Added notebook solving the following: Implement an algorithm to determine if a string has all unique characters.

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Donne Martin 2015-04-30 17:45:08 -04:00
parent 4b0cd10ce3
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{
"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
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"file_extension": ".py",
"mimetype": "text/x-python",
"name": "python",
"nbconvert_exporter": "python",
"pygments_lexer": "ipython2",
"version": "2.7.9"
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"nbformat": 4,
"nbformat_minor": 0
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