Question: You are given an array (which will have a length of at least 3, but could be very large) containing integers. The array is either entirely comprised of odd integers or entirely comprised of even integers except for a single integer N. Write a method that takes the array as an argument and returns this "outlier" N.
Examples
[2, 4, 0, 100, 4, 11, 2602, 36]
Should return: 11 (the only odd number)
[160, 3, 1719, 19, 11, 13, -21]
Should return: 160 (the only even number)
Problem: I can find the target number. It is "6" in this instance. But when i try to write it, it is written 3 times. I did not understand why it is not written once.
namespace sayitoplamları
{
class Program
{
static void Main(string[] args)
{
int[] integers;
integers = new int[] {1,3,5,6};
int even = 0;
int odd = 0;
foreach (var num in integers)
{
if (num%2==0)
{
even += 1;
}
else
{
odd += 1;
}
if (even>1)
{
foreach (var item in integers)
{
if (item%2!=0)
{
Console.WriteLine(item);
}
}
}
else if (odd>1)
{
foreach (var item in integers)
{
if (item%2==0)
{
Console.WriteLine(item);
}
}
}
}
}
}
}```
Since we are doing this assignment for a good grade :)
We really don't need to count both odd and even elements and can just count odd once as it more convenient - just add up all remainders ( odd % 2 = 1, even % 2 = 0). If we would need number of even elements it would be array.Length - countOdd.
int countOdd = 0;
for (int i = 0; i < array.Length; i++) // or foreach
{
countOdd += array[i] % 2;
}
If you feel that trick with counting is too complicated and only needed for A+ mark on the assignment than variant of your original code is fine:
int countOdd = 0;
foreach (int element in array)
{
if (element % 2 == 1)
countOdd ++; // which is exactly countOdd += 1, just shorter.
}
Now for the second part - finding the outlier. We don't even need to know if it is Odd or Even but rather just reminder from "% 2". We now know how many odd element - which could be either 1 (than desired remainder is 1 - we are looking for odd element) or more than one (then desired remainder is 0 as we are looking for even one).
int desiredRemainder = countOdd == 1 ? 1 : 0;
Now all its left is to check which element has desired remainder and return from our method:
foreach (int element in array)
{
if (element % 2 == desiredRemainder)
{
return element;
}
}
Please note that assignment asks for "return" and not "print". Printing the value does not allow it to be used by the caller of the method unlike returning it. Using early return also saves some time to iterate the rest of array.
Complete code:
int FindOutlier(int[] array)
{
int desiredReminder = array.Sum(x => x % 2) == 1 ? 1 : 0;
return array.First(x => x % 2 == desiredReminder);
}
Alternative solution would be to iterate array only once - since we need to return first odd or first even number as soon as we figure out which one repeats similar how your tried with nested foreach. Notes since we are guaranteed that outlier present and is only one we don't need to try to remember first number - just current is ok.
int FindOutlier(int[] array)
{
int? odd = null; // you can use separate int + bool instead
int? even = null;
int countOdd = 0;
int countEven = 0;
foreach (int element in array)
{
bool isEven = element % 2 == 0;
if (isEven)
{
countEven++;
even = element;
}
else
{
countOdd++;
odd = element;
}
if (countEven > 1 && odd.HasValue)
return odd.Value;
if (countOdd > 1 && even.HasValue)
return even.Value;
}
throw new Exception("Value must be found before");
}
And if you ok to always iterate whole list code will be simple as again we'd need to only keep track of count of odd numbers (or even, just one kind) and return would not need to check if we found the value yet as we know that both type of numbers are present in the list:
int FindOutlier(int[] array)
{
int odd = 0;
int even = 0;
int countEven = 0;
foreach (int element in array)
{
if (element % 2 == 0)
{
countEven++;
even= element;
}
else
{
odd = element;
}
}
return countEven > 1 ? odd : even;
}
You are mixing everything into a single algorithm. In most cases it's a better idea to have separate steps. In your case, the steps could be:
Count odds and evens
Find the outlier
Print the result
That way you can make sure that the output does not interfere with the loops.
class Program
{
static void Main(string[] args)
{
int[] integers;
integers = new int[] {1,3,5,6};
int even = 0;
int odd = 0;
// Count odds and evens
foreach (var num in integers)
{
if (num%2==0)
{
even += 1;
}
else
{
odd += 1;
}
}
// Find the outlier
var outlier = 0;
foreach (var item in integers)
{
if (even == 1 && item%2==0)
{
outlier = item;
break;
}
if (odd == 1 && item%2!=0)
{
outlier = item;
break;
}
}
// Print the result
Console.WriteLine(outlier);
}
}
You could even make this separate methods. It also makes sense due to SoC (separation of concerns) and if you want to achieve testability (unit tests).
