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Loop nested loops

I would like to create a nested loop where the nesting depth determined by the size on an array. For example I have an array of integers and I would like to check all tuples where . Is there a way to do this simply in c# (or any other language)?
The only easy idea I had was to just multiply the numbers and do a for loop up to that number but unfortunately the product reaches the int limit. Also in this case I have no way to add extra conditions on each level.
Here's a small example and the product approach:
int[] a = new int[]{2, 3, 2}; //we have an array like this. In the description above it is a_1, a_2, a_3
void f(int[] i) //I have a function that works on an array input
{...}
bool check(int[] i) //I have a checker function, with an array parameter too
{...}
//And I would like to perform this function on all these arrays
//f({0, 0, 0})
//f({0, 0, 1})
//010
//011
//020
//021
//100
//101
//110
//111
//120
//f({1, 2, 1})
//But if let's say check({1, 0}) is false then don't perform the f function on {1, 0, 0} and {1, 0, 1}
//A simple implementation if we know the size of the array a is the following
for(int i1=0;i1<a[0];i1++)
{
if(!check({i1}))
continue;
for(int i2=0;i2<a[1];i2++)
{
if(!check({i1, i2}))
continue;
for(int i3=0;i3<a[2];i3++)
{
if(!check({i1, i2, i3}))
continue;
f({i1, i2, i3});
}
}
}
//But this obviously fails as we have no idea apriori of the size of the array a
//An alternative I have is the following:
int prod = 1;
foreach(int x in a)
{
prod *= x;
}
for(int c=0;c<prod;c++)
{
int d=c;
int[] i = new int[a.Length];
for(int l=0;l<a.Length;l++)
{
i[l]=d%(a[l]);
d /= a[l];
}
f(i);
}
//But the problem with this implementation is that in my case prod is larger than the int limit. Also this loops through all the incorrect cases too, where the check function can highly reduce the number of cases to calculate.

I've managed to solve the problem. The idea is that increasing the i value by one is really easy and we can check the condition that we don't overstep the a values and don't violate the check function easily. Below is a code that works
int[] a = ...;
void f(int[] i){...}
bool check(int[] i){...}
int n = a.Length;
int[] i = new int[n];
while (true) //keep increasing the coordinates of i, while we can
{
for (int l = 0; l < n; l++)
{
int[] il = copyFirst(i, l);
while(!check(il)) //check for all first few coordinates if it is correct, skip if incorrect.
//There is a way to improve this even further, as we don't have to check the first few
// coords if it was correct before, so should only care about the recently changed section
{
i = increase(a, i, l);
if (i == null)
{
return;
}
else
{
il = copyFirst(i, l);
}
}
}
f(i);
i = increase(a, i, n-1);
if (i == null) return;
}
int[] copyFirst(int[] i, int l) //this is just a small helper function to copy the first few elements of i, to use in the check
{
int[] ret = new int[l];
for (int k = 0; k < l; k++)
{
ret[k] = i[k];
}
return ret;
}
int[] increase(int[] a, int[] i, int l) //this results in the increased vector and in null if we've reached the end
{
for (int k = l; k >= 0; k--)
{
i[k] = i[k] + 1;
if (i[k] >= a[k])
{
i[k] = 0;
}
else
{
return i;
}
}
return null;
}

Shifting A single element in an array

I want to shift one element in an array to the right each time whilst leaving the original elements in their specific order in C#.
Ok so I've been asked to reword the code I can understand why so here we go:
I might have a number 48390
//the ar elements have been commented out to show that we never know what ar contains but only the that I will always want to shift; ar[4]
int[] ar = new int[5];
//ar[0] = 4
//ar[1] = 8
//ar[2] = 3
//ar[3] = 9
//ar[4] = 0
while(ar != 04839)
{
Shift code
}
I might input 5 numbers 48390 if you notice its the same number but one digit is out. I want a while loop to rotate that 4 ar[1] to shift until the number forms 04839
I hope this makes sense. I am posting this question because most pages posting information about shifting based on shifting all elements to the right and I only really want to shift one specific element.
Thanks for looking.
edit: I should have been more specific. What if you don't know what each of the array elements could be? So I couldn't depend on "0" as an anchor. as another set of numbers might include another number for example "00238."

This method will give you a sequence of arrays made by inserting a single element into (between) each position in a given array:
public static IEnumerable<T[]> InsertElementBetweenAllPositions<T>(
T[] array, T element)
{
int newLength = array.Length + 1;
for (int i = 0; i < newLength; i++)
{
T[] rtn = new T[newLength];
rtn[i] = element;
Array.Copy(array, 0, rtn, 0, i);
Array.Copy(array, i, rtn, i + 1, array.Length - i);
yield return rtn;
}
}
For your example, you might call it as
foreach (int[] arr in InsertElementBetweenAllPositions(new[] { 6, 7, 8, 9 }, 0))
{
foreach (int i in arr)
Console.Write(i + " ");
Console.WriteLine();
}

How about this:
List<int> l = new List<int>(){0,6,7,8,9};
for (int i=1;i<5;i++)
{
l.Remove(0);
l.Insert(i, 0);
}

What's in your example is a swap, which can be implemented like:
private void Swap(ref int[] array, int index1, int index2)
{
int temp = array[index1];
array[index1] = array[index2];
array[index2] = temp;
}
Calling Swap(ref source, 0, 1) would exchange the first and second element. What you want then is:
for (int i = 0; i < a.Length-1; i++)
{
Swap(ref a, i, i+1);
}
This "bubbles" the first element up to the last position in each iteration.

