]}, child elt: {self.H[i]}'\n    \n    def min_element(self):\n        return self.H[1]\n\n    ## bubble_up function at index\n    ## WARNING: this function has been cut and paste for the next problem as well \n    def bubble_up(self, index):\n        assert index >= 1\n        if index == 1: \n            return \n        parent_index = index // 2\n        if self.H[parent_index] < self.H[index]:\n            return \n        else:\n            self.H[parent_index], self.H[index] = self.H[index], self.H[parent_index]\n            self.bubble_up(parent_index)\n    \n    ## bubble_down function at index\n    ## WARNING: this function has been cut and paste for the next problem as well \n    def bubble_down(self, index):\n        assert index >= 1 and index < len(self.H)\n        lchild_index = 2 * index\n        rchild_index = 2 * index + 1\n        # set up the value of left child to the element at that index if valid, or else make it +Infinity\n        lchild_value = self.H[lchild_index] if lchild_index < len(self.H) else float('inf')\n        # set up the value of right child to the element at that index if valid, or else make it +Infinity\n        rchild_value = self.H[rchild_index] if rchild_index < len(self.H) else float('inf')\n        # If the value at the index is lessthan or equal to the minimum of two children, then nothing else to do\n        if self.H[index] <= min(lchild_value, rchild_value):\n            return \n        # Otherwise, find the index and value of the smaller of the two children.\n        # A useful python trick is to compare \n        min_child_value, min_child_index = min ((lchild_value, lchild_index), (rchild_value, rchild_index))\n        # Swap the current index with the least of its two children\n        self.H[index], self.H[min_child_index] = self.H[min_child_index], self.H[index]\n        # Bubble down on the minimum child index\n        self.bubble_down(min_child_index)\n        \n        \n    # Function: heap_insert\n    # Insert elt into heap\n    # Use bubble_up/bubble_down function\n    def insert(self, elt):\n        \n        self.H.append(elt)\n        assert self.H[self.size()] == elt\n        self.bubble_up(self.size())\n        \n    # Function: heap_delete_min\n    # delete the smallest element in the heap. Use bubble_up/bubble_down\n    def delete_min(self):\n        \n        if self.size() > 1:\n            self.H[1] = self.H.pop(-1)\n            self.bubble_down(1)\n        else:\n            self.H.pop(-1)\n",
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      "source": "h = MinHeap()\nprint('Inserting: 5, 2, 4, -1 and 7 in that order.')\nh.insert(5)\nprint(f'\\t Heap = {h}')\nassert(h.min_element() == 5)\nh.insert(2)\nprint(f'\\t Heap = {h}')\nassert(h.min_element() == 2)\nh.insert(4)\nprint(f'\\t Heap = {h}')\nassert(h.min_element() == 2)\nh.insert(-1)\nprint(f'\\t Heap = {h}')\nassert(h.min_element() == -1)\nh.insert(7)\nprint(f'\\t Heap = {h}')\nassert(h.min_element() == -1)\nh.satisfies_assertions()\n\nprint('Deleting minimum element')\nh.delete_min()\nprint(f'\\t Heap = {h}')\nassert(h.min_element() == 2)\nh.delete_min()\nprint(f'\\t Heap = {h}')\nassert(h.min_element() == 4)\nh.delete_min()\nprint(f'\\t Heap = {h}')\nassert(h.min_element() == 5)\nh.delete_min()\nprint(f'\\t Heap = {h}')\nassert(h.min_element() == 7)\n# Test delete_max on heap of size 1, should result in empty heap.\nh.delete_min()\nprint(f'\\t Heap = {h}')\nprint('All tests passed: 10 points!')\n",
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          "text": "Inserting: 5, 2, 4, -1 and 7 in that order.\n\n\t Heap = [5]\n\n\t Heap = [2, 5]\n\n\t Heap = [2, 5, 4]\n\n\t Heap = [-1, 2, 4, 5]\n\n\t Heap = [-1, 2, 4, 5, 7]\n\nDeleting minimum element\n\n\t Heap = [2, 5, 4, 7]\n\n\t Heap = [4, 5, 7]\n\n\t Heap = [5, 7]\n\n\t Heap = [7]\n\n\t Heap = []\n\nAll tests passed: 10 points!\n"