You want a method anyway and use the return statement, as mentioned in the assignment
Write a method that takes the array as an argument and returns this "outlier" N.
Maybe you want to check the following code, which uses separate methods:
class Program
{
static void Main(string[] args)
{
int[] integers;
integers = new int[] { 1, 3, 5, 6 };
var outlier = FindOutlier(integers);
Console.WriteLine(outlier);
}
private static int FindOutlier(int[] integers)
{
if (integers.Length < 3) throw new ArgumentException("Need at least 3 elements.");
bool isEvenOutlier = IsEvenOutlier(integers);
var outlier = FindOutlier(integers, isEvenOutlier ? 0:1);
return outlier;
}
static bool IsEvenOutlier(int[] integers)
{
// Count odds and evens
int even = 0;
int odd = 0;
foreach (var num in integers)
{
if (num % 2 == 0)
{
even += 1;
}
else
{
odd += 1;
}
}
// Check which one occurs only once
if (even == 1) return true;
if (odd == 1) return false;
throw new ArgumentException("No outlier in data.", nameof(integers));
}
private static int FindOutlier(int[] integers, int remainder)
{
foreach (var item in integers)
{
if (item % 2 == remainder)
{
return item;
}
}
throw new ArgumentException("No outlier in argument.", nameof(integers));
}
}
The reason it is printing '6' 3 times is because you iterate over the inetegers inside a loop where you iterate over the integers. Once with num and again with item.
Consider what happens when you run it against [1,3,5,6].
num=1, you get odd=1 and even=0. Neither of the if conditions are true so we move to the next iteration.
num=3, you get odd=2 and even=0. odd > 1 so we iterate over the integers (with item) again and print '6' when item=6.
num=5, you get odd=3 and even=0. Again, this will print '6' when item=6.
num=6, you get odd=3 and even=1. Even though we incremented the even count, we still have odd > 3 so will print '6' when item=6.
It is best to iterate over the values once and store the last even and last odd numbers along with the count (it is more performant as well).
You can then print the last odd number if even > 0 (as there will only be one odd number, which will be the last one) or the last even number if odd > 0.
I.e.
namespace sayitoplamları
{
class Program
{
static void Main(string[] args)
{
int[] integers;
integers = new int[] { 1, 3, 5, 6 };
int even = 0;
int? lastEven = null;
int odd = 0;
int? lastOdd = null;
foreach (var num in integers)
{
if (num % 2 == 0)
{
even += 1;
lastEven = num;
}
else
{
odd += 1;
lastOdd = num;
}
}
if (even > 1)
{
Console.WriteLine(lastOdd);
}
else if (odd > 1)
{
Console.WriteLine(lastEven);
}
}
}
}
because you didn't close the the first foreach loop before printing...
namespace sayitoplamları
{
class Program
{
static void Main(string[] args)
{
int[] integers;
integers = new int[] {1,3,5,6};
int even = 0;
int odd = 0;
foreach (var num in integers)
{
if (num%2==0)
{
even += 1;
}
else
{
odd += 1;
}//you need to close it here
if (even>1)
{
foreach (var item in integers)
{
if (item%2!=0)
{
Console.WriteLine(item);
}
}
}
else if (odd>1)
{
foreach (var item in integers)
{
if (item%2==0)
{
Console.WriteLine(item);
}
}
}
}
}//not here
}
}
Related
I am trying to join the next array element together to single element from array if the element of the array is less than length 4. It should add to the next element index.
Another logic is, if the next consecutive array length is also less then 4 char then it joins the next array element also up to 3 times in total. I want to implement this. It's getting complex for me I am not understand this logic.
This is the code, here it has array on titleList, now we have to use the above logic in this array list.
