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c# finding even or odd numbers

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
}
}

Finding how many times an instance happens in a list

I have a list and my goal is to determine how many times the values in that list goes above a certain value.
For instance if my list is:
List = {0, 0, 3, 3, 4, 0, 4, 4, 4}
Id like to know that there were two instances where my values in the list were greater than 2 and stayed above 2. So in this case there were 2 instances, since it dropped below 2 at one point and went above it again.
private void Report_GeneratorButton_Click(object sender, EventArgs e)
{
//Lists
var current = _CanDataGraph._DataPoints[CanDataGraph.CurveTag.Current].ToList();
var SOC = _CanDataGraph._DataPoints[CanDataGraph.CurveTag.Soc].ToList();
var highcell = _CanDataGraph._DataPoints[CanDataGraph.CurveTag.HighestCell].ToList();
var lowcell = _CanDataGraph._DataPoints[CanDataGraph.CurveTag.LowestCell].ToList();
//Seperates current list into charging, discharging, and idle
List<double> charging = current.FindAll(i => i > 2);
List<double> discharging = current.FindAll(i => i < -2);
List<double> idle = current.FindAll(i => i < 2 && i > -2);
//High cell
List<double> overcharged = highcell.FindAll(i => i > 3.65);
int ov = overcharged.Count;
if (ov > 1)
{
Console.WriteLine("This Battery has gone over Voltage!");
}
else
{
Console.WriteLine("This battery has never been over Voltage.");
}
//Low cell
List<double> overdischarged = lowcell.FindAll(i => i > 3.65);
int lv = overdischarged.Count;
if (lv > 1)
{
Console.WriteLine("This Battery has been overdischarged!");
}
else
{
Console.WriteLine("This battery has never been overdischarged.");
}
//Each value is 1 second
int chargetime = charging.Count;
int dischargetime = discharging.Count;
int idletime = idle.Count;
Console.WriteLine("Charge time: " + chargetime + "s" + "\n" + "Discharge time: " + dischargetime + "s" + "\n" + "Idle time: " + idletime);
}
My current code is this and outputs:
This battery has never been over Voltage.
This battery has never been overdischarged.
Charge time: 271s
Discharge time: 0s
Idle time: 68

There are a great many ways to solve this problem; my suggestion is that you break it down into a number of smaller problems and then write a simple method that solves each problem.
Here's a simpler problem: given a sequence of T, give me back a sequence of T with "doubled" items removed:
public static IEnumerable<T> RemoveDoubles<T>(
this IEnumerable<T> items)
{
T previous = default(T);
bool first = true;
foreach(T item in items)
{
if (first || !item.Equals(previous)) yield return item;
previous = item;
first = false;
}
}
Great. How is this helpful? Because the solution to your problem is now:
int count = myList.Select(x => x > 2).RemoveDoubles().Count(x => x);
Follow along.
If you have myList as {0, 0, 3, 3, 4, 0, 4, 4, 4} then the result of the Select is {false, false, true, true, true, false, true, true, true}.
The result of the RemoveDoubles is {false, true, false, true}.
The result of the Count is 2, which is the desired result.
Try to use off-the-shelf parts when you can. If you cannot, try to solve a simple, general problem that gets you what you need; now you have a tool you can use for other tasks that require you to remove duplicates in a sequence.

This solution should achieve the desired result.
List<int> lsNums = new List<int>() {0, 0, 3, 3, 4, 0, 4, 4, 4} ;
public void MainFoo(){
int iChange = GetCritcalChangeNum(lsNums, 2);
Console.WriteLine("Critical change = %d", iChange);
}
public int GetCritcalChangeNum(List<int> lisNum, int iCriticalThreshold) {
int iCriticalChange = 0;
int iPrev = 0;
lisNum.ForEach( (int ele) => {
if(iPrev <= iCriticalThreshold && ele > iCriticalThreshold){
iCriticalChange++;
}
iPrev = ele;
});
return iCriticalChange;
}

