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How do I select multiple values in an array?

int[] numbers = new int[]
{
1, 2, 33, 44, 55, 68, 77, 96, 100
};
if (numbers[] % 2 == 0)
{
Console.WriteLine("EVEN");
}
I want to make the code say even when the value in the array is an even number.
How could I do this?

Your question is vague one:
...the code say even when the value in the array is an even number
You may want to query numbers (with a help of Linq):
If array contains at least one even value:
using System.Linq;
...
if (numbers.Any(item => item % 2 == 0))
Console.WriteLine("EVEN");
If array values are all even:
if (numbers.All(item => item % 2 == 0))
Console.WriteLine("EVEN");
Obtain even values (let's have an int[] even array):
int[] even = numbers
.Where(item => item % 2 == 0)
.ToArray();
if (even.Length > 0)
Console.WriteLine($"EVEN: {string.Join(", ", even)}");
Finally, you may want to scan the numbers array and do something on each even item:
foreach (int item in numbers) {
if (item % 2 == 0) {
Console.WriteLine("EVEN:");
}
}

You need loop
for (int i = 0; i < numbers.Length; i++)
{
if (numbers[i] % 2 == 0)
{
Console.WriteLine("EVEN");
}
}

I would do it:
var evenNumbers = numbers.Where(x => x % 2 == 0).ToArray();

Pick a varying number of item combinations from a List

Assume I have a list of integers of any length, for an example I have the list of 1,3,5 and 7.
I would like an algorithm to pick a combination of X elements from the list.
For example, X = 1 would return:
1
3
5
7
x = 2 would return:
1 + 1
1 + 3
1 + 5
1 + 7
3 + 3
3 + 5
3 + 7
5 + 5
5 + 7
7 + 7
var listOfInts = new List<int> { 1, 3, 5, 7 };
var combinedInts = new List<int>();
// x = 1 solution
// This is only picking one item from the list.
for (int i = 0; i < listOfInts.Count(); i++)
{
combinedInts.Add(listOfInts[i]);
}
// x = 2 solution
// This is how to pick two. I wrap it around another for loop.
for (int i = 0; i < listOfInts.Count(); i++)
{
for (int j = i; j < listOfInts.Count(); j++)
{
combinedInts.Add(listOfInts[i] + listOfInts[j]);
}
}
// x = 3 solution
// If I go up another level I have to wrap it around another for loop. This solution won't scale.
for (int i = 0; i < listOfInts.Count(); i++)
{
for (int j = i; j < listOfInts.Count(); j++)
{
for (int k = j; k < listOfInts.Count(); k++)
{
combinedInts.Add(listOfInts[i] + listOfInts[j] + listOfInts[k]);
}
}
}
This solution doesn't scale as I have to continually wrap around another for loop for each number of element I'm picking. For example X = 7 would need 7-nested for loops. Is there a better way to write this method that doesn't involve nesting for loops?

You can use the following to get combinations of the sequences:
public static class LinqHelper
{
public static IEnumerable<IEnumerable<T>> Combinations<T>(this IEnumerable<T> elements, int? k = null)
{
if (!k.HasValue)
k = elements.Count();
return k == 0 ? new[] { new T[0] } :
elements.SelectMany((e, i) => elements.Skip(i).Combinations(k - 1).Select(c => (new[] { e }).Concat(c)));
}
}
var list = new List<int> { 1, 3, 5, 7 };
int x = 2; //Change to 3, 4, 5, etc
var result = list.Combinations(x);
Yields:
1 1
1 3
1 5
1 7
3 3
3 5
3 7
5 7
7 7
To get the sum of each one, you'd aggregate the result:
var result = list.Combinations(x).Select(g => g.Aggregate((left, right) => left + right));
Which produces:
2
4
6
8
6
8
10
10
12
14

