Develop Reference

How to get every other character in a string in c#

so I'm working on this problem: https://www.hackerrank.com/challenges/30-review-loop/problem (it's in C#)
and so far I'm just trying to break it down piece by piece, and so far I was able to get it to show every other character, but I'm not sure how to concatenate each letter into a new string.
My code for the problem is as follows I've commented out the two for loops, because I felt like there was a more elegant solution to this than what I had, but I didn't want to lose where I was in case another path proved to be more challenging.
using System;
using System.Collections.Generic;
using System.IO;
class Solution {
static void Main(String[] args) {
/* Enter your code here. Read input from STDIN. Print output to STDOUT. Your class should be named Solution */
int inputQTY = Int32.Parse(Console.ReadLine());
string input = Console.ReadLine(); // The example gives us the first word to be Hacker then the next word Rank on the next line, so the outputs would be Hce akr, and Rn ak respectively.
int strLen = input.Length;
char[] inputCharArray = input.ToCharArray();
string output = "";
/*
for (int j = 0; j < inputQTY; j++){
for (int i = 0; i < strLen; i++) {
if (j % 2 == 0 && i % 2 == 0) {
Console.WriteLine(inputCharArray[i]);
output = new string (new char[] {inputCharArray[i]});
Console.WriteLine(output);
Console.WriteLine("This is i: {0}", i);
Console.WriteLine("This is j: {0}", j);
Console.WriteLine("--------------");
Console.WriteLine("");
}
else {
Console.WriteLine("This is the next j part hopefully: {0}", j);
}
}
}*/
}
}
Like I understand I need to first step through the word, grab every other letter, then step through the word again and grab the remaining letters, then concatenate those letters into words, and concatenate those words into a sentence, so j would be the loop giving me the two words where I is the loop getting the two words put together..... but I'm having a difficult time wrapping my head around where I'm going wrong here. On top of this, I feel like there's another approach entirely that I'm missing, using commands I may not even know about.
Anyhoo any help is appreciated, hopefully I won't be so green after this. Thanks!
Ok so I ended up solving it with the following code, thank you for all your help everyone!
I ended up solving it with the following code (in C#):
using System;
using System.Collections.Generic;
using System.IO;
using System.Text;
class Solution {
static void Main(String[] args) {
/* Enter your code here. Read input from STDIN. Print output to STDOUT. Your class should be named Solution */
int count = Int32.Parse(Console.ReadLine());
for (int k = 0; k < count; k++) {
char[] word = Console.ReadLine().ToCharArray();
StringBuilder sb1 = new StringBuilder();
StringBuilder sb2 = new StringBuilder();
for (int i = 0; i < word.Length; i+=2) {
sb1.Append(word[i]);
}
for (int j = 1; j < word.Length; j+=2) {
sb2.Append(word[j]);
}
Console.WriteLine(sb1 + " " + sb2);
}
}
}

LINQ version, updated to fix index error:
output = $"{new string(s.Where((x,i) => i % 2 == 0).ToArray())} {new string(s.Where((x,i) => i % 2 != 0).ToArray())}";
To explain, you're grabbing every character whose index in the string is evenly divisible by 2 and printing it, then every character in the string whose index is not evenly divisible by 2 and printing it.
Update:
Since I was asked for further explanation. First, here's the full code that runs successfully in the HackerRank challenge:
using System;
using System.Collections.Generic;
using System.Linq;
class Solution
{
static void Main(String[] args)
{
List<string> tests = new List<string>();
var testCount = int.Parse(Console.ReadLine());
for (var i = 0; i < testCount; i++)
{
tests.Add(Console.ReadLine());
}
foreach (var s in tests)
{
Console.WriteLine($"{new string(s.Where((x, i) => i % 2 == 0).ToArray())} {new string(s.Where((x, i) => i % 2 != 0).ToArray())}");
}
}
}
Regarding what each section of the code does:
i % 2 == 0
This is a test to see if a number is evenly divisible by two, or an even number.
s.Where((x,i) => i % 2 == 0)
This says, for the array of characters that make up the string 's', return all characters (the result is an IEnumerable) where that character's index (location in the string) is an even number.
new string(s.Where((x,i) => i % 2 == 0).ToArray())
This says to take that IEnumerable of characters with even numbered indexes and return them to an Array of characters. Then, create a new string out of that array of characters.
For the odd numbers, it's the same, but you use != 0 in the mod.