You can create an extension method as shown below.
public static class ListExtensions
{
public static int InstanceCount(this List<double> list, Predicate<double> predicate)
{
int instanceCount = 0;
bool instanceOccurring = false;
foreach (var item in list)
{
if (predicate(item))
{
if (!instanceOccurring)
{
instanceCount++;
instanceOccurring = true;
}
}
else
{
instanceOccurring = false;
}
}
return instanceCount;
}
}
And use your newly created method like this
current.InstanceCount(p => p > 2)

public static int CountOverLimit(IEnumerable<double> items, double limit)
{
int overLimitCount = 0;
bool isOverLimit = false;
foreach (double item in items)
{
if (item > limit)
{
if (!isOverLimit)
{
overLimitCount++;
isOverLimit = true;
}
}
else if (isOverLimit)
{
isOverLimit = false;
}
}
return overLimitCount;
}

Here's a fairly concise and readable solution. Hopefully this helps. If the limit is variable, just put it in a function and take the list and the limit as parameters.
int [] array = new int [9]{0, 0, 3, 1, 4, 0, 4, 4, 4};
List<int> values = array.ToList();
int overCount = 0;
bool currentlyOver2 = false;
for (int i = 0; i < values.Count; i++)
{
if (values[i] > 2)
{
if (!currentlyOver2)
overCount++;
currentlyOver2 = true;
}
else
currentlyOver2 = false;
}

Another way to do this using System.Linq is to walk through the list, selecting both the item itself and it's index, and return true for each item where the item is greater than value and the previous item is less than or equal to value, and then select the number of true results. Of course there's a special case for index 0 where we don't check the previous item:
public static int GetSpikeCount(List<int> items, int threshold)
{
return items?
.Select((item, index) =>
index == 0
? item > threshold
: item > threshold && items[index - 1] <= threshold)
.Count(x => x == true) // '== true' is here for readability, but it's not necessary
?? 0; // return '0' if 'items' is null
}
Sample usage:
private static void Main()
{
var myList = new List<int> {0, 0, 3, 3, 4, 0, 4, 4, 4};
var count = GetSpikeCount(myList, 2);
// count == 2
}

Can an arbitrary-length parameter array be turned into nested loops?

Is there anyway to achieve the following in C# (or any other ,Net language)?
public double nestedParamArrayLoop(function delegatedFunction, LoopControllers loopControllers)
{
double total = 0;
NestedLoopControllers loopControllers = new NestedLoopControllers(loopController, loopMaxes);
foreach(LoopController loopController in loopControllers);
{
nestedfor (loopController)
{
// this line results in one or more loopControllers being passed in
total += delegatedFunction(loopController);
}
}
return total;
}
public double delegatedFunction(params int[] arguments)
{
// dummy function to compute product of values
long product = 1;
for (int i = 0; i < arguments.Count ; i++)
product *= arguments[i];
return product;
}
Where delegatedFunction is called with a variable number of parameters, according to the number of controllers in the array loopControllers? Each loopController would contain a start value, a max value and an increment value (i.e. template a for loop).
The syntax above doesn't quite work as I'm not sure any exists to capture this paradigm. But the idea is that you can specify an arbitrary number of nested loops and then the nesting is done for you by the compiler (or the runtime). So it's a kind of templated nesting where you define the loop conditions for an arbitrary number of loops and the environment constructs the loops for you.
For example
NestedParamsArrayLoop(delegatedFunction, loopContoller1); results in iterated calls to delegatedFunction(values for loopValue1);
NestedParamsArrayLoop(delegatedFunction, loopContoller1, loopController2); results in
iterated calls to delegatedFunction(values for loopValue1, values for loopValue2);
NestedParamsArrayLoop(delegatedFunction, values for loopContoller1, values for values for loopController2, loopController3); results in
iterated calls to delegatedFunction(loopValue1, values for loopValue2, values for loopValue3);
The goal of this is to avoid writing separate functions with different numbers of arguments but where the actual guts of the logic is common across them.
I hope I've done a decent job of explaining this but if not please ask!