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      "source": "class TopKHeap:\n    \n    # The constructor of the class to initialize an empty data structure\n    def __init__(self, k):\n        self.k = k\n        self.A = []\n        self.H = MinHeap()\n        \n    def size(self): \n        return len(self.A) + (self.H.size())\n    \n    def get_jth_element(self, j):\n        assert 0 <= j < self.k-1\n        assert j < self.size()\n        return self.A[j]\n    \n    def satisfies_assertions(self):\n        # is self.A sorted\n        for i in range(len(self.A) -1 ):\n            assert self.A[i] <= self.A[i+1], f'Array A fails to be sorted at position {i}, {self.A[i], self.A[i+1]}'\n        # is self.H a heap (check min-heap property)\n        self.H.satisfies_assertions()\n        # is every element of self.A less than or equal to each element of self.H\n        for i in range(len(self.A)):\n            assert self.A[i] <= self.H.min_element(), f'Array element A[{i}] = {self.A[i]} is larger than min heap element {self.H.min_element()}'\n        \n    # Function : insert_into_A\n    # This is a helper function that inserts an element `elt` into `self.A`.\n    # whenever size is < k,\n    #       append elt to the end of the array A\n    # Move the element that you just added at the very end of \n    # array A out into its proper place so that the array A is sorted.\n    # return the \"displaced last element\" jHat (None if no element was displaced)\n    def insert_into_A(self, elt):\n        k = self.k\n        print(\"k = \", k)\n#         assert(self.size() < k)\n        self.A.append(elt)\n        j = len(self.A)-1\n        while (j >= 1 and self.A[j] < self.A[j-1]):\n            # Swap A[j] and A[j-1]\n            (self.A[j], self.A[j-1]) = (self.A[j-1], self.A[j])\n            j = j -1 \n        return\n    # bug here. it should be self.k (or add k = self.k)\n    # bug assert(self.size() < k) ????\n    \n            \n    # Function: insert -- insert an element into the data structure.\n    # Code to handle when self.size < self.k is already provided\n    def insert(self, elt):\n        size = self.size()\n        # If we have fewer than k elements, handle that in a special manner\n        if size <= self.k:\n            self.insert_into_A(elt)\n            return \n        \n        self.insert_into_A(elt)\n        self.H.insert(self.A.pop(-1))\n        \n    # Function: Delete top k -- delete an element from the array A\n    # In particular delete the j^{th} element where j = 0 means the least element.\n    # j must be in range 0 to self.k-1\n    def delete_top_k(self, j):\n        k = self.k\n        assert self.size() > k # we need not handle the case when size is less than or equal to k\n        assert j >= 0\n        assert j < self.k\n    \n        self.A.pop(j)\n        self.A.append(self.H.min_element())\n        self.H.delete_min()\n    \n    ",
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      "source": "h = TopKHeap(5)\n# Force the array A\nh.A = [-10, -9, -8, -4, 0]\n# Force the heap to this heap\n[h.H.insert(elt) for elt in  [1, 4, 5, 6, 15, 22, 31, 7]]\n\nprint('Initial data structure: ')\nprint('\\t A = ', h.A)\nprint('\\t H = ', h.H)\n\n# Insert an element -2\nprint('Test 1: Inserting element -2')\nh.insert(-2)\nprint('\\t A = ', h.A)\nprint('\\t H = ', h.H)\n# After insertion h.A should be [-10, -9, -8, -4, -2]\n# After insertion h.H should be [None, 0, 1, 5, 4, 15, 22, 31, 7, 6]\nassert h.A == [-10,-9,-8,-4,-2]\nassert h.H.min_element() == 0 , 'Minimum element of the heap is no longer 0'\nh.satisfies_assertions()\n\nprint('Test2: Inserting element -11')\nh.insert(-11)\nprint('\\t A = ', h.A)\nprint('\\t H = ', h.H)\nassert h.A == [-11, -10, -9, -8, -4]\nassert h.H.min_element() == -2\nh.satisfies_assertions()\n\nprint('Test 3 delete_top_k(3)')\nh.delete_top_k(3)\nprint('\\t A = ', h.A)\nprint('\\t H = ', h.H)\nh.satisfies_assertions()\nassert h.A == [-11,-10,-9,-4,-2]\nassert h.H.min_element() == 0\nh.satisfies_assertions()\n\nprint('Test 4 delete_top_k(4)')\nh.delete_top_k(4)\nprint('\\t A = ', h.A)\nprint('\\t H = ', h.H)\nassert h.A == [-11, -10, -9, -4, 0]\nh.satisfies_assertions()\n\nprint('Test 5 delete_top_k(0)')\nh.delete_top_k(0)\nprint('\\t A = ', h.A)\nprint('\\t H = ', h.H)\nassert h.A == [-10, -9, -4, 0, 1]\nh.satisfies_assertions()\n\nprint('Test 6 delete_top_k(1)')\nh.delete_top_k(1)\nprint('\\t A = ', h.A)\nprint('\\t H = ', h.H)\nassert h.A == [-10, -4, 0, 1, 4]\nh.satisfies_assertions()\nprint('All tests passed - 15 points!')\n",
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          "text": "Initial data structure: \n\n\t A =  [-10, -9, -8, -4, 0]\n\n\t H =  [1, 4, 5, 6, 15, 22, 31, 7]\n\nTest 1: Inserting element -2\n\nk =  5\n\n\t A =  [-10, -9, -8, -4, -2]\n\n\t H =  [0, 1, 5, 4, 15, 22, 31, 7, 6]\n\nTest2: Inserting element -11\n\nk =  5\n\n\t A =  [-11, -10, -9, -8, -4]\n\n\t H =  [-2, 0, 5, 4, 1, 22, 31, 7, 6, 15]\n\nTest 3 delete_top_k(3)\n\n\t A =  [-11, -10, -9, -4, -2]\n\n\t H =  [0, 1, 5, 4, 15, 22, 31, 7, 6]\n\nTest 4 delete_top_k(4)\n\n\t A =  [-11, -10, -9, -4, 0]\n\n\t H =  [1, 4, 5, 6, 15, 22, 31, 7]\n\nTest 5 delete_top_k(0)\n\n\t A =  [-10, -9, -4, 0, 1]\n\n\t H =  [4, 6, 5, 7, 15, 22, 31]\n\nTest 6 delete_top_k(1)\n\n\t A =  [-10, -4, 0, 1, 4]\n\n\t H =  [5, 6, 22, 7, 15, 31]\n\nAll tests passed - 15 points!\n"
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      "source": "## Problem 2: Heap data structure to mantain/extract median (instead of minimum/maximum key)\n\nWe have seen how min-heaps can efficiently extract the smallest element efficiently and maintain the least element as we insert/delete elements. Similarly, max-heaps can maintain the largest element. In this exercise, we combine both to maintain the \"median\" element.\n\nThe median is the middle element of a list of numbers. \n- If the list has size $n$ where $n$ is odd, the median is the $(n-1)/2^{th}$ element where $0^{th}$ is least and $(n-1)^{th}$ is the maximum. \n- If $n$ is even, then we designate the median the average of the $(n/2-1)^{th}$ and $(n/2)^{th}$ elements.\n\n\n#### Example \n\n- List is $[-1, 5, 4, 2, 3]$ has size $5$, the median is the $2^{nd}$ element (remember again least element is designated as $0^{th}$) which is $3$.\n- List is $[-1, 3, 2, 1 ]$ has size $4$. The median element is the average of  $1^{st}$ element (1) and $2^{nd}$ element (2) which is  $1.5$.\n\n## Maintaining median using two heaps.\n\nThe data will be maintained as the union of the elements in two heaps $H_{\\min}$ and $H_{\\max}$, wherein $H_{\\min}$ is a min-heap and $H_{\\max}$ is a max-heap.  We will maintain the following invariant:\n  - The max element of  $H_{\\max}$ will be less than or equal to the min element of  $H_{\\min}$. \n  - The sizes of $H_{max}$ and $H_{min}$ are equal (if number of elements in the data structure is even) or $H_{max}$ may have one less element than $H_{min}$ (if the number of elements in the data structure is odd).\n  \n\n\n## (A)  Design algorithm for insertion.\n\nSuppose, we have the current data split between $H_{max}$ and $H_{min}$ and we wish to insert an element $e$ into the data structure, describe the algorithm you will use to insert. Your algorithm must decide which of the two heaps will $e$ be inserted into and how to maintain the size balance condition.\n\nDescribe the algorithm below and the overall complexity of an insert operation. This part will not be graded.",