#foreach (var title in randomtitle)
{
</span>#titleList.ElementAt(title)</span>
}
#code {
[Parameter]
public string theTitle { get; set; }
private string[] titleList = Array.Empty<string>();
protected override void OnParametersSet()
{
if (!string.IsNullOrEmpty(theTitle))
{
titleList = theTitle.Split(" ");
}
}
private Random random = new();
private IEnumerable<int> randomtitle =>
Enumerable.Range(0, titleList.Count() - 1) // { 0, 1, 2, 3 } generate sequence
.OrderBy(x => random.Next()) // { 3, 1, 0, 2 } random shuffle
.Take(2) // { 3, 1 } pick two
.ToList();
}
I think you are looking for a method that does following:
take a collection of strings(like a string[])
iterate each string and check if it's length is greater than or equal 4
if so, everything is fine, take it as it is
if not, append the next to the current string and check their length afterwards
if they are still not greater than or equal 4 append the next one, but max 3 times
Then this should work (not tested well though):
Demo: https://dotnetfiddle.net/2uHREv
static string[] MergeItems(IList<string> all, int minLength, int maxGroupCount)
{
List<string> list = new List<string>(all.Count);
for(int i = 0; i < all.Count; i++)
{
string current = all[i];
if(i == all.Count - 1)
{
list.Add(current);
break;
}
if (current.Length < minLength)
{
for (int ii = 1; ii < maxGroupCount && i + ii < all.Count; ii++)
{
int nextIndex = i + ii;
string next = all[nextIndex];
current = current + next;
if (current.Length >= minLength || ii+1 == maxGroupCount)
{
list.Add(current);
i = nextIndex;
break;
}
}
}
else
{
list.Add(current);
}
}
return list.ToArray();
}
I've been going though www.testdome.com to test my skills and opened a list of public questions. One of the practice questions was:
Implement function CountNumbers that accepts a sorted array of
integers and counts the number of array elements that are less than
the parameter lessThan.
For example, SortedSearch.CountNumbers(new int[] { 1, 3, 5, 7 }, 4)
should return 2 because there are two array elements less than 4.
And my answer was:
using System;
public class SortedSearch
{
public static int CountNumbers(int[] sortedArray, int lessThan)
{
int count = 0;
int l = sortedArray.Length;
for (int i = 0; i < l; i++) {
if (sortedArray [i] < lessThan)
count++;
}
return count;
}
public static void Main(string[] args)
{
Console.WriteLine(SortedSearch.CountNumbers(new int[] { 1, 3, 5, 7 }, 4));
}
}
It seems that I've failed on two counts:
Performance test when sortedArray contains lessThan: Time limit exceeded
and
Performance test when sortedArray doesn't contain lessThan: Time limit exceeded
To be honest I'm not sure what to optimize there? Maybe I'm using a wrong method and there is a similar way to speed up the calculation?
If someone could point out my mistake or explain what I'm going wrong, I'd really appreciate it!
Because the array is sorted, you can stop counting as soon as you reach or exceed the lessThan parameter.
else break would probably do it.
Does it have to be really a loop? You could do Lambda exp for that
public static int CountNumbers(int[] sortedArray, int lessThan)
{
return sortedArray.ToList().Where(x=>x < lessThan).Count();
}
Harold's answer and approach is spot on.
Find below another code sample in case you're practicing for technical interviews. It handles cases when the array is null or empty, when lessThan is presented in the array (including duplicates), etc.
private static int CountNumbers(int[] sortedArray, int lessThan)
{
if (sortedArray == null)
{
throw new ArgumentNullException("Sorted array cannot be null.");
}
if (sortedArray.Length == 0)
{
throw new ArgumentException("Sorted array cannot be empty.");
}
int start = 0;
int end = sortedArray.Length;
int middle = int.MinValue;
while (start < end)
{
middle = (start + end) / 2;
if (sortedArray[middle] == lessThan)
{
break; // Found the "lessThan" number in the array, we can stop and move left
}
else if (sortedArray[middle] < lessThan)
{
start = middle + 1;
}
else
{
end = middle - 1;
}
}
// Adjust the middle pointer based on the "current" and "lessThan" numbers in the sorted array
while (middle >= 0 && sortedArray[middle] >= lessThan)
{
middle--;
}
// +1 because middle is calculated through 0-based (e.g. start)
return middle + 1;
}
I am trying to output the odd and even numbers of an array, i have got it all working but for some reason when i create a foreach loop and output each number that is odd/even it starts with a zero?