I think this is pretty much what you want to do.
Start with a LoopController definition:
public class LoopController : IEnumerable<int>
{
public int Start;
public int End;
public int Increment;
private IEnumerable<int> Enumerate()
{
var i = this.Start;
while (i <= this.End)
{
yield return i;
i += this.Increment;
}
}
public IEnumerator<int> GetEnumerator()
{
return this.Enumerate().GetEnumerator();
}
IEnumerator IEnumerable.GetEnumerator()
{
return this.GetEnumerator();
}
}
Now you can define NestedParamArrayLoop like so:
public double NestedParamArrayLoop(Func<int[], double> delegatedFunction, List<LoopController> loopControllers)
{
double total = 0;
foreach (LoopController loopController in loopControllers)
{
total += delegatedFunction(loopController.ToArray());
}
return total;
}
Now the rest is easy:
void Main()
{
var loopControllers = new List<LoopController>()
{
new LoopController() { Start = 4, End = 10, Increment = 2 },
new LoopController() { Start = 17, End = 19, Increment = 1 },
};
Console.WriteLine(NestedParamArrayLoop(DelegatedFunction, loopControllers));
}
public double DelegatedFunction(params int[] arguments)
{
long product = 1;
for (int i = 0; i < arguments.Count(); i++)
product *= arguments[i];
return product;
}
You could even define NestedParamArrayLoop as this:
public double NestedParamArrayLoop(Func<int[], double> delegatedFunction, List<LoopController> loopControllers)
{
return
loopControllers
.Select(lc => delegatedFunction(lc.ToArray()))
.Sum();
}
Is this more like what you're after?
public double NestedParamArrayLoop(Func<int[], double> delegatedFunction, List<LoopController> loopControllers)
{
Func<IEnumerable<int>, IEnumerable<IEnumerable<int>>> getAllSubsets = null;
getAllSubsets = xs =>
(xs == null || !xs.Any())
? Enumerable.Empty<IEnumerable<int>>()
: xs.Skip(1).Any()
? getAllSubsets(xs.Skip(1))
.SelectMany(ys => new[] { ys, xs.Take(1).Concat(ys) })
: new[] { Enumerable.Empty<int>(), xs.Take(1) };
double total = 0;
foreach (LoopController loopController in loopControllers)
{
foreach (var subset in getAllSubsets(loopController))
{
total += delegatedFunction(subset.ToArray());
}
}
return total;
}

Since the OP asked for a solution in any .NET language, I wrote one in F# (translating #Enigmativity's answer). No NestedLoopController class required:
[[ 4 .. 2 .. 10 ]; [ 17 .. 1 .. 19 ]]
|> Seq.map (fun args ->
(1L, args)
||> Seq.fold (fun a x -> a * int64 x))
|> Seq.sum
|> printfn "%d"
You can probably translate this to C# LINQ in a relatively straightforward way…

I think what you really need is what is called cartesian product of multiple sets. Here is good article from Eric Lippert about doing that with arbitrary number of sets in C#. So create function like this (I won't explain it because I cannot do this better than Eric did in his article):
static IEnumerable<IEnumerable<T>> CartesianProduct<T>(params IEnumerable<T>[] sources) {
IEnumerable<IEnumerable<T>> result = new[] { Enumerable.Empty<T>() };
foreach (var source in sources) {
var tmp = source;
result = result.SelectMany(
seq => tmp,
(seq, item) => seq.Concat(new[] { item }));
}
return result;
}
Then use like this:
foreach (var n in CartesianProduct(Enumerable.Range(1, 4), Enumerable.Range(1, 4))) {
Console.WriteLine(String.Join(", ", n));
// in your case: delegatedFunction(n);
}
outputs
1, 1
1, 2
1, 3
1, 4
2, 1
2, 2
2, 3
2, 4
3, 1
3, 2
3, 3
3, 4
4, 1
4, 2
4, 3
4, 4
It's easy to replace Enumerable.Range with your LoopController, Enumerable.Range is used just as an example.