I used this simple method of appending to two StringBuilder objects
var sb1 = new StringBuilder();
var sb2 = new StringBuilder();
int i = 0;
foreach (char c in input)
{
var sb = (i % 2 == 0 ? sb1 : sb2);
sb.Append(c);
i = i + 1;
}
output = sb1.ToString() + " " + sb2.ToString();

this is the long way..
int count = int.Parse(Console.ReadLine());
for(int k = 0; k < count; k++){
char[] inputChars = Console.ReadLine().ToCharArray();
char[] evenChars = new char[inputChars.Length % 2 == 0 ? inputChars.Length / 2 : (inputChars.Length + 1) / 2];
char[] oddChars = new char[inputChars.Length - evenChars.Length];
int evenIndex=0,oddIndex = 0;
for(int i = 0; i < inputChars.Length;i++)
if(i % 2 == 0)
evenChars[evenIndex++] = inputChars[i];
else
oddChars[oddIndex++] = inputChars[i];
Console.WriteLine(string.Format("{0} {1}",string.Concat(evenChars),string.Concat(oddChars)));
}
an alternative..
int count = int.Parse(Console.ReadLine());
for(int k = 0; k < count; k++){
string input = Console.ReadLine();
Enumerable.Range(0, input.Length)
.OrderBy(o => o % 2 != 0)
.Select(o => {
if(o == 1)
Console.Write(" ");
Console.Write(input[o]);
return input[o];
}).ToArray();
Console.Write("\n");
}

Appending to StringBuilder while there is nothing left

I have to the task to rearrange the words in a sentence backwards, but i am able to do it only for the first letter.Example: Fun exam right.What i have until now:
var sentance = Console.Readline().Split(' ');
var rearrangedSentence = new StringBuilder();
for(int i = 0,i<sentance.Lenght,i++)
{
rearrangedSentence.Append(sentance[i].Last());//this gives me "nmt"
}
My question is how to make this loop repeat itself while there is nothing left.
Any help will be greatly appriciated :)
EDIT: Question is
I mean if i have the sentence "Fun exam right" the result should be :nmtuahFxgeir . We first take the last chars of each word append that results in "nmt" then take the next one and add them resulting in "nmtuah" and so on

When you use sentance[i].Last(), you are only picking up the last element of your array.
EDIT: As per your updated requirements, you can use this code.
//Get the sentence array
var sentence = Console.ReadLine().Split(' ');
var rearrangedSentence = new StringBuilder();
//Get the length of longest word in array
int loopLength = sentence.OrderBy(n => n.Length).Last().Length;
int x = 0;
// Run for the length of longest word
for (int i = loopLength-1; i >=0 ; i--)
{
// need to pick up an element at every run for each element.
for (var j = 0; j < sentence.Length; j++)
{
//Picking the position of item to be picked up
int val = sentence[j].Length - (x + 1);
// If index not out of bounds
if(val >= 0 && val <= sentence[j].Length)
{
// Pick the character and append to stringbuilder.
rearrangedSentence.Append(sentence[j][val]);
}
}
// Next letter should be n-1, then n-2.
// Increase this. Val will decrease
x++;
}
Console.WriteLine(rearrangedSentence.ToString());
Console.ReadLine();

Find all possible combinations of word with and without hyphens

For a string that may have zero or more hyphens in it, I need to extract all the different possibilities with and without hyphens.
For example, the string "A-B" would result in "A-B" and "AB" (two possibilities).
The string "A-B-C" would result in "A-B-C", "AB-C", "A-BC" and "ABC" (four possibilities).
The string "A-B-C-D" would result in "A-B-C-D", "AB-C-D", "A-BC-D", "A-B-CD", "AB-CD", "ABC-D", "A-BCD" and "ABCD" (eight possibilities).
...etc, etc.
I've experimented with some nested loops but haven't been able to get anywhere near the desired result. I suspect I need something recursive unless there is some simple solution I am overlooking.
NB. This is to build a SQL query (shame that SQL Server does't have MySQL's REGEXP pattern matching).
Here is one attempt I was working on. This might work if I do this recursively.
string keyword = "A-B-C-D";
List<int> hyphens = new List<int>();
int pos = keyword.IndexOf('-');
while (pos != -1)
{
hyphens.Add(pos);
pos = keyword.IndexOf('-', pos + 1);
}
for (int i = 0; i < hyphens.Count(); i++)
{
string result = keyword.Substring(0, hyphens[i]) + keyword.Substring(hyphens[i] + 1);
Response.Write("<p>" + result);
}
A B C D are words of varying length.

