problems.ipynb

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   "cell_type": "markdown",
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   "source": [
    "# Problem Statements\n",
    "\n",
    "These are some problems that comes in handy. Mostly uses logic.\n",
    "\n",
    "**Note:** These problems might have more than one solution, but each solution has its own advantages and disadvantages. You might want to use some AI Model to iterate on them and understand them better.\n",
    "\n",
    "Feel free to create a PR suggesting better solutions!"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Fibonacci Series"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 7,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "1, 2, 3, 5, 8, 13, 21, 34, 55, 89, "
     ]
    }
   ],
   "source": [
    "# init the first two numbers\n",
    "a = 0\n",
    "b = 1\n",
    "\n",
    "new = 0\n",
    "\n",
    "for i in range(10): # upto first 10 numbers\n",
    "    new = a+b # F(new) = F(n-1) + F(n-2)\n",
    "    print(new, end=\", \")\n",
    "    a, b = b, new \n",
    "    # swap the numbers so that it progresses. 0 becomes 1, 1 becomes 2 and so on."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 8,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "[0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1597, 2584]\n"
     ]
    }
   ],
   "source": [
    "def fibo(n):\n",
    "    if n in (0, 1):\n",
    "        return n\n",
    "    return fibo(n-1) + fibo(n-2)\n",
    "\n",
    "print([fibo(i) for i in range(19)])"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Find if Prime"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "First method is the Sieve of Eratosthenes:\n",
    "\n",
    "This is very inefficient as compared to the second method."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 9,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "not prime\n"
     ]
    }
   ],
   "source": [
    "def prime(n):\n",
    "    sep = (2, 3, 5, 7, 11, 13, 17) # using lists will be slower\n",
    "    if 0 in [n % i for i in sep]:\n",
    "        print('not prime')\n",
    "    else:\n",
    "        print(\"prime\")\n",
    "\n",
    "\n",
    "prime(60342232138)  "
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Second method is trying to find `num % i` for every `i` till `num`."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 10,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "not prime\n"
     ]
    }
   ],
   "source": [
    "def prime_while(num):\n",
    "    i = 2\n",
    "\n",
    "    while i < num:\n",
    "        if num % i == 0:\n",
    "            break\n",
    "        i +=1\n",
    "    \n",
    "    print(\"not prime\") if i < num else print(\"prime\")\n",
    "\n",
    "prime_while(27)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Finding number of strings, digits, char, etc."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "I have certain string `colemak@459!!!`\n",
    "\n",
    "I want to find:\n",
    "- special characters (specified)\n",
    "- char\n",
    "- digits"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 11,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "7 3 4\n"
     ]
    }
   ],
   "source": [
    "string = 'colemak@459!!!'\n",
    "\n",
    "count_digit = 0\n",
    "count_char = 0\n",
    "count_special = 0\n",
    "\n",
    "for i in string:\n",
    "    if i.isdigit():\n",
    "        count_digit += 1\n",
    "    elif i in (\"@\", \"!\"):\n",
    "        count_special += 1\n",
    "    else:\n",
    "        count_char += 1\n",
    "\n",
    "print(count_char, count_digit, count_special)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Dictionary of a string and corresponding index\n",
    "Write a python program to create a dictionary from a string `‘w3resource’` such that each individual character makes a key and its index value for first occurrence makes the corresponding value in dictionary.\n",
    "\n",
    "e.g. `{‘w’:0,’3’:1,’r’:2,’e’:3,’s’:4,’o’:5,’u’:6,’c’:8}`"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 12,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "{'w': 0, '3': 1, 'r': 2, 'e': 3, 's': 4, 'o': 5, 'u': 6, 'c': 8}\n"
     ]
    }
   ],
   "source": [
    "w = 'w3resource'\n",
    "\n",
    "my_dic = {}\n",
    "count = 0\n",
    "for i in w:\n",
    "    if i not in my_dic.keys():\n",
    "        my_dic[i] = count\n",
    "    count +=1\n",
    "\n",
    "print(my_dic)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Loop Pattern\n",
    "Write the program to print the following pattern using loops only.\n",
    "```\n",
    "1\n",
    "1 2\n",
    "1 2 3\n",
    "1 2 3 4\n",
    "```\n",
    "\n",
    "Thanks to @Jonak-Adipta-Kalita for showing me a better method, which is as follows:"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 13,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "\n",
      "1 \n",
      "1 2 \n",
      "1 2 3 \n",
      "1 2 3 4 "
     ]
    }
   ],
   "source": [
    "for e in range(1, 4+1):\n",
    "    print(\"\")\n",
    "    for i in range(1, e+1): \n",
    "        print(i, end=\" \")"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Palindrome\n",
    "\n",
    "Write a program to take a number from the user and find out whether the entered number is palindrome or not .\n",
    "For example if entered number is 5225 it is palindrome . If you read number from\n",
    "right side it will be same"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 14,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "def a palindrome\n"
     ]
    }
   ],
   "source": [
    "num = \"5225\"\n",
    "\n",
    "if num[::-1] == num: # this works because a string is a \"list of characters\".\n",
    "    print(\"def a palindrome\")\n",
    "else:\n",
    "    print(\"you might want to double check the string, sir!\")"
   ]
  }
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