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Finding the longest word (and write it out) in string without split, distinct and foreach [duplicate]

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Finding longest word in string
(4 answers)
Closed 3 years ago.
I got an assignment to make a method to find the longest word in a string without split, distinct and foreach.
I was able to split the words and count the length but I am stuck on how can I actually compare and write them out.
static void Main(string[] args)
{
String s1 = "Alex has 2 hands.";
longestWord(s1);
Console.
}
static void longestWord(String s1)
{
char emp = ' ';
int count = 0;
char[] ee = s1.ToCharArray();
for (int i = 0; i < ee.Length; i++)
{
if (ee[i] == emp || ee[i] == '.')
{
count++;
Console.Write(" " + (count-1));
Console.WriteLine();
count = 0;
}
else
{
Console.Write(ee[i]);
count++;
}
}
}
The output right now looks like this:
Alex 4
has 3
2 1
hands 5
I am pretty sure I would be able to get only the longest number to show by comparing count before reset with temp int but how to write out the word with it.
Or if there is a easier way which probably is.

You are already on a good way. Instead of directly printing the words, store the length and position of the longest word and print it at the end. Like so:
static void longestWord(String s1)
{
char emp = ' ';
int longestLength = 0;
int longestStart = 0;
int currentStart = 0;
for (int i = 0; i < s1.Length; i++)
{
if (s1[i] == emp || s1[i] == '.')
{
// calculate the current word length
int currentLength = i - currentStart;
// test if this is longer than the currently longest
if(currentLength > longestLength)
{
longestLength = currentLength;
longestStart = currentStart;
}
// a new word starts at the next character
currentStart = i + 1;
}
}
// print the longest word
Console.WriteLine($"Longest word has length {longestLength}: \"{s1.Substring(longestStart, longestLength)}\"");
}
There is no need for the .ToCharArray(). You can access the string directly.

I will question whether you are actually supposed to treat "2" as a word and count it at all. Using regular expressions will allow you to approach the problem using a LINQ one-liner:
static void Main(string[] args)
{
String s1 = "Alex has 2 hands.";
var word = longestWord(s1);
Console.WriteLine(word);
//Console.ReadLine();
}
static string longestWord(string s1) {
return Regex.Matches(s1,"[A-Za-z]+") // find all sequences containing alphabetical characters, you can add numberas as well: [A-Za-z0-9]
.Cast<Match>() // cast results to Enumberable<Match> so we can apply LINQ to it
.OrderByDescending(m => m.Length) // luckily for us, Match comes with Length, so we just sort our results by it
.Select(m => m.Value) // instead of picking up everything, we only want the actual word
.First(); // since we ordered by descending - first item will be the longest word
}

You can store for every word the chars in new list of chars (list for dynamic length)
and if the new word is longer of the prev long word convert it to string.
If you have two word in same length it will take the first.
If you want the last change the "maxLength < count" to "maxLength <= count"
static string longestWord(String s1)
{
char emp = ' ';
int count = 0;
int maxLength = 0;
string maxWord = string.Empty;
List<char> newWord = new List<char>();
char[] ee = s1.ToCharArray();
for (int i = 0; i < ee.Length; i++)
{
if (ee[i] == emp || ee[i] == '.')
{
if (maxLength < count)
{
maxLength = count;
maxWord = new string(newWord.ToArray());
}
count = 0;
newWord = new List<char>();
}
else
{
newWord.Add(ee[i]);
count++;
}
}
return maxWord;
}