Combination Algorithm

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

in C#, how do I order items in a list where the "largest" values are in the middle of the list

I have been stumped on this one for a while. I want to take a List and order the list such that the Products with the largest Price end up in the middle of the list. And I also want to do the opposite, i.e. make sure that the items with the largest price end up on the outer boundaries of the list.
Imagine a data structure like this.. 1,2,3,4,5,6,7,8,9,10
In the first scenario I need to get back 1,3,5,7,9,10,8,6,4,2
In the second scenario I need to get back 10,8,6,4,2,1,3,5,7,9
The list may have upwards of 250 items, the numbers will not be evenly distributed, and they will not be sequential, and I wanted to minimize copying. The numbers will be contained in Product objects, and not simple primitive integers.
Is there a simple solution that I am not seeing?
Any thoughts.
So for those of you wondering what I am up to, I am ordering items based on calculated font size. Here is the code that I went with...
The Implementation...
private void Reorder()
{
var tempList = new LinkedList<DisplayTag>();
bool even = true;
foreach (var tag in this) {
if (even)
tempList.AddLast(tag);
else
tempList.AddFirst(tag);
even = !even;
}
this.Clear();
this.AddRange(tempList);
}
The Test...
[TestCase(DisplayTagOrder.SmallestToLargest, Result=new[]{10,14,18,22,26,30})]
[TestCase(DisplayTagOrder.LargestToSmallest, Result=new[]{30,26,22,18,14,10})]
[TestCase(DisplayTagOrder.LargestInTheMiddle, Result = new[] { 10, 18, 26, 30, 22, 14 })]
[TestCase(DisplayTagOrder.LargestOnTheEnds, Result = new[] { 30, 22, 14, 10, 18, 26 })]
public int[] CalculateFontSize_Orders_Tags_Appropriately(DisplayTagOrder sortOrder)
{
list.CloudOrder = sortOrder;
list.CalculateFontSize();
var result = (from displayTag in list select displayTag.FontSize).ToArray();
return result;
}
The Usage...
public void CalculateFontSize()
{
GetMaximumRange();
GetMinimunRange();
CalculateDelta();
this.ForEach((displayTag) => CalculateFontSize(displayTag));
OrderByFontSize();
}
private void OrderByFontSize()
{
switch (CloudOrder) {
case DisplayTagOrder.SmallestToLargest:
this.Sort((arg1, arg2) => arg1.FontSize.CompareTo(arg2.FontSize));
break;
case DisplayTagOrder.LargestToSmallest:
this.Sort(new LargestFirstComparer());
break;
case DisplayTagOrder.LargestInTheMiddle:
this.Sort(new LargestFirstComparer());
Reorder();
break;
case DisplayTagOrder.LargestOnTheEnds:
this.Sort();
Reorder();
break;
}
}

The appropriate data structure is a LinkedList because it allows you to efficiently add to either end:
LinkedList<int> result = new LinkedList<int>();
int[] array = { 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 };
Array.Sort(array);
bool odd = true;
foreach (var x in array)
{
if (odd)
result.AddLast(x);
else
result.AddFirst(x);
odd = !odd;
}
foreach (int item in result)
Console.Write("{0} ", item);
No extra copying steps, no reversing steps, ... just a small overhead per node for storage.