Take a look at your sample cases. Have you noticed a pattern?
With 1 hyphen there are 2 possibilities.
With 2 hyphens there are 4 possibilities.
With 3 hyphens there are 8 possibilities.
The number of possibilities is 2n.
This is literally exponential growth, so if there are too many hyphens in the string, it will quickly become infeasible to print them all. (With just 30 hyphens there are over a billion combinations!)
That said, for smaller numbers of hyphens it might be interesting to generate a list. To do this, you can think of each hyphen as a bit in a binary number. If the bit is 1, the hyphen is present, otherwise it is not. So this suggests a fairly straightforward solution:
Split the original string on the hyphens
Let n = the number of hyphens
Count from 2n - 1 down to 0. Treat this counter as a bitmask.
For each count begin building a string starting with the first part.
Concatenate each of the remaining parts to the string in order, preceded by a hyphen only if the corresponding bit in the bitmask is set.
Add the resulting string to the output and continue until the counter is exhausted.
Translated to code we have:
public static IEnumerable<string> EnumerateHyphenatedStrings(string s)
{
string[] parts = s.Split('-');
int n = parts.Length - 1;
if (n > 30) throw new Exception("too many hyphens");
for (int m = (1 << n) - 1; m >= 0; m--)
{
StringBuilder sb = new StringBuilder(parts[0]);
for (int i = 1; i <= n; i++)
{
if ((m & (1 << (i - 1))) > 0) sb.Append('-');
sb.Append(parts[i]);
}
yield return sb.ToString();
}
}
Fiddle: https://dotnetfiddle.net/ne3N8f

You should be able to track each hyphen position, and basically say its either there or not there. Loop through all the combinations, and you got all your strings. I found the easiest way to track it was using a binary, since its easy to add those with Convert.ToInt32
I came up with this:
string keyword = "A-B-C-D";
string[] keywordSplit = keyword.Split('-');
int combinations = Convert.ToInt32(Math.Pow(2.0, keywordSplit.Length - 1.0));
List<string> results = new List<string>();
for (int j = 0; j < combinations; j++)
{
string result = "";
string hyphenAdded = Convert.ToString(j, 2).PadLeft(keywordSplit.Length - 1, '0');
// Generate string
for (int i = 0; i < keywordSplit.Length; i++)
{
result += keywordSplit[i] +
((i < keywordSplit.Length - 1) && (hyphenAdded[i].Equals('1')) ? "-" : "");
}
results.Add(result);
}

This works for me:
Func<IEnumerable<string>, IEnumerable<string>> expand = null;
expand = xs =>
{
if (xs != null && xs.Any())
{
var head = xs.First();
if (xs.Skip(1).Any())
{
return expand(xs.Skip(1)).SelectMany(tail => new []
{
head + tail,
head + "-" + tail
});
}
else
{
return new [] { head };
}
}
else
{
return Enumerable.Empty<string>();
}
};
var keyword = "A-B-C-D";
var parts = keyword.Split('-');
var results = expand(parts);
I get:
ABCD
A-BCD
AB-CD
A-B-CD
ABC-D
A-BC-D
AB-C-D
A-B-C-D

I've tested this code and it is working as specified in the question. I stored the strings in a List<string>.
string str = "AB-C-D-EF-G-HI";
string[] splitted = str.Split('-');
List<string> finalList = new List<string>();
string temp = "";
for (int i = 0; i < splitted.Length; i++)
{
temp += splitted[i];
}
finalList.Add(temp);
temp = "";
for (int diff = 0; diff < splitted.Length-1; diff++)
{
for (int start = 1, limit = start + diff; limit < splitted.Length; start++, limit++)
{
int i = 0;
while (i < start)
{
temp += splitted[i++];
}
while (i <= limit)
{
temp += "-";
temp += splitted[i++];
}
while (i < splitted.Length)
{
temp += splitted[i++];
}
finalList.Add(temp);
temp = "";
}
}

I'm not sure your question is entirely well defined (i.e. could you have something like A-BCD-EF-G-H?). For "fully" hyphenated strings (A-B-C-D-...-Z), something like this should do:
string toParse = "A-B-C-D";
char[] toParseChars = toPase.toCharArray();
string result = "";
string binary;
for(int i = 0; i < (int)Math.pow(2, toParse.Length/2); i++) { // Number of subsets of an n-elt set is 2^n
binary = Convert.ToString(i, 2);
while (binary.Length < toParse.Length/2) {
binary = "0" + binary;
}
char[] binChars = binary.ToCharArray();
for (int k = 0; k < binChars.Length; k++) {
result += toParseChars[k*2].ToString();
if (binChars[k] == '1') {
result += "-";
}
}
result += toParseChars[toParseChars.Length-1];
Console.WriteLine(result);
}
The idea here is that we want to create a binary word for each possible hyphen. So, if we have A-B-C-D (three hyphens), we create binary words 000, 001, 010, 011, 100, 101, 110, and 111. Note that if we have n hyphens, we need 2^n binary words.
Then each word maps to the output you desire by inserting the hyphen where we have a '1' in our word (000 -> ABCD, 001 -> ABC-D, 010 -> AB-CD, etc). I didn't test the code above, but this is at least one way to solve the problem for fully hyphenated words.
Disclaimer: I didn't actually test the code