find number with no pair in array

I am having trouble with a small bit of code, which in a random size array, with random number pairs, except one which has no pair.
I need to find that number which has no pair.
arLength is the length of the array.
but i am having trouble actually matching the pairs, and finding the one which has no pair..
for (int i = 0; i <= arLength; i++)
{ // go through the array one by one..
var number = nArray[i];
// now search through the array for a match.
for (int e = 0; e <= arLength; e++)
{
if (e != i)
{
}
}
}
I have also tried this :
var findValue = nArray.Distinct();
I have searched around, but so far, i haven't been able to find a method for this.
This code is what generates the array, but this question isn't about this part of the code, only for clarity.
Random num = new Random();
int check = CheckIfOdd(num.Next(1, 1000000));
int counter = 1;
while (check <= 0)
{
if (check % 2 == 0)
{
check = CheckIfOdd(num.Next(1, 1000000)); ;
}
counter++;
}
int[] nArray = new int[check];
int arLength = 0;
//generate arrays with pairs of numbers, and one number which does not pair.
for (int i = 0; i < check; i++)
{
arLength = nArray.Length;
if (arLength == i + 1)
{
nArray[i] = i + 1;
}
else
{
nArray[i] = i;
nArray[i + 1] = i;
}
i++;
}

You can do it using the bitwise operator ^, and the complexity is O(n).
Theory
operator ^ aka xor has the following table:
So suppose you have only one number without pair, all the pairs will get simplified because they are the same.
var element = nArray[0];
for(int i = 1; i < arLength; i++)
{
element = element ^ nArray[i];
}
at the end, the variable element will be that number without pair.

Distict will give you back the array with distinct values. it will not find the value you need.
You can GroupBy and choose the values with Count modulo 2 equals 1.
var noPairs = nArray.GroupBy(i => i)
.Where(g => g.Count() % 2 == 1)
.Select(g=> g.Key);

You can use a dictionary to store the number of occurrences of each value in the array. To find the value without pairs, look for a (single) number of occurrences smaller than 2.
using System.Linq;
int[] data = new[] {1, 2, 3, 4, 5, 3, 2, 4, 1};
// key is the number, value is its count
var numberCounts = new Dictionary<int, int>();
foreach (var number in data) {
if (numberCounts.ContainsKey(number)) {
numberCounts[number]++;
}
else {
numberCounts.Add(number, 1);
}
}
var noPair = numberCounts.Single(kvp => kvp.Value < 2);
Console.WriteLine(noPair.Key);
Time complexity is O(n) because you traverse the array only a single time and then traverse the dictionary a single time. The same dictionary can also be used to find triplets etc.
.NET Fiddle

An easy and fast way to do this is with a Frequency Table. Keep a dictionary with as key your number and as value the number of times you found it. This way you only have to run through your array once.
Your example should work too with some changes. It will be a lot slower if you have a big array.
for (int i = 0; i <= arLength; i++)
{
bool hasMatch = false;
for (int e = 0; e <= arLength; e++)
{
if (nArray[e] == nArray[i])//Compare the element, not the index.
{
hasMatch = true;
}
}
//if hasMatch == false, you found your item.
}

All you have to do is to Xor all the numbers:
int result = nArray.Aggregate((s, a) => s ^ a);
all items which has pair will cancel out: a ^ a == 0 and you'll have the distinc item: 0 ^ 0 ^ ...^ 0 ^ distinct ^ 0 ^ ... ^0 == distinct

Because you mentioned you like short and simple in a comment, how about getting rid of most of your other code as well?
var total = new Random().Next(500000) * 2 + 1;
var myArray = new int[total];
for (var i = 1; i < total; i+=2)
{
myArray[i] = i;
myArray[i -1] = i;
}
myArray[total - 1] = total;
Then indeed use Linq to get what you are looking for. Here is a slight variation, returning the key of the item in your array:
var key = myArray.GroupBy(t => t).FirstOrDefault(g=>g.Count()==1)?.Key;