C# Iterator version
(Very simple code to satisfy all conditions.)
One function to rule them all! Doesn't use intermediate storage collection (see yield keyword). Orders the large numbers either to the middle, or to the sides depending on the argument. It's implemented as a C# iterator
// Pass forward sorted array for large middle numbers,
// or reverse sorted array for large side numbers.
//
public static IEnumerable<long> CurveOrder(long[] nums) {
if (nums == null || nums.Length == 0)
yield break; // Nothing to do.
// Move forward every two.
for (int i = 0; i < nums.Length; i+=2)
yield return nums[i];
// Move backward every other two. Note: Length%2 makes sure we're on the correct offset.
for (int i = nums.Length-1 - nums.Length%2; i >= 0; i-=2)
yield return nums[i];
}
Example Usage
For example with array long[] nums = { 1,2,3,4,5,6,7,8,9,10,11 };
Start with forward sort order, to bump high numbers into the middle.
Array.Sort(nums); //forward sort
// Array argument will be: { 1,2,3,4,5,6,7,8,9,10,11 };
long[] arrLargeMiddle = CurveOrder(nums).ToArray();
Produces: 1 3 5 7 9 11 10 8 6 4 2
Or, Start with reverse sort order, to push high numbers to sides.
Array.Reverse(nums); //reverse sort
// Array argument will be: { 11,10,9,8,7,6,5,4,3,2,1 };
long[] arrLargeSides = CurveOrder(nums).ToArray();
Produces: 11 9 7 5 3 1 2 4 6 8 10
Significant namespaces are:
using System;
using System.Collections.Generic;
using System.Linq;
Note: The iterator leaves the decision up to the caller about whether or not to use intermediate storage. The caller might simply be issuing a foreach loop over the results instead.
Extension Method Option
Optionally change the static method header to use the this modifier public static IEnumerable<long> CurveOrder(this long[] nums) { and put it inside a static class in your namespace;
Then call the order method directly on any long[ ] array instance like so:
Array.Reverse(nums); //reverse sort
// Array argument will be: { 11,10,9,8,7,6,5,4,3,2,1 };
long[] arrLargeSides = nums.CurveOrder().ToArray();
Just some (unneeded) syntactic sugar to mix things up a bit for fun. This can be applied to any answers to your question that take an array argument.

I might go for something like this
static T[] SortFromMiddleOut<T, U>(IList<T> list, Func<T, U> orderSelector, bool largestInside) where U : IComparable<U>
{
T[] sortedArray = new T[list.Count];
bool add = false;
int index = (list.Count / 2);
int iterations = 0;
IOrderedEnumerable<T> orderedList;
if (largestInside)
orderedList = list.OrderByDescending(orderSelector);
else
orderedList = list.OrderBy(orderSelector);
foreach (T item in orderedList)
{
sortedArray[index] = item;
if (add)
index += ++iterations;
else
index -= ++iterations;
add = !add;
}
return sortedArray;
}
Sample invocations:
int[] array = { 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 };
int[] sortedArray = SortFromMiddleOut(array, i => i, false);
foreach (int item in sortedArray)
Console.Write("{0} ", item);
Console.Write("\n");
sortedArray = SortFromMiddleOut(array, i => i, true);
foreach (int item in sortedArray)
Console.Write("{0} ", item);
With it being generic, it could be a list of Foo and the order selector could be f => f.Name or whatever you want to throw at it.

The fastest (but not the clearest) solution is probably to simply calculate the new index for each element:
Array.Sort(array);
int length = array.Length;
int middle = length / 2;
int[] result2 = new int[length];
for (int i = 0; i < array.Length; i++)
{
result2[middle + (1 - 2 * (i % 2)) * ((i + 1) / 2)] = array[i];
}