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      "source": "## (B) Design algorithm for finding the median.\n\nImplement an algorithm for finding the median given the heaps $H_{\\min}$ and $H_{\\max}$. What is its complexity?",
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      "source": "## (C) Implement the algorithm\n\nComplete the implementation for maxheap data structure.\nFirst complete the implementation of MaxHeap.  You can cut and paste relevant parts from previous problems although we do not really recommend doing that. A better solution would have been to write a single implementation that could have served as min/max heap based on a flag. ",
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      "source": "class MaxHeap:\n    def __init__(self):\n        self.H = [None]\n        \n    def size(self):\n        return len(self.H)-1\n    \n    def __repr__(self):\n        return str(self.H[1:])\n        \n    def satisfies_assertions(self):\n        for i in range(2, len(self.H)):\n            assert self.H[i] <= self.H[i//2],  f'Maxheap property fails at position {i//2}, parent elt: {self.H[i//2]}, child elt: {self.H[i]}'\n    \n    def max_element(self):\n        return self.H[1]\n    \n    def bubble_up(self, index):\n        # recycle bubble_up in min-heap, modify to Funcitonal programming style\n        assert index >= 1\n        if index == 1: \n            return self.H\n        parent_index = index // 2\n        if self.H[parent_index] > self.H[index]:\n            return self.H\n        else:\n            self.H[parent_index], self.H[index] = self.H[index], self.H[parent_index]\n            return self.bubble_up(parent_index)        \n            \n    \n    def bubble_down(self, index):\n        # recycle bubble_down in min-heap\n        assert index >= 1 and index < len(self.H)\n        lchild_index = 2 * index\n        rchild_index = 2 * index + 1\n        # set up the value of left child to the element at that index if valid, or else make it +Infinity\n        lchild_value = self.H[lchild_index] if lchild_index < len(self.H) else -float('inf')\n        # set up the value of right child to the element at that index if valid, or else make it +Infinity\n        rchild_value = self.H[rchild_index] if rchild_index < len(self.H) else -float('inf')\n        # If the value at the index is lessthan or equal to the minimum of two children, then nothing else to do\n        if self.H[index] >= max(lchild_value, rchild_value):\n            return self.H\n        # Otherwise, find the index and value of the smaller of the two children.\n        # A useful python trick is to compare \n        min_child_value, min_child_index = max ((lchild_value, lchild_index), (rchild_value, rchild_index))\n        # Swap the current index with the least of its two children\n        self.H[index], self.H[min_child_index] = self.H[min_child_index], self.H[index]\n        # Bubble down on the minimum child index\n        return self.bubble_down(min_child_index)        \n        \n        \n    \n    # Function: insert\n    # Insert elt into minheap\n    # Use bubble_up/bubble_down function\n    def insert(self, elt):\n        #(recycle)\n        \n        self.H.append(elt)\n        assert self.H[self.size()] == elt\n        # use bubble_up\n        self.bubble_up(self.size())        \n        \n    # Function: delete_max\n    # delete the largest element in the heap. Use bubble_up/bubble_down\n    def delete_max(self):\n        #(recycle)\n         \n        if self.size() > 1:\n            self.H[1] = self.H.pop(-1)\n            self.bubble_down(1)\n        else:\n            self.H.pop(-1)       \n        ",
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      "cell_type": "code",