There is also a purely iterative way to do this. It requires a great deal more thought and complexity, but can be made very efficient. The basic idea is to simulate the same nested loops, but track the iterations of each nested loop as an array of loop counters, which are iterated forward in the same manner as the original nested loop code. Here is a fully working example:
var listOfInts = new List<int> { 1, 3, 5, 7 };
var combinedInts = new List<int>();
var numInts = listOfInts.Count;
var numElements = 5; // number of "nested loops", or ints selected in each combination
var loopCounters = new int[numElements]; // make one loop counter for each "nested loop"
var lastCounter = numElements - 1; // iterate the right-most counter by default
// maintain current sum in a variable for efficiency, since most of the time
// it is changing only by the value of one loop counter change.
var tempSum = listOfInts[0] * numElements;
// we are finished when the left/outer-most counter has looped past number of ints
while (loopCounters[0] < numInts) {
// you can use this to verify the output is iterating correctly:
// Console.WriteLine(string.Join(",", loopCounters.Select(x => listOfInts[x])) + ": " + loopCounters.Select(x => listOfInts[x]).Sum() + "; " + tempSum);
combinedInts.Add(tempSum);
tempSum -= listOfInts[loopCounters[lastCounter]];
loopCounters[lastCounter]++;
if (loopCounters[lastCounter] < numInts) tempSum += listOfInts[loopCounters[lastCounter]];
// if last element reached in inner-most counter, increment previous counter(s).
while (lastCounter > 0 && loopCounters[lastCounter] == numInts) {
lastCounter--;
tempSum -= listOfInts[loopCounters[lastCounter]];
loopCounters[lastCounter]++;
if (loopCounters[lastCounter] < numInts) tempSum += listOfInts[loopCounters[lastCounter]];
}
// if a previous counter was advanced, reset all future counters to same
// starting number to start iteration forward again.
while (lastCounter < numElements - 1) {
lastCounter++;
if (loopCounters[lastCounter] < numInts) tempSum -= listOfInts[loopCounters[lastCounter]];
loopCounters[lastCounter] = loopCounters[lastCounter - 1];
if (loopCounters[lastCounter] < numInts) tempSum += listOfInts[loopCounters[lastCounter]];
}
}
At the end of the iteration, combinedInts should contains a list of all sum combinations, similar to the original code or the other recursive solutions. If you are working with small sets, and small combinations, then this level of efficiency is unnecessary and you should prefer a recursive solution which is easier to reason about correctness. I present this as an alternative way to think about the problem. Cheers!

This works for me:
Func<IEnumerable<int>, int, IEnumerable<IEnumerable<int>>> generate = null;
generate = (xs, n) =>
(xs == null || !xs.Any())
? Enumerable.Empty<IEnumerable<int>>()
: n == 1
? xs.Select(x => new [] { x })
: xs.SelectMany(x => generate(xs, n - 1).Select(ys => ys.Concat(new [] { x })));
int[] array = { 1, 3, 5, 7, };
var results =
generate(array, 3)
.Select(xs => String.Join("+", xs));
With this call I get:
1+1+1, 3+1+1, 5+1+1, 7+1+1, 1+3+1, 3+3+1, 5+3+1, 7+3+1, 1+5+1, 3+5+1, 5+5+1, 7+5+1, 1+7+1, 3+7+1, 5+7+1, 7+7+1, 1+1+3, 3+1+3, 5+1+3, 7+1+3, 1+3+3, 3+3+3, 5+3+3, 7+3+3, 1+5+3, 3+5+3, 5+5+3, 7+5+3, 1+7+3, 3+7+3, 5+7+3, 7+7+3, 1+1+5, 3+1+5, 5+1+5, 7+1+5, 1+3+5, 3+3+5, 5+3+5, 7+3+5, 1+5+5, 3+5+5, 5+5+5, 7+5+5, 1+7+5, 3+7+5, 5+7+5, 7+7+5, 1+1+7, 3+1+7, 5+1+7, 7+1+7, 1+3+7, 3+3+7, 5+3+7, 7+3+7, 1+5+7, 3+5+7, 5+5+7, 7+5+7, 1+7+7, 3+7+7, 5+7+7,7+7+7