how to optimize FirstOrDefault Statement in linq

I have a two linq statements that the first one takes 25 ms and the second one takes 1100 ms second in a loop of 100,000.
I have replaced FirstAll with ElementAt and even used foreach to get the first element, but still takes the same time.
Is there any faster way to get the first element ?
I have considered few other questions but still couldn't find any solution to solve this problem.
var matches = (from subset in MyExtensions.SubSetsOf(List1)
where subset.Sum() <= target
select subset).OrderByDescending(i => i.Sum());
var match = matches.FirstOrDefault(0);
Also tried:
foreach (var match in matches)
{
break;
}
Or even:
var match = matches.ElementAt(0);
Any comments would be appreciated.
EDIT: here is the code for SubSetOf
public static class MyExtensions
{
public static IEnumerable<IEnumerable<T>> SubSetsOf<T>(this IEnumerable<T> source)
{
// Deal with the case of an empty source (simply return an enumerable containing a single, empty enumerable)
if (!source.Any())
return Enumerable.Repeat(Enumerable.Empty<T>(), 1);
// Grab the first element off of the list
var element = source.Take(1);
// Recurse, to get all subsets of the source, ignoring the first item
var haveNots = SubSetsOf(source.Skip(1));
// Get all those subsets and add the element we removed to them
var haves = haveNots.Select(set => element.Concat(set));
// Finally combine the subsets that didn't include the first item, with those that did.
return haves.Concat(haveNots);
}
}

You call Sum twice - it's bad. Precalc it:
var matches = MyExtensions.SubSetsOf(List1)
.Select(subset => new { subset, Sum = subset.Sum() })
.Where(o => o.Sum < target).OrderByDescending(i => i.Sum);
var match = matches.FirstOrDefault();
var subset = match != null ? match.subset : null;

As Jason said, it's a subset sum problem - the option of Knapsack Problem where weight is equal to value. The simplest solution - generate all subsets and check thiers sum, but this algorithm has horrible complexity. So, our optimization does not matter.
You shoul use dynamic programmig to solve this problem:
Let assume a two-dimensional array D(i, c) - maximal sum of i elements that is less or equal to c. N - is amount of elements (list size). W - max sum (your target).
D(0,c) = 0 for every c, because you have no elements :)
And changing c from 1 to W and i from 1 to N let's compute
D(i,c) = max(D(i-1,c),D(i-1,c-list[i])+list[i]).
To restore subset, we must store array of parents and set them during calculations.
Another examples are here.
Whole code:
class Program
{
static void Main(string[] args)
{
var list = new[] { 11, 2, 4, 6 };
var target = 13;
var n = list.Length;
var result = KnapSack(target, list, n);
foreach (var item in result)
{
Console.Write(item + " ");
}
}
private static List<int> KnapSack(int target, int[] val, int n)
{
var d = new int[n + 1, target + 1];
var p = new int[n + 1, target + 1];
for (var i = 0; i <= n; i++)
{
for (var c = 0; c <= target; c++)
{
p[i, c] = -1;
}
}
for (int i = 0; i <= n; i++)
{
for (int c = 0; c <= target; c++)
{
if (i == 0 || c == 0)
{
d[i, c] = 0;
}
else
{
var a = d[i - 1, c];
if (val[i - 1] <= c)
{
var b = val[i - 1] + d[i - 1, c - val[i - 1]];
if (a > b)
{
d[i, c] = a;
p[i, c] = p[i - 1, c];
}
else
{
d[i, c] = b;
p[i, c] = i - 1;
}
}
else
{
d[i, c] = a;
p[i, c] = p[i - 1, c];
}
}
}
}
//sum
//Console.WriteLine(d[n, target);
//restore set
var resultSet = new List<int>();
var m = n;
var s = d[n, target];
var t = p[m, s];
while (t != -1)
{
var item = val[t];
resultSet.Add(item);
m--;
s -= item;
t = p[m, s];
}
return resultSet;
}
}

It looks like the general problem you are trying solve is to find the subset of numbers with the largest sum less than target. The execution time of your linq function is a symptom of your solution. That is a well known and much researched problem called the 'knapsack problem'. I believe your specific variant would fall into the 'bounded knapsack problem' class with the weight being equal to the value. I would start by researching that. The solution you have implemented, brute forcing every possible subset, is known as the 'naive' solution. I am pretty sure it is the worst performing of all possible solutions.