The way it is setup that the user inputs 10 numbers with a comma in between each letter (Starts as a string) and then it removes the comma and converts the numbers into an int and places it in an int array, to then be used in the other class. I input 1,2,3,4,5,6,7,8,9,1 to test it.
Odds(ai.Odds);
Evens(ai.Evens);
Console.ReadKey();
}
public static void Odds(int[] odds)
{
Console.WriteLine("\nOdds:");
foreach (var odd in odds)
{
Console.Write(odd + " ");
}
}
public static void Evens(int[] evens)
{
Console.WriteLine("\nEvens:");
foreach (var even in evens)
{
Console.Write(even + " ");
}
}
and in the separate class doing the magic:
public int[] Odds
{
get
{
int count = 0;
int[] odds = new int[NumArray.Length];
string[] oddNums = {"1", "3", "5", "7", "9"};
foreach (int num in NumArray)
{
foreach (string oddNum in oddNums)
{
if (num.ToString().EndsWith(oddNum))
{
odds[count++] = num;
}
}
}
Array.Sort(odds);
return RemoveDuplicates(odds);
}
}
public int[] Evens
{
get
{
int count = 0;
int[] evens = new int[NumArray.Length];
string[] evenNum = {"0", "2", "4", "6", "8"};
foreach (int num in NumArray)
{
foreach (string oddNum in evenNum)
{
if (num.ToString().EndsWith(oddNum))
{
evens[count++] = num;
}
}
}
Array.Sort(evens);
return RemoveDuplicates(evens);
}
}
Here is the RemoveDuplicates Method:
private int[] RemoveDuplicates(int[] numbers)
{
if (numbers.Length == 0)
{
return numbers;
}
int[] uniques = new int[numbers.Length];
int previousVal = numbers[0];
uniques[0] = numbers[0];
int count = 1;
for (int i = 1; i < numbers.Length; i++)
{
if (numbers[i] == previousVal) continue;
previousVal = numbers[i];
uniques[count++] = numbers[i];
}
Array.Resize(ref uniques, count);
return uniques;
}
as an output i get(vvv), but i don't want the 0's at the beginning of them both, i don't want a 0 at all
but i don't want the 0's at the beginning of them both, i don't want a 0 at all
Then you shouldn't create an array with 0s in, which is what you're doing.
Before you call Array.Sort(odds), your array will have all the odd numbers from NumArray, followed by a bunch of 0 values (as many as there are even values in NumArray), because when you construct an array, it is automatically filled with the default value of the element type for each element. You'll only have no 0 values if the whole of NumArray is odd.
You only want as many elements as you need, so it would be better to use List<int>, and just add to that when you need to. You can call ToArray() to create an array at the end, if you really need to.
When you're assigning values to the evens and odds arrays, you're using [count++] as the index. This means that it's never assigning index 0 and it'll have a default value of 0 from when the array was initialised.
Length = input Long(can be 2550, 2880, 2568, etc)
List<long> = {618, 350, 308, 300, 250, 232, 200, 128}
The program takes a long value, for that particular long value we have to find the possible combination from the above list which when added give me a input result(same value can be used twice). There can be a difference of +/- 30.
Largest numbers have to be used most.
Ex:Length = 868
For this combinations can be
Combination 1 = 618 + 250
Combination 2 = 308 + 232 + 200 +128
Correct Combination would be Combination 1
But there should also be different combinations.
public static void Main(string[] args)
{
//subtotal list
List<int> totals = new List<int>(new int[] { 618, 350, 308, 300, 250, 232, 200, 128 });
// get matches
List<int[]> results = KnapSack.MatchTotal(2682, totals);
// print results
foreach (var result in results)
{
Console.WriteLine(string.Join(",", result));
}
Console.WriteLine("Done.");
}
internal static List<int[]> MatchTotal(int theTotal, List<int> subTotals)
{
List<int[]> results = new List<int[]>();
while (subTotals.Contains(theTotal))
{
results.Add(new int[1] { theTotal });
subTotals.Remove(theTotal);
}
if (subTotals.Count == 0)
return results;
subTotals.Sort();
double mostNegativeNumber = subTotals[0];
if (mostNegativeNumber > 0)
mostNegativeNumber = 0;
if (mostNegativeNumber == 0)
subTotals.RemoveAll(d => d > theTotal);
for (int choose = 0; choose <= subTotals.Count; choose++)
{
IEnumerable<IEnumerable<int>> combos = Combination.Combinations(subTotals.AsEnumerable(), choose);
results.AddRange(from combo in combos where combo.Sum() == theTotal select combo.ToArray());
}
return results;
}
public static class Combination
{
public static IEnumerable<IEnumerable<T>> Combinations<T>(this IEnumerable<T> elements, int choose)
{
return choose == 0 ?
