35_RGB_Element_Swap_Google.ipynb

{
 "cells": [
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "# For this problem, your goal is to sort an array of 0, 1 and 2's but you must do this in place, in linear time and without any extra space (such as creating an extra array). This is called the Dutch national flag sorting problem. \n",
    "For example, if the input array is [2,0,0,1,2,1] then your program should output [0,0,1,1,2,2] and the algorithm should \n",
    "run in O(n) time."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "The solution to this algorithm will require 3 pointers to iterate throughout the array, swapping the necessary elements.\n",
    "\n",
    "(1) Create a low pointer at the beginning of the array and a high pointer at the end of the array.\n",
    "\n",
    "(2) Create a mid pointer that starts at the beginning of the array and iterates through each element.\n",
    "\n",
    "(3) If the element at arr[mid] is a 2, then swap arr[mid] and arr[high] and decrease the high pointer by 1.\n",
    "\n",
    "(4) If the element at arr[mid] is a 0, then swap arr[mid] and arr[low] and increase the low and mid pointers by 1.\n",
    "\n",
    "(5) If the element at arr[mid] is a 1, don't swap anything and just increase the mid pointer by 1."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 2,
   "metadata": {},
   "outputs": [
    {
     "data": {
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      "text/plain": [
       "<IPython.core.display.Image object>"
      ]
     },
     "execution_count": 2,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "from IPython.display import Image\n",
    "Image(filename=\"dutch.jpg\")"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 3,
   "metadata": {},
   "outputs": [],
   "source": [
    "def swap(arr, i1, i2):\n",
    "  temp = arr[i1]\n",
    "  arr[i1] = arr[i2]\n",
    "  arr[i2] = temp\n",
    "\n",
    "def dutchNatFlag(arr):\n",
    "  \n",
    "  low = 0\n",
    "  mid = 0\n",
    "  high = len(arr) - 1\n",
    "  \n",
    "  # one pass through the array swapping\n",
    "  # the necessary elements in place\n",
    "  while mid <= high:\n",
    "    if arr[mid] == 0:\n",
    "      swap(arr, low, mid)\n",
    "      low += 1\n",
    "      mid += 1\n",
    "    elif arr[mid] == 2:\n",
    "      swap(arr, mid, high)\n",
    "      high -= 1\n",
    "    elif arr[mid] == 1:\n",
    "      mid += 1\n",
    "      \n",
    "  return arr"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 5,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "[0, 0, 0, 1, 1, 2, 2, 2]\n"
     ]
    }
   ],
   "source": [
    "print(dutchNatFlag([2,2,2,0,0,0,1,1])) "
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "# Given an array of strictly the characters ‘R’, ‘G’, and ‘B’, segregate the values of the array so that all the Rs come first, the Gs come second, and the Bs come last. You can only swap elements of the array.  Do this in linear time and in-place.\n",
    "\n",
    "For example, given the array ['G', 'B', 'R', 'R', 'B', 'R', 'G'], it should become ['R', 'R', 'R', 'G', 'G', 'B', 'B']."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 16,
   "metadata": {},
   "outputs": [],
   "source": [
    "\"\"\"\n",
    "Treat ‘R’,’G’ and ‘B’ as numbers. The problem can be solved by sorting this array based on certain order. \n",
    "We can use Quick Sort to achieve that. And the idea is that we keep three pointers lo <= mid <= hi such that ‘G’ grows \n",
    "from lo, ‘B’ grows from hi and ‘B’ grows from mid and swap w/ lo to make some room. Such technique to partition the \n",
    "array into 3 parts is called 3-Way Quick Select. It feels like normal Quick Select except segregate array into 3 parts.\n",
    "\"\"\"\n",
    "\n",
    "def swap_rgb_array(arr):\n",
    "    def swap(i, j):\n",
    "        tmp = arr[i]\n",
    "        arr[i] = arr[j]\n",
    "        arr[j] = tmp\n",
    "    lo = mid = 0\n",
    "    hi = len(arr) - 1\n",
    "    while mid <= hi:\n",
    "        if arr[mid] == 'R':\n",
    "            swap(lo, mid)\n",
    "            lo += 1\n",
    "            mid += 1\n",
    "        elif arr[mid] == 'B':\n",
    "            swap(mid, hi)\n",
    "            hi -= 1\n",
    "        else:\n",
    "            # arr[mid] == 'G'\n",
    "            mid += 1\n",
    "    return arr"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 17,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "['R', 'R', 'R', 'G', 'G', 'B', 'B']\n"
     ]
    }
   ],
   "source": [
    "mylist=['G', 'B', 'R', 'R', 'B', 'R', 'G']\n",
    "print(swap_rgb_array(mylist))"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": []
  }
 ],
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  "kernelspec": {
   "display_name": "Anaconda 3",
   "language": "python",
   "name": "python3"
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  "language_info": {
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    "name": "ipython",
    "version": 3
   },
   "file_extension": ".py",
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   "name": "python",
   "nbconvert_exporter": "python",
   "pygments_lexer": "ipython3",
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