More efficient way to get all indexes of a character in a string

Instead of looping through each character to see if it's the one you want then adding the index your on to a list like so:
var foundIndexes = new List<int>();
for (int i = 0; i < myStr.Length; i++)
{
if (myStr[i] == 'a')
foundIndexes.Add(i);
}

You can use String.IndexOf, see example below:
string s = "abcabcabcabcabc";
var foundIndexes = new List<int>();
long t1 = DateTime.Now.Ticks;
for (int i = s.IndexOf('a'); i > -1; i = s.IndexOf('a', i + 1))
{
// for loop end when i=-1 ('a' not found)
foundIndexes.Add(i);
}
long t2 = DateTime.Now.Ticks - t1; // read this value to see the run time

I use the following extension method to yield all results:
public static IEnumerable<int> AllIndexesOf(this string str, string searchstring)
{
int minIndex = str.IndexOf(searchstring);
while (minIndex != -1)
{
yield return minIndex;
minIndex = str.IndexOf(searchstring, minIndex + searchstring.Length);
}
}
usage:
IEnumerable<int> result = "foobar".AllIndexesOf("o"); // [1,2]
Side note to a edge case: This is a string approach which works for one or more characters. In case of "fooo".AllIndexesOf("oo") the result is just 1 https://dotnetfiddle.net/CPC7D2

How about
string xx = "The quick brown fox jumps over the lazy dog";
char search = 'f';
var result = xx.Select((b, i) => b.Equals(search) ? i : -1).Where(i => i != -1);

The raw iteration is always better & most optimized.
Unless it's a bit complex task, you never really need to seek for a better optimized solution...
So I would suggest to continue with :
var foundIndexes = new List<int>();
for (int i = 0; i < myStr.Length; i++)
if (myStr[i] == 'a') foundIndexes.Add(i);

If the string is short, it may be more efficient to search the string once and count up the number of times the character appears, then allocate an array of that size and search the string a second time, recording the indexes in the array. This will skip any list re-allocations.
What it comes down to is how long the string is and how many times the character appears. If the string is long and the character appears few times, searching it once and appending indicies to a List<int> will be faster. If the character appears many times, then searching the string twice (once to count, and once to fill an array) may be faster. Exactly where the tipping point is depends on many factors that can't be deduced from your question.
If you need to search the string for multiple different characters and get a list of indexes for those characters separately, it may be faster to search through the string once and build a Dictionary<char, List<int>> (or a List<List<int>> using character offsets from \0 as the indicies into the outer array).
Ultimately, you should benchmark your application to find bottlenecks. Often the code that we think will perform slowly is actually very fast, and we spend most of our time blocking on I/O or user input.

public static List<int> GetSubstringLocations(string text, string searchsequence)
{
try
{
List<int> foundIndexes = new List<int> { };
int i = 0;
while (i < text.Length)
{
int cindex = text.IndexOf(searchsequence, i);
if (cindex >= 0)
{
foundIndexes.Add(cindex);
i = cindex;
}
i++;
}
return foundIndexes;
}
catch (Exception ex) { }
return new List<int> { };
}

public static String[] Split(this string s,char c = '\t')
{
if (s == null) return null;
var a = new List<int>();
int i = s.IndexOf(c);
if (i < 0) return new string[] { s };
a.Add(i);
for (i = i+1; i < s.Length; i++) if (s[i] == c) a.Add(i);
var result = new string[a.Count +1];
int startIndex = 0;
result[0] = s.Remove(a[0]);
for(i=0;i<a.Count-1;i++)
{
result[i + 1] = s.Substring(a[i] + 1, a[i + 1] - a[i] - 1);
}
result[a.Count] = s.Substring(a[a.Count - 1] + 1);
return result;
}