new[] { new T[0] } :
elements.SelectMany((element, i) =>
elements.Skip(i + 1).Combinations(choose - 1).Select(combo => (new[] { element }).Concat(combo)));
}
}
I Have used the above code, can it be more simplified, Again here also i get unique values. A value can be used any number of times. But the largest number has to be given the most priority.
I have a validation to check whether the total of the sum is greater than the input value. The logic fails even there..
The algorithm you have shown assumes that the list is sorted in ascending order. If not, then you shall first have to sort the list in O(nlogn) time and then execute the algorithm.
Also, it assumes that you are only considering combinations of pairs and you exit on the first match.
If you want to find all combinations, then instead of "break", just output the combination and increment startIndex or decrement endIndex.
Moreover, you should check for ranges (targetSum - 30 to targetSum + 30) rather than just the exact value because the problem says that a margin of error is allowed.
This is the best solution according to me because its complexity is O(nlogn + n) including the sorting.
V4 - Recursive Method, using Stack structure instead of stack frames on thread
It works (tested in VS), but there could be some bugs remaining.
static int Threshold = 30;
private static Stack<long> RecursiveMethod(long target)
{
Stack<long> Combination = new Stack<long>(establishedValues.Count); //Can grow bigger, as big as (target / min(establishedValues)) values
Stack<int> Index = new Stack<int>(establishedValues.Count); //Can grow bigger
int lowerBound = 0;
int dimensionIndex = lowerBound;
long fail = -1 * Threshold;
while (true)
{
long thisVal = establishedValues[dimensionIndex];
dimensionIndex++;
long afterApplied = target - thisVal;
if (afterApplied < fail)
lowerBound = dimensionIndex;
else
{
target = afterApplied;
Combination.Push(thisVal);
if (target <= Threshold)
return Combination;
Index.Push(dimensionIndex);
dimensionIndex = lowerBound;
}
if (dimensionIndex >= establishedValues.Count)
{
if (Index.Count == 0)
return null; //No possible combinations
dimensionIndex = Index.Pop();
lowerBound = dimensionIndex;
target += Combination.Pop();
}
}
}
Maybe V3 - Suggestion for Ordered solution trying every combination
Although this isn't chosen as the answer for the related question, I believe this is a good approach - https://stackoverflow.com/a/17258033/887092(, otherwise you could try the chosen answer (although the output for that is only 2 items in set being summed, rather than up to n items)) - it will enumerate every option including multiples of the same value. V2 works but would be slightly less efficient than an ordered solution, as the same failing-attempt will likely be attempted multiple times.
V2 - Random Selection - Will be able to reuse the same number twice
I'm a fan of using random for "intelligence", allowing the computer to brute force the solution. It's also easy to distribute - as there is no state dependence between two threads trying at the same time for example.
static int Threshold = 30;
public static List<long> RandomMethod(long Target)
{
List<long> Combinations = new List<long>();
Random rnd = new Random();
//Assuming establishedValues is sorted
int LowerBound = 0;
long runningSum = Target;
while (true)
{
int newLowerBound = FindLowerBound(LowerBound, runningSum);
if (newLowerBound == -1)
{
//No more beneficial values to work with, reset
runningSum = Target;
Combinations.Clear();
LowerBound = 0;
continue;
}
LowerBound = newLowerBound;
int rIndex = rnd.Next(LowerBound, establishedValues.Count);
long val = establishedValues[rIndex];
runningSum -= val;
Combinations.Add(val);
if (Math.Abs(runningSum) <= 30)
return Combinations;
}
}
static int FindLowerBound(int currentLowerBound, long runningSum)
{
//Adjust lower bound, so we're not randomly trying a number that's too high
for (int i = currentLowerBound; i < establishedValues.Count; i++)
{
//Factor in the threshold, because an end aggregate which exceeds by 20 is better than underperforming by 21.
