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docs: add kth largest elements lab
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‎Basic Algorithms/Faster Divide & Conquer Algorithms/Kth Largest Element.ipynb‎

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"source": [
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"### Problem Statement\n",
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"Given an unsorted array `Arr` with `n` positive integers. Find the $k^{th}$ smallest element in the given array, using Divide & Conquer approach. \n",
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"\n",
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"**Input**: Unsorted array `Arr` and an integer `k` where $1 \\leq k \\leq n$ <br>\n",
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"**Output**: The $k^{th}$ smallest element of array `Arr`<br>\n",
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"\n",
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"\n",
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"**Example 1**<br>\n",
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"Arr = `[6, 80, 36, 8, 23, 7, 10, 12, 42, 99]`<br>\n",
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"k = `10`<br>\n",
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"Output = `99`<br>\n",
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"\n",
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"**Example 2**<br>\n",
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"Arr = `[6, 80, 36, 8, 23, 7, 10, 12, 42, 99]`<br>\n",
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"k = `5`<br>\n",
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"Output = `12`<br>\n",
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"\n",
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"---\n",
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"\n",
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"### The Pseudocode - `fastSelect(Arr, k)`\n",
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"1. Break `Arr` into $\\frac{n}{5}$ (actually it is $\\left \\lceil{\\frac{n}{5}} \\right \\rceil $) groups, namely $G_1, G_2, G_3...G_{\\frac{n}{5}}$\n",
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"\n",
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"\n",
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"2. For each group $G_i, \\forall 1 \\leq i \\leq \\frac{n}{5} $, do the following:\n",
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" - Sort the group $G_i$\n",
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" - Find the middle position i.e., median $m_i$ of group $G_i$\n",
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" - Add $m_i$ to the set of medians **$S$**\n",
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"\n",
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"\n",
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"3. The set of medians **$S$** will become as $S = \\{m_1, m_2, m_3...m_{\\frac{n}{5}}\\}$. The \"good\" `pivot` element will be the median of the set **$S$**. We can find it as $pivot = fastSelect(S, \\frac{n}{10})$. \n",
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"\n",
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"\n",
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"4. Partition the original `Arr` into three sub-arrays - `Arr_Less_P`, `Arr_Equal_P`, and `Arr_More_P` having elements less than `pivot`, equal to `pivot`, and bigger than `pivot` **respectively**.\n",
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"\n",
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"\n",
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"5. Recurse based on the **sizes of the three sub-arrays**, we will either recursively search in the small set, or the big set, as defined in the following conditions:\n",
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" - If `k <= length(Arr_Less_P)`, then return `fastSelect(Arr_Less_P, k)`. This means that if the size of the \"small\" sub-array is at least as large as `k`, then we know that our desired $k^{th}$ smallest element lies in this sub-array. Therefore recursively call the same function on the \"small\" sub-array. <br><br>\n",
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" \n",
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" - If `k > (length(Arr_Less_P) + length(Arr_Equal_P))`, then return `fastSelect(Arr_More_P, (k - length(Arr_Less_P) - length(Arr_Equal_P)))`. This means that if `k` is more than the size of \"small\" and \"equal\" sub-arrays, then our desired $k^{th}$ smallest element lies in \"bigger\" sub-array. <br><br>\n",
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" \n",
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" - Return `pivot` otherwise. This means that if the above two cases do not hold true, then we know that $k^{th}$ smallest element lies in the \"equal\" sub-array.\n",
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" \n",
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"---\n",
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"### Exercise - Write the function definition here"
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"source": [
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"def fastSelect(Arr, k): # k is an index\n",
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" n = len(Arr) # length of the original array\n",
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" \n",
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" if(k>0 and k <= n): # k should be a valid index \n",
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" # Helper variables\n",
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" setOfMedians = []\n",
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" Arr_Less_P = []\n",
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" Arr_Equal_P = []\n",
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" Arr_More_P = []\n",
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" i = 0\n",
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" \n",
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" # Step 1 - Break Arr into groups of size 5\n",
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" # Step 2 - For each group, sort and find median (middle). Add the median to setOfMedians\n",
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" while (i < n // 5): # n//5 gives the integer quotient of the division \n",
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" median = findMedian(Arr, 5*i, 5) # find median of each group of size 5\n",
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" setOfMedians.append(median) \n",
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" i += 1\n",
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"\n",
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" # If n is not a multiple of 5, then a last group with size = n % 5 will be formed\n",
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" if (5*i < n): \n",
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" median = findMedian(Arr, 5*i, n % 5)\n",
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" setOfMedians.append(median)\n",
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" \n",
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" # Step 3 - Find the median of setOfMedians\n",
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" if (len(setOfMedians) == 1): # Base case for this task\n",
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" pivot = setOfMedians[0]\n",
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" elif (len(setOfMedians)>1):\n",
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" pivot = fastSelect(setOfMedians, (len(setOfMedians)//2))\n",
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" \n",
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" # Step 4 - Partition the original Arr into three sub-arrays\n",
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" for element in Arr:\n",
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" if (element<pivot):\n",
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" Arr_Less_P.append(element)\n",
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" elif (element>pivot):\n",
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" Arr_More_P.append(element)\n",
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" else:\n",
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" Arr_Equal_P.append(element)\n",
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" \n",
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" # Step 5 - Recurse based on the sizes of the three sub-arrays\n",
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" if (k <= len(Arr_Less_P)):\n",
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" return fastSelect(Arr_Less_P, k)\n",
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" \n",
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" elif (k > (len(Arr_Less_P) + len(Arr_Equal_P))):\n",
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" return fastSelect(Arr_More_P, (k - len(Arr_Less_P) - len(Arr_Equal_P)))\n",
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" \n",
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" else:\n",
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" return pivot \n",
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"\n",
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"# Helper function\n",
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"def findMedian(Arr, start, size): \n",
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" myList = [] \n",
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" for i in range(start, start + size): \n",
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" myList.append(Arr[i]) \n",
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" \n",
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" # Sort the array \n",
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" myList.sort() \n",
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" \n",
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" # Return the middle element \n",
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" return myList[size // 2] "
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"<span class=\"graffiti-highlight graffiti-id_dsq4qxt-id_29dh0dm\"><i></i><button>Show Solution</button></span>"
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"source": [
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"### Test - Let's test your function"
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]
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"text": [
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"12\n"
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]
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}
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],
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"source": [
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"Arr = [6, 80, 36, 8, 23, 7, 10, 12, 42]\n",
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"k = 5\n",
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"print(fastSelect(Arr, k)) # Outputs 12"
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]
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"text": [
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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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"Arr = [5, 2, 20, 17, 11, 13, 8, 9, 11]\n",
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"k = 5\n",
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"print(fastSelect(Arr, k)) # Outputs 11"
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]
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"name": "stdout",
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"text": [
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"99\n"
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]
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}
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],
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"source": [
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"Arr = [6, 80, 36, 8, 23, 7, 10, 12, 42, 99]\n",
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"k = 10\n",
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"print(fastSelect(Arr, k)) # Outputs 99"
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]
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