      "source": "h = MaxHeap()\nprint('Inserting: 5, 2, 4, -1 and 7 in that order.')\nh.insert(5)\nprint(f'\\t Heap = {h}')\nassert(h.max_element() == 5)\nh.insert(2)\nprint(f'\\t Heap = {h}')\nassert(h.max_element() == 5)\nh.insert(4)\nprint(f'\\t Heap = {h}')\nassert(h.max_element() == 5)\nh.insert(-1)\nprint(f'\\t Heap = {h}')\nassert(h.max_element() == 5)\nh.insert(7)\nprint(f'\\t Heap = {h}')\nassert(h.max_element() == 7)\nh.satisfies_assertions()\n\nprint('Deleting maximum element')\nh.delete_max()\nprint(f'\\t Heap = {h}')\nassert(h.max_element() == 5)\nh.delete_max()\nprint(f'\\t Heap = {h}')\nassert(h.max_element() == 4)\nh.delete_max()\nprint(f'\\t Heap = {h}')\nassert(h.max_element() == 2)\nh.delete_max()\nprint(f'\\t Heap = {h}')\nassert(h.max_element() == -1)\n# Test delete_max on heap of size 1, should result in empty heap.\nh.delete_max()\nprint(f'\\t Heap = {h}')\nprint('All tests passed: 5 points!')\n",
      "metadata": {
        "deletable": false,
        "editable": false,
        "nbgrader": {
          "cell_type": "code",
          "checksum": "cae564e9f64ed9a114536344fbb66866",
          "grade": true,
          "grade_id": "cell-0f15a9f966d07a21",
          "locked": true,
          "points": 5,
          "schema_version": 3,
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        "scrolled": true
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      "execution_count": 8,
      "outputs": [
        {
          "name": "stdout",
          "output_type": "stream",
          "text": "Inserting: 5, 2, 4, -1 and 7 in that order.\n\n\t Heap = [5]\n\n\t Heap = [5, 2]\n\n\t Heap = [5, 2, 4]\n\n\t Heap = [5, 2, 4, -1]\n\n\t Heap = [7, 5, 4, -1, 2]\n\nDeleting maximum element\n\n\t Heap = [5, 2, 4, -1]\n\n\t Heap = [4, 2, -1]\n\n\t Heap = [2, -1]\n\n\t Heap = [-1]\n\n\t Heap = []\n\nAll tests passed: 5 points!\n"
        }
      ]
    },
    {
      "cell_type": "code",
      "source": "class MedianMaintainingHeap:\n    def __init__(self):\n        self.hmin = MinHeap()\n        self.hmax = MaxHeap()\n        \n    def satisfies_assertions(self):\n        if self.hmin.size() == 0:\n            assert self.hmax.size() == 0\n            return \n        if self.hmax.size() == 0:\n            assert self.hmin.size() == 1\n            return \n        # 1. min heap min element >= max heap max element\n        assert self.hmax.max_element() <= self.hmin.min_element(), f'Failed: Max element of max heap = {self.hmax.max_element()} > min element of min heap {self.hmin.min_element()}'\n        # 2. size of max heap must be equal or one lessthan min heap.\n        s_min = self.hmin.size()\n        s_max = self.hmax.size()\n        assert (s_min == s_max or s_max  == s_min -1 ), f'Heap sizes are unbalanced. Min heap size = {s_min} and Maxheap size = {s_max}'\n    \n    def __repr__(self):\n        return 'Maxheap:' + str(self.hmax) + ' Minheap:'+str(self.hmin)\n    \n    def get_median(self):\n        if self.hmin.size() == 0:\n            assert self.hmax.size() == 0, 'Sizes are not balanced'\n            assert False, 'Cannot ask for median from empty heaps'\n        if self.hmax.size() == 0:\n            assert self.hmin.size() == 1, 'Sizes are not balanced'\n            return self.hmin.min_element()\n        \n        if self.hmax.size() == self.hmin.size():\n            return (self.hmax.max_element() + self.hmin.min_element())/2\n        else:\n            return self.hmin.min_element()\n        \n    \n    # function: balance_heap_sizes\n    # ensure that the size of hmax == size of hmin or size of hmax +1 == size of hmin\n    # If the condition above does not hold, move the max element from max heap into the min heap or\n    # vice versa as needed to balance the sizes.