Find numbers in List<int> with specified difference

For a given a space separated list of numbers, what is the most effecient way of counting the total pairs of numbers which have a difference of N.
e.g. command line in put would be:
5 2
where 5 is the count of numbers to follow and 2 is the difference required
1 5 3 4 2
the 5 numbers to be considered
Output should be
3
because (5,3), (4,2) and (3,1) all have a diff of 2
I can get this algorithm to work, but is there a more efficient way of doing this if you have large sets of numbers to work with? I have incluced three comparison options and the second one should be better than the third but is there something I'm forgetting which could make it much quicker?
private static void Difference()
{
string[] firstInput = SplitInput(Console.ReadLine());
int numberOfNumbers = int.Parse(firstInput[0]);
int diffOfNumbers = int.Parse(firstInput[1]);
string[] secondInput = SplitInput(Console.ReadLine());
List<int> numbers = secondInput.Select(x => Int32.Parse(x)).ToList();
int possibleCombinations = 0;
// Option 1
foreach (int firstNumber in numbers)
{
List<int> compareTo = numbers.GetRange(numbers.IndexOf(firstNumber) + 1, numbers.Count - numbers.IndexOf(firstNumber) - 1);
foreach (int secondNumber in compareTo)
{
int diff = firstNumber - secondNumber;
if (Math.Abs(diff) == diffOfNumbers)
{
possibleCombinations++;
}
}
}
// Option 2
foreach (int firstNumber in numbers)
{
if (numbers.Contains(firstNumber + diffOfNumbers))
{
possibleCombinations++;
}
}
// Option 3
foreach (int firstNumber in numbers)
{
foreach (int secondNumber in numbers)
{
int diff = firstNumber - secondNumber;
if(Math.Abs(diff) == diffOfNumbers)
{
possibleOptions++;
}
}
}
Console.WriteLine(string.Format("Possible number of options are: {0}", possibleCombinations));
Console.ReadLine();
}
private static string[] SplitInput(string input)
{
return input.Split(new char[1] { ' ' }, StringSplitOptions.RemoveEmptyEntries);
}

If duplicate numbers are not allowed or to be ignored (only count unique pairs), you could use a HashSet<int>:
HashSet<int> myHashSet = ...
int difference = ...
int count;
foreach (int number in myHashSet)
{
int counterpart = number - difference;
if (myHashSet.Contains(counterpart))
{
count++;
}
}

Given the constraints of the problem, where N is the "count of numbers to follow" [1..N], and M is the difference (N=5 and M=2 in the example), why not just return N - M ?

This is done easily with LINQ, allowing for duplicates:
var dict = numbers.GroupBy(n => n).ToDictionary(g => g.Key, g => g.Count());
return dict.Keys.Where(n => dict.ContainsKey(difference-n)).Select(n => dict[difference - n]).Sum();
In the first line we create a dictionary where the keys are the distinct numbers in the input list (numbers) and the values are how many times they appear.
In the second, for each distinct number in the list (equivalent to the keys of the dictioanry) we look to see if the dictionary contains a key for the target number. If so, we add the number of times that target number appeared, which we previously stored as the value for that key. If not we add 0. Finally we sum it all up.
Note in theory this could cause arithmetic overflows if there's no bound other than Int.MinValue and Int.MaxValue on the items in the list. To get around this we need to do a "safe" check, which first makes sure that the difference won't be out of bounds before we try to calculate it. That might look like:
int SafeGetCount(int difference, int number, Dictionary<int,int> dict)
{
if(difference < 0 && number < 0 && int.MinValue - difference > number)
return 0;
if(difference > 0 && number > 0 && int.MaxValue - difference < number)
return 0;
return dict.ContainsKey(difference-number) ? dict[difference - number] : 0;
}
Update
There are a couple of things note entirely clear from your question, like whether you actually want to count duplicate pairs multiple times, and does swapping the numbers count as two different pairs. e.g. if (1,4) is a pair, is (4,1)? My answer above assumes that the answer to both of those questions is yes.
If you don't want to count duplicate pairs multiple times, then go with the HashSet solution from other answers. If you do want to count duplicate pairs but don't want to count twice by swapping the values in the pair, you have to get slightly more complex. E.g.:
var dict = numbers.GroupBy(n => n).ToDictionary(g => g.Key, g => g.Count());
var sum = dict.Keys.Where(n => n*2 != difference)
.Where(n => dict.ContainsKey(difference-n))
.Select(n => dict[difference - n]).Sum()/2;
if(n%2 == 0)
{
sum += dict.ContainsKey(n/2) ? dict[n/2] : 0
}
return sum;

