update

4a. Hash table.ipynb

@@ -49,9 +49,9 @@
    "source": [
     "Thus the hash table as follows after the additions:\n",
     "\n",
-    "| Location | 00   | 11 | 22 | 33 | 44 | 55 | 66   | 77 | 88 | 99 |\n",
-    "| -------- | ---- | -- | -- | -- | -- | -- | ---- | -- | -- | -- |\n",
-    "| Elements | 3030 | –  | –  | 99 | 22 | 55 | 1818 | –  | 44 | –  |\n",
+    "| Location | 0  | 1 | 2 | 3 | 4 | 5 | 6  | 7 | 8 | 9 |\n",
+    "| -------- | -- | - | - | - | - | - | -- | - | - | - |\n",
+    "| Elements | 30 | – | – | 9 | 2 | 5 | 18 | – | 4 | – |\n",
     "\n",
     "When we want to check if the hash table contains an element $x$, we search for it at the location $7x \\bmod 10$. For example, we will search for the element $4$ at the location $7 \\cdot 4 \\bmod 10 = 8$."
    ]

Exercises/Week 4/26 - No pair.ipynb

@@ -0,0 +1,279 @@
+{
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "# No pair"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "You are given a list containing $n$ integers. Every number has exactly two occurrences on the list, except one number that occurs only once. Your task is to find this number.\n",
+    "\n",
+    "The time complexity of the algorithm should be $O(n)$.\n",
+    "\n",
+    "In a file `nopair.py`, implement a function `find` that returns the desired number."
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": null,
+   "metadata": {},
+   "outputs": [],
+   "source": [
+    "def find(t):\n",
+    "    # TODO\n",
+    "\n",
+    "if __name__ == \"__main__\":\n",
+    "    print(find([2,1,3,2,3])) # 1\n",
+    "    print(find([5,5,9])) # 9\n",
+    "    print(find([1,2,3,4,1,3,4])) # 2\n",
+    "    print(find([8])) # 8\n",
+    "    print(find([7,1,7,4,4,5,1])) # 5"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Attempt 1"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 1,
+   "metadata": {},
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "{2: 2, 1: 1, 3: 2}\n",
+      "1\n",
+      "{5: 2, 9: 1}\n",
+      "9\n",
+      "{1: 2, 2: 1, 3: 2, 4: 2}\n",
+      "2\n",
+      "{8: 1}\n",
+      "8\n",
+      "{7: 2, 1: 2, 4: 2, 5: 1}\n",
+      "5\n"
+     ]
+    }
+   ],
+   "source": [
+    "def find(t):\n",
+    "    d = {}\n",
+    "    \n",
+    "    for i in t:\n",
+    "        if i not in d: d[i] = 0 # noqa: E701\n",
+    "        d[i] += 1\n",
+    "    \n",
+    "    keys = list(d.keys())\n",
+    "    counts = list(d.values())\n",
+    "    \n",
+    "    print(d)\n",
+    "    return keys[counts.index(1)]\n",
+    "        \n",
+    "\n",
+    "if __name__ == \"__main__\":\n",
+    "    print(find([2,1,3,2,3])) # 1\n",
+    "    print(find([5,5,9])) # 9\n",
+    "    print(find([1,2,3,4,1,3,4])) # 2\n",
+    "    print(find([8])) # 8\n",
+    "    print(find([7,1,7,4,4,5,1])) # 5"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Attempt 2"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 10,
+   "metadata": {},
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "1\n",
+      "9\n",
+      "2\n",
+      "8\n",
+      "5\n"
+     ]
+    }
+   ],
+   "source": [
+    "def find(t):\n",
+    "    d = set()\n",
+    "    \n",
+    "    for i in t:\n",
+    "        if i not in d: \n",
+    "            d.add(i) \n",
+    "        else:\n",
+    "            d.remove(i)\n",
+    "    \n",
+    "    return next(iter(d))\n",
+    "        \n",
+    "\n",
+    "if __name__ == \"__main__\":\n",
+    "    print(find([2,1,3,2,3])) # 1\n",
+    "    print(find([5,5,9])) # 9\n",
+    "    print(find([1,2,3,4,1,3,4])) # 2\n",
+    "    print(find([8])) # 8\n",
+    "    print(find([7,1,7,4,4,5,1])) # 5"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 11,
+   "metadata": {},
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "1\n",
+      "9\n",
+      "2\n",
+      "8\n",
+      "5\n"
+     ]
+    }
+   ],
+   "source": [
+    "def find(t):\n",
+    "    d = set()\n",
+    "    \n",
+    "    for i in t:\n",
+    "        d.add(i) if i not in d else d.remove(i)\n",
+    "        \n",
+    "    return next(iter(d))\n",
+    "        \n",
+    "\n",
+    "if __name__ == \"__main__\":\n",
+    "    print(find([2,1,3,2,3])) # 1\n",
+    "    print(find([5,5,9])) # 9\n",
+    "    print(find([1,2,3,4,1,3,4])) # 2\n",
+    "    print(find([8])) # 8\n",
+    "    print(find([7,1,7,4,4,5,1])) # 5"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Solution"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "The following solution counts the occurrences of each number using a dictionary `count`, and then finds the number with count one."
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": null,
+   "metadata": {},
+   "outputs": [],
+   "source": [
+    "def find(t):\n",
+    "    count = {}\n",
+    " \n",
+    "    for x in t:\n",
+    "        if x not in count:\n",
+    "            count[x] = 0\n",
+    "        count[x] += 1\n",
+    " \n",
+    "    for x in t:\n",
+    "        if count[x] == 1:\n",
+    "            return x"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "Here is another solution based on the xor operator `^` that operates on the binary representations of the numbers. When all the numbers are xored together, the result has one bit at positions that had an odd number of one bits in the collection of numbers. Thus the result is exactly the number that has no pair."
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": null,
+   "metadata": {},
+   "outputs": [],
+   "source": [
+    "def find(t):\n",
+    "    result = 0\n",
+    "    for x in t:\n",
+    "        result ^= x\n",
+    "    return result"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 20,
+   "metadata": {},
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "10\n",
+      "0\n",
+      "2\n",
+      "6\n",
+      "2\n",
+      "7\n"
+     ]
+    }
+   ],
+   "source": [
+    "result = 0\n",
+    "result ^= 10\n",
+    "print(result)\n",
+    "result ^= 10\n",
+    "print(result)\n",
+    "result ^= 2\n",
+    "print(result)\n",
+    "result ^= 4\n",
+    "print(result)\n",
+    "result ^= 4\n",
+    "print(result)\n",
+    "result ^= 5\n",
+    "print(result)"
+   ]
+  }
+ ],
+ "metadata": {
+  "kernelspec": {
+   "display_name": "Python 3",
+   "language": "python",
+   "name": "python3"
+  },
+  "language_info": {
+   "codemirror_mode": {
+    "name": "ipython",
+    "version": 3
+   },
+   "file_extension": ".py",
+   "mimetype": "text/x-python",
+   "name": "python",
+   "nbconvert_exporter": "python",
+   "pygments_lexer": "ipython3",
+   "version": "3.12.3"
+  }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 2
+}