What's a good way for figuring out all possible words of a given length

I'm trying to create an algorithm in C# which produces the following output strings:
AAAA
AAAB
AAAC
...and so on...
ZZZX
ZZZY
ZZZZ
What is the best way to accomplish this?
public static IEnumerable<string> GetWords()
{
//Perform algorithm
yield return word;
}

well, if the length is a constant 4, then this would handle it:
public static IEnumerable<String> GetWords()
{
for (Char c1 = 'A'; c1 <= 'Z'; c1++)
{
for (Char c2 = 'A'; c2 <= 'Z'; c2++)
{
for (Char c3 = 'A'; c3 <= 'Z'; c3++)
{
for (Char c4 = 'A'; c4 <= 'Z'; c4++)
{
yield return "" + c1 + c2 + c3 + c4;
}
}
}
}
}
if the length is a parameter, this recursive solution would handle it:
public static IEnumerable<String> GetWords(Int32 length)
{
if (length <= 0)
yield break;
for (Char c = 'A'; c <= 'Z'; c++)
{
if (length > 1)
{
foreach (String restWord in GetWords(length - 1))
yield return c + restWord;
}
else
yield return "" + c;
}
}

There's always the obligatory LINQ implementation. Most likely rubbish performance, but since when did performance get in the way of using cool new features?
var letters = "ABCDEFGHIJKLMNOPQRSTUVWXYZ".ToCharArray();
var sequence = from one in letters
from two in letters
from three in letters
from four in letters
orderby one, two, three, four
select new string(new[] { one, two, three, four });
'sequence' will now be an IQueryable that contains AAAA to ZZZZ.
Edit:
Ok, so it was bugging me that it should be possible to make a sequence of configurable length with a configurable alphabet using LINQ. So here it is. Again, completely pointless but it was bugging me.
public void Nonsense()
{
var letters = new[]{"A","B","C","D","E","F",
"G","H","I","J","K","L",
"M","N","O","P","Q","R","S",
"T","U","V","W","X","Y","Z"};
foreach (var val in Sequence(letters, 4))
Console.WriteLine(val);
}
private IQueryable<string> Sequence(string[] alphabet, int size)
{
// create the first level
var sequence = alphabet.AsQueryable();
// add each subsequent level
for (var i = 1; i < size; i++)
sequence = AddLevel(sequence, alphabet);
return from value in sequence
orderby value
select value;
}
private IQueryable<string> AddLevel(IQueryable<string> current, string[] characters)
{
return from one in current
from character in characters
select one + character;
}
The call to the Sequence method produces the same AAAA to ZZZZ list as before but now you can change the dictionary used and how long the produced words will be.

Just a coment to Garry Shutler, but I want code coloring:
You really don't need to make it IQuaryable, neither the sort, so you can remove the second method. One step forwad is to use Aggregate for the cross product, it end up like this:
IEnumerable<string> letters = new[]{
"A","B","C","D","E","F",
"G","H","I","J","K","L",
"M","N","O","P","Q","R","S",
"T","U","V","W","X","Y","Z"};
var result = Enumerable.Range(0, 4)
.Aggregate(letters, (curr, i) => curr.SelectMany(s => letters, (s, c) => s + c));
foreach (var val in result)
Console.WriteLine(val);
Anders should get a Nobel prize for the Linq thing!

GNU Bash!
{a..z}{a..z}{a..z}{a..z}

Python!
(This is only a hack, dont' take me too seriously :-)
# Convert a number to the base 26 using [A-Z] as the cyphers
def itoa26(n):
array = []
while n:
lowestDigit = n % 26
array.append(chr(lowestDigit + ord('A')))
n /= 26
array.reverse()
return ''.join(array)
def generateSequences(nChars):
for n in xrange(26**nChars):
string = itoa26(n)
yield 'A'*(nChars - len(string)) + string
for string in generateSequences(3):
print string