\n    # This function could be called from insert/delete_median methods\n    def balance_heap_sizes(self):\n        \n        if self.hmax.size() +1 == self.hmin.size() or self.hmax.size() == self.hmin.size(): # explicit is better\n            return self.hmax, self.hmin\n        elif self.hmax.size() > self.hmin.size():\n            self.hmin.insert(self.hmax.max_element())\n            self.hmax.delete_max()\n            return self.balance_heap_sizes()\n            \n        else:\n            self.hmax.insert(self.hmin.min_element())\n            self.hmin.delete_min()\n            return self.balance_heap_sizes()\n            \n        \n    \n    def insert(self, elt):\n        # Handle the case when either heap is empty\n        if self.hmin.size() == 0:\n            # min heap is empty -- directly insert into min heap\n            self.hmin.insert(elt)\n            return \n        if self.hmax.size() == 0:\n            # max heap is empty -- this better happen only if min heap has size 1.\n            assert self.hmin.size() == 1\n            if elt > self.hmin.min_element():\n                # Element needs to go into the min heap\n                current_min = self.hmin.min_element()\n                self.hmin.delete_min()\n                self.hmin.insert(elt)\n                self.hmax.insert(current_min)\n                # done!\n            else:\n                # Element goes into the max heap -- just insert it there.\n                self.hmax.insert(elt)\n            return \n        # Now assume both heaps are non-empty\n        \n        self.hmin.insert(elt)\n        self.balance_heap_sizes() \n        \n        \n    def delete_median(self):\n        self.hmax.delete_max()\n        self.balance_heap_sizes()",
      "metadata": {
        "deletable": false,
        "nbgrader": {
          "cell_type": "code",
          "checksum": "216b84cc8d77dc51531a57706c092f6d",
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          "grade_id": "cell-91adccbf6dbb6a77",
          "locked": false,
          "schema_version": 3,
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      "execution_count": 9,
      "outputs": []
    },
    {
      "cell_type": "code",
      "source": "m = MedianMaintainingHeap()\nprint('Inserting 1, 5, 2, 4, 18, -4, 7, 9')\n\nm.insert(1)\nprint(m)\nprint(m.get_median())\nm.satisfies_assertions()\nassert m.get_median() == 1,  f'expected median = 1, your code returned {m.get_median()}'\n\nm.insert(5)\nprint(m)\nprint(m.get_median())\nm.satisfies_assertions()\nassert m.get_median() == 3,  f'expected median = 3.0, your code returned {m.get_median()}'\n\nm.insert(2)\nprint(m)\nprint(m.get_median())\nm.satisfies_assertions()\n\nassert m.get_median() == 2,  f'expected median = 2, your code returned {m.get_median()}'\nm.insert(4)\nprint(m)\nprint(m.get_median())\nm.satisfies_assertions()\nassert m.get_median() == 3,  f'expected median = 3, your code returned {m.get_median()}'\n\nm.insert(18)\nprint(m)\nprint(m.get_median())\nm.satisfies_assertions()\nassert m.get_median() == 4,  f'expected median = 4, your code returned {m.get_median()}'\n\nm.insert(-4)\nprint(m)\nprint(m.get_median())\nm.satisfies_assertions()\nassert m.get_median() == 3,  f'expected median = 3, your code returned {m.get_median()}'\n\nm.insert(7)\nprint(m)\nprint(m.get_median())\nm.satisfies_assertions()\nassert m.get_median() == 4, f'expected median = 4, your code returned {m.get_median()}'\n\nm.insert(9)\nprint(m)\nprint(m.get_median())\nm.satisfies_assertions()\nassert m.get_median()== 4.5, f'expected median = 4.5, your code returned {m.get_median()}'\n\nprint('All tests passed: 15 points')\n",
      "metadata": {
        "deletable": false,
        "editable": false,
        "nbgrader": {
          "cell_type": "code",
          "checksum": "422671072f7ffc9a81991e5eb5ad5ba3",
          "grade": true,