if ((establishedValues[i] - Threshold) < runningSum)
{
return i;
}
}
return -1;
}
V1 - Ordered selection - Will not be able to reuse the same number twice
Add this very handy extension function (uses a binary algorithm to find all combinations):
//Make sure you put this in a static class inside System namespace
public static IEnumerable<List<T>> EachCombination<T>(this List<T> allValues)
{
var collection = new List<List<T>>();
for (int counter = 0; counter < (1 << allValues.Count); ++counter)
{
List<T> combination = new List<T>();
for (int i = 0; i < allValues.Count; ++i)
{
if ((counter & (1 << i)) == 0)
combination.Add(allValues[i]);
}
if (combination.Count == 0)
continue;
yield return combination;
}
}
Use the function
static List<long> establishedValues = new List<long>() {618, 350, 308, 300, 250, 232, 200, 128, 180, 118, 155};
//Return is a list of the values which sum to equal the target. Null if not found.
List<long> FindFirstCombination(long target)
{
foreach (var combination in establishedValues.EachCombination())
{
//if (combination.Sum() == target)
if (Math.Abs(combination.Sum() - target) <= 30) //Plus or minus tolerance for difference
return combination;
}
return null; //Or you could throw an exception
}
Test the solution
var target = 858;
var result = FindFirstCombination(target);
bool success = (result != null && result.Sum() == target);
//TODO: for loop with random selection of numbers from the establishedValues, Sum and test through FindFirstCombination
Consider this List<string>
List<string> data = new List<string>();
data.Add("Text1");
data.Add("Text2");
data.Add("Text3");
data.Add("Text4");
The problem I had was: how can I get every combination of a subset of the list?
Kinda like this:
#Subset Dimension 4
Text1;Text2;Text3;Text4
#Subset Dimension 3
Text1;Text2;Text3;
Text1;Text2;Text4;
Text1;Text3;Text4;
Text2;Text3;Text4;
#Subset Dimension 2
Text1;Text2;
Text1;Text3;
Text1;Text4;
Text2;Text3;
Text2;Text4;
#Subset Dimension 1
Text1;
Text2;
Text3;
Text4;
I came up with a decent solution which a think is worth to share here.
Similar logic as Abaco's answer, different implementation....
foreach (var ss in data.SubSets_LB())
{
Console.WriteLine(String.Join("; ",ss));
}
public static class SO_EXTENSIONS
{
public static IEnumerable<IEnumerable<T>> SubSets_LB<T>(
this IEnumerable<T> enumerable)
{
List<T> list = enumerable.ToList();
ulong upper = (ulong)1 << list.Count;
for (ulong i = 0; i < upper; i++)
{
List<T> l = new List<T>(list.Count);
for (int j = 0; j < sizeof(ulong) * 8; j++)
{
if (((ulong)1 << j) >= upper) break;
if (((i >> j) & 1) == 1)
{
l.Add(list[j]);
}
}
yield return l;
}
}
}
I think, the answers in this question need some performance tests. I'll give it a go. It is community wiki, feel free to update it.
void PerfTest()
{
var list = Enumerable.Range(0, 21).ToList();
var t1 = GetDurationInMs(list.SubSets_LB);
var t2 = GetDurationInMs(list.SubSets_Jodrell2);
var t3 = GetDurationInMs(() => list.CalcCombinations(20));
Console.WriteLine("{0}\n{1}\n{2}", t1, t2, t3);
}
long GetDurationInMs(Func<IEnumerable<IEnumerable<int>>> fxn)
{
fxn(); //JIT???
var count = 0;
var sw = Stopwatch.StartNew();
foreach (var ss in fxn())
{
count = ss.Sum();
}
return sw.ElapsedMilliseconds;
}
OUTPUT:
1281
1604 (_Jodrell not _Jodrell2)
6817
Jodrell's Update
I've built in release mode, i.e. optimizations on. When I run via Visual Studio I don't get a consistent bias between 1 or 2, but after repeated runs LB's answer wins, I get answers approaching something like,
1190
1260
more
but if I run the test harness from the command line, not via Visual Studio, I get results more like this
987
879
still more
EDIT
I've accepted the performance gauntlet, what follows is my amalgamation that takes the best of all answers. In my testing, it seems to have the best performance yet.