how about sorting the list then iterating over it.
int PairsWithMatchingDifferenceCount(
IEnumerable<int> source,
int difference)
{
var ordered = source.OrderBy(i => i).ToList();
var count = ordered.Count;
var result = 0;
for (var i = 0; i < count - 1; i++)
{
for (var j = i + 1; j < count; j++)
{
var d = Math.Abs(ordered[j] - ordered[i]);
if (d == difference)
{
result++;
}
else if (d > difference)
{
break;
}
}
}
return result;
}
so, as per the example you would call it like this,
PairsWithMatchingDifferenceCount(Enumerable.Range(1, 5), 2);
but, if the sequence generation is a simple as the question suggests why not just.
var m = 5;
var n = 2;
var result = Enumerable.Range(n + 1, m - n)
.Select(x => Tuple.Create(x, x - n)).Count();
or indeed,
var result = m - n;

LINQ get x amount of elements from a list

I have a query which I get as:
var query = Data.Items
.Where(x => criteria.IsMatch(x))
.ToList<Item>();
This works fine.
However now I want to break up this list into x number of lists, for example 3. Each list will therefore contain 1/3 the amount of elements from query.
Can it be done using LINQ?

You can use PLINQ partitioners to break the results into separate enumerables.
var partitioner = Partitioner.Create<Item>(query);
var partitions = partitioner.GetPartitions(3);
You'll need to reference the System.Collections.Concurrent namespace. partitions will be a list of IEnumerable<Item> where each enumerable returns a portion of the query.

I think something like this could work, splitting the list into IGroupings.
const int numberOfGroups = 3;
var groups = query
.Select((item, i) => new { item, i })
.GroupBy(e => e.i % numberOfGroups);

You can use Skip and Take in a simple for to accomplish what you want
var groupSize = (int)Math.Ceiling(query.Count() / 3d);
var result = new List<List<Item>>();
for (var j = 0; j < 3; j++)
result.Add(query.Skip(j * groupSize).Take(groupSize).ToList());

If the order of the elements doesn't matter using an IGrouping as suggested by Daniel Imms is probably the most elegant way (add .Select(gr => gr.Select(e => e.item)) to get an IEnumerable<IEnumerable<T>>).
If however you want to preserve the order you need to know the total number of elements. Otherwise you wouldn't know when to start the next group. You can do this with LINQ but it requires two enumerations: one for counting and another for returning the data (as suggested by Esteban Elverdin).
If enumerating the query is expensive you can avoid the second enumeration by turning the query into a list and then use the GetRange method:
public static IEnumerable<List<T>> SplitList<T>(List<T> list, int numberOfRanges)
{
int sizeOfRanges = list.Count / numberOfRanges;
int remainder = list.Count % numberOfRanges;
int startIndex = 0;
for (int i = 0; i < numberOfRanges; i++)
{
int size = sizeOfRanges + (remainder > 0 ? 1 : 0);
yield return list.GetRange(startIndex, size);
if (remainder > 0)
{
remainder--;
}
startIndex += size;
}
}
static void Main()
{
List<int> list = Enumerable.Range(0, 10).ToList();
IEnumerable<List<int>> result = SplitList(list, 3);
foreach (List<int> values in result)
{
string s = string.Join(", ", values);
Console.WriteLine("{{ {0} }}", s);
}
}
The output is:
{ 0, 1, 2, 3 }
{ 4, 5, 6 }
{ 7, 8, 9 }