          "grade_id": "cell-dbf8d2bf71617a5e",
          "locked": true,
          "points": 15,
          "schema_version": 3,
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          "task": false
        }
      },
      "execution_count": 10,
      "outputs": [
        {
          "name": "stdout",
          "output_type": "stream",
          "text": "Inserting 1, 5, 2, 4, 18, -4, 7, 9\n\nMaxheap:[] Minheap:[1]\n\n1\n\nMaxheap:[1] Minheap:[5]\n\n3.0\n\nMaxheap:[1] Minheap:[2, 5]\n\n2\n\nMaxheap:[2, 1] Minheap:[4, 5]\n\n3.0\n\nMaxheap:[2, 1] Minheap:[4, 5, 18]\n\n4\n\nMaxheap:[2, 1, -4] Minheap:[4, 5, 18]\n\n3.0\n\nMaxheap:[2, 1, -4] Minheap:[4, 5, 18, 7]\n\n4\n\nMaxheap:[4, 2, -4, 1] Minheap:[5, 7, 18, 9]\n\n4.5\n\nAll tests passed: 15 points\n"
        }
      ]
    },
    {
      "cell_type": "markdown",
      "source": "## Solutions to Manually Graded Portions\n\n### Problem 1 A\nIn order to insert a new element `j`, we will first  distinguish between two cases:\n- $j < A[k-1]$ : In this case $j$ belongs to the array $A$.\n  - First, let $j' = A[k-1]$. \n  - Replace $A[k-1]$ by $j$.\n  - Perform an insertion to move $j$ into its correct place in the sorted array $A$.\n  - Insert $j'$ into the heap using heap insert.\n- $j \\geq A[k-1]$: In this case, $j$ belongs to the heap $H$.\n  - Insert $j$ into the heap using heap-insert.\n  \nIn terms of $k, n$, the worst case complexity is $\\Theta(k + \\log(n))$ for each insertion operation.\n\n### Problem 1B \n\n- First, in order to delete the index j from array, move elements from j+1 .. k-1 left one position.\n- Insert the minimum heap element at position $k-1$ of the array A.\n- Delete the element at index 1 of the heap.\n\nOverall complexity = $\\Theta(k + \\log(n))$ in the worst case.\n\n### Problem 2 A \n\nLet $a$ be the largest element in $H_{\\max}$ and $b$ be the least element in $H_{\\min}$. \n - If $elt < a$, then we insert the new element into $H_{\\max}$.\n - If $elt >= a$, then we insert the new element into $H_{\\min}$. \n \n If the size of $H_{\\max}$ and $H_{\\min}$ differ by 2, then \n - If $H_{\\max}$ is larger then, extract the largest element from $H_{\\max}$ andd insert into $H_{\\min}$.\n - If $H_{\\min}$ is larger then,  extract the least element from $H_{\\min}$ andd insert into $H_{\\max}$.\n \n The overall complexity is $\\Theta(\\log(n))$.\n \n### Problem 2 B \n \n If sizes of heaps are the same, then median is the average of maximum element of the max heap and minimum element of the minheap.\n\nOtherwise, the median is simply the minimum elemment of the min-heap. \n\nOverall complexity is $\\Theta(1)$.\n\n",
      "metadata": {
        "deletable": false,
        "editable": false,
        "nbgrader": {
          "cell_type": "markdown",
          "checksum": "0cf7c180c39d286859661169519c2bc1",
          "grade": false,
          "grade_id": "cell-f4cd81bc93540969",
          "locked": true,
          "schema_version": 3,
          "solution": false,
          "task": false
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    },
    {
      "cell_type": "markdown",
      "source": "## That's all folks",
      "metadata": {
        "deletable": false,
        "editable": false,
        "nbgrader": {
          "cell_type": "markdown",
          "checksum": "b2b4532b793f0c869c0cabd5f78b6e82",
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          "grade_id": "cell-486292f54f89bff9",
          "locked": true,
          "schema_version": 3,
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        }
      }
    },
    {
      "cell_type": "code",
      "source": "print(\"Thanks For Reading - Efekan Alipasha\")\n",
      "metadata": {
        "trusted": true
      },
      "execution_count": null,
      "outputs": []
    }
  ]
}