Something like this?
public IEnumerable<int> SortToMiddle(IEnumerable<int> input)
{
var sorted = new List<int>(input);
sorted.Sort();
var firstHalf = new List<int>();
var secondHalf = new List<int>();
var sendToFirst = true;
foreach (var current in sorted)
{
if (sendToFirst)
{
firstHalf.Add(current);
}
else
{
secondHalf.Add(current);
}
sendToFirst = !sendToFirst;
}
//to get the highest values on the outside just reverse
//the first list instead of the second
secondHalf.Reverse();
return firstHalf.Concat(secondHalf);
}
For your specific (general) case (assuming unique keys):
public static IEnumerable<T> SortToMiddle<T, TU>(IEnumerable<T> input, Func<T, TU> getSortKey)
{
var sorted = new List<TU>(input.Select(getSortKey));
sorted.Sort();
var firstHalf = new List<TU>();
var secondHalf = new List<TU>();
var sendToFirst = true;
foreach (var current in sorted)
{
if (sendToFirst)
{
firstHalf.Add(current);
}
else
{
secondHalf.Add(current);
}
sendToFirst = !sendToFirst;
}
//to get the highest values on the outside just reverse
//the first list instead of the second
secondHalf.Reverse();
sorted = new List<TU>(firstHalf.Concat(secondHalf));
//This assumes the sort keys are unique - if not, the implementation
//needs to use a SortedList<TU, T>
return sorted.Select(s => input.First(t => s.Equals(getSortKey(t))));
}
And assuming non-unique keys:
public static IEnumerable<T> SortToMiddle<T, TU>(IEnumerable<T> input, Func<T, TU> getSortKey)
{
var sendToFirst = true;
var sorted = new SortedList<TU, T>(input.ToDictionary(getSortKey, t => t));
var firstHalf = new SortedList<TU, T>();
var secondHalf = new SortedList<TU, T>();
foreach (var current in sorted)
{
if (sendToFirst)
{
firstHalf.Add(current.Key, current.Value);
}
else
{
secondHalf.Add(current.Key, current.Value);
}
sendToFirst = !sendToFirst;
}
//to get the highest values on the outside just reverse
//the first list instead of the second
secondHalf.Reverse();
return(firstHalf.Concat(secondHalf)).Select(kvp => kvp.Value);
}

Simplest solution - order the list descending, create two new lists, into the first place every odd-indexed item, into the other every even indexed item. Reverse the first list then append the second to the first.

Okay, I'm not going to question your sanity here since I'm sure you wouldn't be asking the question if there weren't a good reason :-)
Here's how I'd approach it. Create a sorted list, then simply create another list by processing the keys in order, alternately inserting before and appending, something like:
sortedlist = list.sort (descending)
biginmiddle = new list()
state = append
foreach item in sortedlist:
if state == append:
biginmiddle.append (item)
state = prepend
else:
biginmiddle.insert (0, item)
state = append
This will give you a list where the big items are in the middle. Other items will fan out from the middle (in alternating directions) as needed:
1, 3, 5, 7, 9, 10, 8, 6, 4, 2
To get a list where the larger elements are at the ends, just replace the initial sort with an ascending one.
The sorted and final lists can just be pointers to the actual items (since you state they're not simple integers) - this will minimise both extra storage requirements and copying.

Maybe its not the best solution, but here's a nifty way...
Let Product[] parr be your array.
Disclaimer It's java, my C# is rusty.
Untested code, but you get the idea.
int plen = parr.length
int [] indices = new int[plen];
for(int i = 0; i < (plen/2); i ++)
indices[i] = 2*i + 1; // Line1
for(int i = (plen/2); i < plen; i++)
indices[i] = 2*(plen-i); // Line2
for(int i = 0; i < plen; i++)
{
if(i != indices[i])
swap(parr[i], parr[indices[i]]);
}
The second case, Something like this?
int plen = parr.length
int [] indices = new int[plen];
for(int i = 0; i <= (plen/2); i ++)
indices[i] = (plen^1) - 2*i;
for(int i = 0; i < (plen/2); i++)
indices[i+(plen/2)+1] = 2*i + 1;
for(int i = 0; i < plen; i++)
{
if(i != indices[i])
swap(parr[i], parr[indices[i]]);
}