You can create an extension method:
public static IList<List<T>> GetChunks<T>(this IList<T> items, int numOfChunks)
{
if (items.Count < numOfChunks)
throw new ArgumentException("The number of elements is lower than the number of chunks");
int div = items.Count / numOfChunks;
int rem = items.Count % numOfChunks;
var listOfLists = new List<T>[numOfChunks];
for (int i = 0; i < numOfChunks; i++)
listOfLists[i] = new List<T>();
int currentGrp = 0;
int currRemainder = rem;
foreach (var el in items)
{
int currentElementsInGrp = listOfLists[currentGrp].Count;
if (currentElementsInGrp == div && currRemainder > 0)
{
currRemainder--;
}
else if (currentElementsInGrp >= div)
{
currentGrp++;
}
listOfLists[currentGrp].Add(el);
}
return listOfLists;
}
then use it like this :
var chunks = query.GetChunks(3);
N.B.
in case of number of elements not divisible by the number of groups, the first groups will be bigger. e.g. [0,1,2,3,4] --> [0,1] - [2,3] - [4]

List collection that replaces numbers with words

I am trying to improve a list collection that i have to replace values that are divisible by two and 10 and replace everything that is divisible by two with dTwo and ten with dTen?
My code works with one divisible statment but not two.
var num = new List<string>();
for (int n = 0; n < 101; n++)
{
num.Add(n % 2 == 0 ? "dTwo" : n.ToString());
num.Add(n % 10 == 0 ? "dTen" : n.ToString());
}

Since any number that is divisible by 10 is also divisible by 2 you have to switch your addition statements, and continue with the next number if you have a number divisible by 10:
var num = new List<string>();
for (int n = 0; n < 101; n++)
{
if( n % 10 == 0)
{
num.Add("dTen");
}
else num.Add(n % 2 == 0 ? "dTwo" : n.ToString());
}

If I can I try avoid using loop controls out side of the defined construct of the actual loop, ie. I prefer to avoid using continue if I can, it sort of feels like using goto statements. For this case, I would go for the plain and simple approach which I believe is readable, maintainable and simple albeit a little more verbose.
You can switch the order of the if/else if statements to change the priority if required, in this case the n % 10 has priority
var num = new List<string>();
for (int n = 0; n < 101; ++n)
{
if (n % 10 == 0)
{
num.Add("dTen");
}
else if (n % 2 == 0)
{
num.Add("dTwo");
}
else
{
num.Add(n.ToString());
}
}

There are two approaches I would take here, the first is verbose, but conveys what you're trying to do in a very readable manner:
var num = new List<string>(101);
for (int i = 0; i < 101 ; i++)
{
if (i == 0)
{
num.Add(i.ToString());
}
else if (i % 10 == 0)
{
num.Add("dTen");
}
else if (i % 2 == 0)
{
num.Add("dTwo");
}
else
{
num.Add(i.ToString());
}
}
The second uses a more concise LINQ-y type approach, like this.
var num = Enumerable.Range(0, 101)
.Select(
n => n == 0 ? n.ToString() :
n % 10 == 0 ? "dTen" :
n % 2 == 0 ? "dTwo" :
n.ToString())
.ToList();
Note that I've also taken into account the 0 edge case, where 0 would otherwise get reported as being divisible by 10.
Which one you go for is largely up to your taste. Personally I'd go for the latter implementation, as it's concise but still conveys the intent of the code. Some very rudimentary tests I've just done shows that it'll execute faster as well.