Inspired by Garry Shutler's answer, I decided to recode his answer in T-SQL.
Say "Letters" is a table with only one field, MyChar, a CHAR(1). It has 26 rows, each an alphabet's letter. So we'd have (you can copy-paste this code on SQL Server and run as-is to see it in action):
DECLARE #Letters TABLE (
MyChar CHAR(1) PRIMARY KEY
)
DECLARE #N INT
SET #N=0
WHILE #N<26 BEGIN
INSERT #Letters (MyChar) VALUES ( CHAR( #N + 65) )
SET #N = #N + 1
END
-- SELECT * FROM #Letters ORDER BY 1
SELECT A.MyChar, B.MyChar, C.MyChar, D.MyChar
FROM #Letters A, Letters B, Letters C, Letters D
ORDER BY 1,2,3,4
The advantages are: It's easily extensible into using capital/lowercase, or using non-English Latin characters (think "Ñ" or cedille, eszets and the like) and you'd still be getting an ordered set, only need to add a collation. Plus SQL Server will execute this slightly faster than LINQ on a single core machine, on multicore (or multiprocessors) the execution can be in parallel, getting even more boost.
Unfortunately, it's stuck for the 4 letters specific case. lassevk's recursive solution is more general, trying to do a general solution in T-SQL would necessarily imply dynamic SQL with all its dangers.

Haskell!
replicateM 4 ['A'..'Z']
Ruby!
('A'*4..'Z'*4).to_a

Simpler Python!
def getWords(length=3):
if length == 0: raise StopIteration
for letter in 'ABCDEFGHIJKLMNOPQRSTUVWXYZ':
if length == 1: yield letter
else:
for partialWord in getWords(length-1):
yield letter+partialWord

This is an recursive version of the same functions in C#:
using System;
using System.Collections.Generic;
using System.Text;
using System.IO;
namespace ConsoleApplication1Test
{
class Program
{
static char[] my_func( char[] my_chars, int level)
{
if (level > 1)
my_func(my_chars, level - 1);
my_chars[(my_chars.Length - level)]++;
if (my_chars[(my_chars.Length - level)] == ('Z' + 1))
{
my_chars[(my_chars.Length - level)] = 'A';
return my_chars;
}
else
{
Console.Out.WriteLine(my_chars);
return my_func(my_chars, level);
}
}
static void Main(string[] args)
{
char[] text = { 'A', 'A', 'A', 'A' };
my_func(text,text.Length);
Console.ReadKey();
}
}
}
Prints out from AAAA to ZZZZ

javascript!
var chars = 4, abc = "ABCDEFGHIJKLMNOPQRSTUVWXYZ", top = 1, fact = [];
for (i = 0; i < chars; i++) { fact.unshift(top); top *= abc.length; }
for (i = 0; i < top; i++)
{
for (j = 0; j < chars; j++)
document.write(abc[Math.floor(i/fact[j]) % abc.length]);
document.write("<br \>\n");
}

Use something which automatically Googles for every single letter combination, then see if there are more ".sz" or ".af" hits then ".com" hits at the five first results ... ;)
Seriously, what you're looking for might be Tries (data structure) though you still need to populate the thing which probably is far harder...

A very simple but awesome code that generates all words of 3 and 4 letters of english language
#include <iostream>
using namespace std;
char alpha[26]={'a','b','c','d','e','f','g','h','i','j','k','l','m','n','o','p','q','r','s','t','u','v','w','x','y','z'}
int main() {
int num;
cin >> num;
if (num == 3) { //all 3 letter words
for (int i = 0; i <= 25; i++) {
for (int o = 0; o <= 25; o++) {
for (int p = 0; p <= 25; p++) {
cout << alpha[i] << alpha[o] << alpha[p] << " ";
}
}
}
}
else if (num == 4) { //all 4 letter words
for (int i = 0; i <= 25; i++) {
for (int o = 0; o <= 25; o++) {
for (int p = 0; p <= 25; p++) {
for (int q = 0; q <= 25; q++) {
cout << alpha[i] << alpha[o] << alpha[p] << alpha[q] << " ";
}
}
}
}
}
else {
cout << "Not more than 4"; //it will take more than 2 hours for generating all 5 letter words
}
}