public static IEnumerable<IEnumerable<T>> SubSets_Jodrell2<T>(
this IEnumerable<T> source)
{
var list = source.ToList();
var limit = (ulong)(1 << list.Count);
for (var i = limit; i > 0; i--)
{
yield return list.SubSet(i);
}
}
private static IEnumerable<T> SubSet<T>(
this IList<T> source, ulong bits)
{
for (var i = 0; i < source.Count; i++)
{
if (((bits >> i) & 1) == 1)
{
yield return source[i];
}
}
}
Same idea again, almost the same as L.B's answer but my own interpretation.
I avoid the use of an internal List and Math.Pow.
public static IEnumerable<IEnumerable<T>> SubSets_Jodrell(
this IEnumerable<T> source)
{
var count = source.Count();
if (count > 64)
{
throw new OverflowException("Not Supported ...");
}
var limit = (ulong)(1 << count) - 2;
for (var i = limit; i > 0; i--)
{
yield return source.SubSet(i);
}
}
private static IEnumerable<T> SubSet<T>(
this IEnumerable<T> source,
ulong bits)
{
var check = (ulong)1;
foreach (var t in source)
{
if ((bits & check) > 0)
{
yield return t;
}
check <<= 1;
}
}
You'll note that these methods don't work with more than 64 elements in the intial set but it starts to take a while then anyhow.
I developed a simple ExtensionMethod for lists:
/// <summary>
/// Obtain all the combinations of the elements contained in a list
/// </summary>
/// <param name="subsetDimension">Subset Dimension</param>
/// <returns>IEnumerable containing all the differents subsets</returns>
public static IEnumerable<List<T>> CalcCombinations<T>(this List<T> list, int subsetDimension)
{
//First of all we will create a binary matrix. The dimension of a single row
//must be the dimension of list
//on which we are working (we need a 0 or a 1 for every single element) so row
//dimension is to obtain a row-length = list.count we have to
//populate the matrix with the first 2^list.Count binary numbers
int rowDimension = Convert.ToInt32(Math.Pow(2, list.Count));
//Now we start counting! We will fill our matrix with every number from 1
//(0 is meaningless) to rowDimension
//we are creating binary mask, hence the name
List<int[]> combinationMasks = new List<int[]>();
for (int i = 1; i < rowDimension; i++)
{
//I'll grab the binary rapresentation of the number
string binaryString = Convert.ToString(i, 2);
//I'll initialize an array of the apropriate dimension
int[] mask = new int[list.Count];
//Now, we have to convert our string in a array of 0 and 1, so first we
//obtain an array of int then we have to copy it inside our mask
//(which have the appropriate dimension), the Reverse()
//is used because of the behaviour of CopyTo()
binaryString.Select(x => x == '0' ? 0 : 1).Reverse().ToArray().CopyTo(mask, 0);
//Why should we keep masks of a dimension which isn't the one of the subset?
// We have to filter it then!
if (mask.Sum() == subsetDimension) combinationMasks.Add(mask);
}
//And now we apply the matrix to our list
foreach (int[] mask in combinationMasks)
{
List<T> temporaryList = new List<T>(list);
//Executes the cycle in reverse order to avoid index out of bound
for (int iter = mask.Length - 1; iter >= 0; iter--)
{
//Whenever a 0 is found the correspondent item is removed from the list
if (mask[iter] == 0)
temporaryList.RemoveAt(iter);
}
yield return temporaryList;
}
}
}
So considering the example in the question:
# Row Dimension of 4 (list.Count)
Binary Numbers to 2^4
# Binary Matrix
0 0 0 1 => skip
0 0 1 0 => skip
[...]
0 1 1 1 => added // Text2;Text3;Text4
[...]
1 0 1 1 => added // Text1;Text3;Text4
1 1 0 0 => skip
1 1 0 1 => added // Text1;Text2;Text4
1 1 1 0 => added // Text1;Text2;Text3
1 1 1 1 => skip
Hope this can help someone :)
If you need clarification or you want to contribute feel free to add answers or comments (which one is more appropriate).