Split a line of strings into strings consisting of two characters

I am reading in a text file which consists of a grid of alpha-numeric values (see below).
IQQQQQ
WG2223
S22228
D22223
Currently these values are looped through and each character sent to a switch case. This switch then reads that character and outputs a given result. The code for this process is as follows.
private void LoadLevel(Stream stream)
{
List<string> lines = new List<string>();
uint width;
using (StreamReader reader = new StreamReader(stream))
{
string line = reader.ReadLine();
width = (uint)line.Length;
while (line != null)
{
lines.Add(line);
line = reader.ReadLine();
}
}
tiles = new Tile[width, lines.Count];
for (int y = 0; y < Height; ++y)
{
for (int x = 0; x < Width; ++x)
{
char type = lines[y][x];
tiles[x, y] = LoadTile(type, x, y);
}
}
}
In this code I retrieve the text file and store each line in a list and then loop through each line and extract each character at a given point in the grid. Rather than extract a single character I would like to extract two characters at the same time and pass this to the LoadTile function.
As an example take the first line of the grid.
IQQQQQ
I would like to split this line into three strings each two characters long and then pass that to LoadTile and then continue looping through the remainder of the grid. However I do not know where to begin to attempt to achieve this task. Any help would be appreciated.

Well first off you'll propably want to change the signature of LoadTile from LoadTile(char, int, int) to LoadTile(string, int, int). Then change the calculation of the width to
width = line.Length/2;
Of course if any line has an odd number of characters you'll lose the last character. Additionally if any line after the first is shorter you'll have exceptions and if it is longer you'll lose the additional data.
Then you can loop through like this and take substrings.
for (int y = 0; y < Height; ++y)
{
for (int x = 0; x < Width; ++x)
{
string type = lines[y].Substring(x*2,2);
tiles[x, y] = LoadTile(type, x, y);
}
}

You can do this:
var str = "IQQQQQ";
var arr = Regex.Matches(str, "..").Cast<Match>().ToArray().Select(u => u.Value);
// arr[0] = "IQ"
// arr[1] = "QQ"
// arr[2] = "QQ"

This is how you get a substring:
var item = source.Substring(i * 2, 2);
You can split to array in one line:
var split = Enumerable.Range(0,3).Select(i=>source.Substring(i * 2, 2)).ToArray();
for (var i = 0; i < 3; i++)
{
var item = split[i];
// ...
}

Using LinQ. Try doing this for every line:
//Converts the string to a List of strings
List<string> stringArray = line.ToCharArray().Select(c=>c.ToString()).ToList();
//Iterates over the list and concats every pair of elements
List<string> result = stringArray.Select((value, index) => new { value, index })
.GroupBy(x => x.index / 2, x => x.value).Select(pair => pair.Aggregate((s1,s2) => s1 + s2)).ToList();

Create a new String with Duplicate Letters

I am trying to create a string that inserts a duplicate letter from the original into the modified. For example, the output of one run would be:
Original word:
stack
Output:
sstack, sttack, staack, stacck, stackk
Does that make sense? I have this so far, and I feel i am close, but I am suing the wrong method to reassemble the string. Any help would be appreciated:
// Use ToCharArray to convert string to array.
char[] array = originalWord.ToCharArray();
// Loop through array.
for (int i = 0; i < array.Length; i++)
{
// Get character from array.
char letter = array[i];
string result = array.ToString();
string result2 = string.Join("", result.Select(x => x + letter));
Console.Write(result2);
}

This should work:
var original = "stack";
for (int i = 0; i < original.Length; i++)
Console.WriteLine(original.Insert(i, original[i].ToString()));

You can use String.Insert to insert a string at a given index into another string.
IEnumerable<string> strings = originalWord
.Select((c, idx) => originalWord.Insert(idx, c.ToString()));

Fixed :
string originalWord = "stack";
// Use ToCharArray to convert string to array.
char[] array = originalWord.ToCharArray();
// Loop through array.
for (int i = 0; i < array.Length; i++)
{
// Get character from array.
char letter = array[i];
string result = originalWord.Insert(i, letter.ToString(CultureInfo.InvariantCulture));
Console.WriteLine(result);
}
The Linq way :
IEnumerable<string> words = originalWord.Select((letter, i) => originalWord.Insert(i, letter.ToString(CultureInfo.InvariantCulture)));

You can use String.Insert() method like;
string s = "stack";
for (int i = 0; i < s.Length; i++)
{
Console.WriteLine (s.Insert(i, s[i].ToString()));
}
Here is a DEMO.
Oh god, already aded 3 answers when I writing it. Damn..