interactive-coding-challenges/arrays-strings/unique_chars.ipynb

{
 "cells": [
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Problem: Implement an algorithm to determine if a string has all unique characters"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Clarifying Questions\n",
    "* Is the string in ASCII (extended?) or Unicode?  \n",
    "    * ASCII extended, which is 256 characters.\n",
    "* Can you use additional data structures?  \n",
    "    * Yes"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Test Cases\n",
    "\n",
    "* \"\" -> True\n",
    "* \"foo\" -> False\n",
    "* \"bar\" -> True"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Algorithm\n",
    "\n",
    "We'll keep a hash map (set) to keep track of unique characters we encounter.  \n",
    "\n",
    "Note:\n",
    "* We could also use a dictionary, but it seems more logical to use a set as it does not contain duplicate elements.\n",
    "* Since the characters are in ASCII, we could potentially use an array of size 128 (or 256 for extended ASCII)\n",
    "\n",
    "Steps:\n",
    "* Scan each character.\n",
    "* For each character:\n",
    "    * If the character does not exist in a hash map, add the character to a hash map.\n",
    "    * Else, return False.\n",
    "* Return True\n",
    "\n",
    "Complexity:\n",
    "* Time: O(n).\n",
    "* Space: Additional O(m), where m is the number of unique characters in the hash map."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Code"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {
    "collapsed": false
   },
   "outputs": [],
   "source": [
    "def unique_chars(string):\n",
    "    chars_set = set()\n",
    "    for char in string:\n",
    "        if char in chars_set:\n",
    "            return False\n",
    "        else:\n",
    "            chars_set.add(char)\n",
    "    return True\n",
    "\n",
    "print(unique_chars(''))\n",
    "print(unique_chars('foo'))\n",
    "print(unique_chars('bar'))"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Algorithm: No Additional Data Structures\n",
    "\n",
    "Since we cannot use additional data structures, this will eliminate the fast lookup O(1) time provided by our hash map.\n",
    "* Scan each character.\n",
    "* For each character:\n",
    "    * Scan all [other] characters in the array\n",
    "        * Exluding the current character from the scan is rather tricky in Python and results in a non-Pythonic solution\n",
    "        * If there is a match, return False\n",
    "* Return True\n",
    "\n",
    "Algorithm Complexity:\n",
    "* Time: O(n^2).\n",
    "* Space: In-place."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Code"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {
    "collapsed": false
   },
   "outputs": [],
   "source": [
    "def unique_chars_alt(string):\n",
    "    for char in string:\n",
    "        if string.count(char) > 1:\n",
    "            return False\n",
    "    return True"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Pythonic Solution(s)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {
    "collapsed": false
   },
   "outputs": [],
   "source": [
    "def unique_chars_alt2(string):\n",
    "    return len(set(string)) == len(string)"
   ]
  }
 ],
 "metadata": {
  "kernelspec": {
   "display_name": "Python 2",
   "language": "python",
   "name": "python2"
  },
  "language_info": {
   "codemirror_mode": {
    "name": "ipython",
    "version": 2
   },
   "file_extension": ".py",
   "mimetype": "text/x-python",
   "name": "python",
   "nbconvert_exporter": "python",
   "pygments_lexer": "ipython2",
   "version": "2.7.9"
  }
 },
 "nbformat": 4,
 "nbformat_minor": 0
}