From the example you need to shift elements around, and the example is a bit confusing over whether you need to loop them around to the start again. I have provided the below example that will loop around to the start - If you do not need to do that, the you can rework the if the statement.
private int[] Shift(int[] a)
{
int zeroPos = Array.IndexOf(a, 0);
int[] rtn = new int[a.Length];
a.CopyTo(rtn, 0);
if (zeroPos + 1 == a.Length)
{
rtn[0] = 0;
for (int i = 0; i < a.Length - 1; i++)
{
rtn[i + 1] = a[i];
}
}
else
{
rtn[zeroPos] = rtn[zeroPos + 1];
rtn[zeroPos + 1] = 0;
}
return rtn;
}

r=ar[0];
for (int i = 0; ar.lenght;i++)
{
ar[i]=ar[i + 1];
}
ar[ar.lenght] = r;

Have you thought about using a LinkedList instead? A linked list data structure is probably more suited to what you are trying to do than an array. The AddFirst, AddLast, AddAfter and AddBefore methods allow you to insert elements into the list in a much more efficient way than re-organizing your array each time.
The disadvantage of linked lists is that you need to read the elements in order. So, it's very efficient for inserting/deleting elements but inefficient for accessing elements randomly.
There is a good overview of LinkedLists here.

Perhaps
int oldLast = ar[ar.Length - 1];
for (int i = ar.Length - 1; i >= 0; i--)
ar[i] = i == 0 ? oldLast : ar[i - 1];
Demo

It is just a permutation of an item , below is the full source code of permutation algorithm.
static List<string> Put(char s1, string list)
{
List<string> str =new List<string>();
for (int i = 0; i < list.Length+1; i++)
{
string s = list.Substring(0, i) + s1.ToString() + list.Substring(i);
str.Add(s);
}
return str;
}
static List<string> Permute(string list,int x)
{
List<string> Result = new List<string>();
if (list.Length == 1)
{
Result.Add(list[0].ToString());
return Result;
}
else
{
char first = list[0];
list = list.Substring(x+1);
List<string> part = Permute(list,0);
foreach (string str in part)
{
List<string> hasBeenPlaced = Put(first, str);
foreach (string str2 in hasBeenPlaced)
{
Result.Add(str2);
}
}
}
return Result;
}
static void Main(string[] args)
{
List<string> per = Permute("abc",0);
for (int i = 0; i < per.Count; i++)
{
Console.WriteLine(per[i]);
}
Console.ReadKey();
}
Now if I add a break after the foreach , your problem has been solved . (it will writes all permuation for just an item which you want , not all of them....)
So change that to :
foreach (string str in part)
{
List<string> hasBeenPlaced = Put(first, str);
foreach (string str2 in hasBeenPlaced)
{
Result.Add(str2);
}
break;
}
Hope to helps you

If you you linq, that's simple :-) But you need a size larger than the array.
ShiftLeft(ar, 1);
private static int[] ShiftLeft(int[] value, int countOfShift = 1)
{
var length = value.Length;
if (countOfShift > length)
{
throw new InvalidOperationException("countOfShift must less then value's length.");
}
var tempList = new List<int>(value);
tempList.RemoveRange(length - countOfShift, countOfShift);
tempList.InsertRange(0, value.Skip(length - countOfShift));
return tempList.ToArray();
}

Find the smallest window of input array that contains all the elements of query array

Problem: Given an input array of integers of size n, and a query array of integers of size k, find the smallest window of input array that contains all the elements of query array and also in the same order.
I have tried below approach.
int[] inputArray = new int[] { 2, 5, 2, 8, 0, 1, 4, 7 };
int[] queryArray = new int[] { 2, 1, 7 };
Will find the position of all query array element in inputArray.
public static void SmallestWindow(int[] inputArray, int[] queryArray)
{
Dictionary<int, HashSet<int>> dict = new Dictionary<int, HashSet<int>>();
int index = 0;
foreach (int i in queryArray)
{
HashSet<int> hash = new HashSet<int>();
foreach (int j in inputArray)
{
index++;
if (i == j)
hash.Add(index);
}
dict.Add(i, hash);
index = 0;
}
// Need to perform action in above dictionary.??
}
I got following dictionary
int 2--> position {1, 3}
int 1 --> position {6}
int 7 --> position {8}
Now I want to perform following step to findout minimum window
Compare int 2 position to int 1 position. As (6-3) < (6-1)..So I will store 3, 6 in a hashmap.
Will compare the position of int 1 and int 7 same like above.
I cannot understand how I will compare two consecutive value of a dictionary. Please help.

The algorithm:
For each element in the query array, store in a map M (V → (I,P)), V is the element, I is an index into the input array, P is the position in the query array. (The index into the input array for some P is the largest such that query[0..P] is a subsequence of input[I..curr])
Iterate through the array.
If the value is the first term in the query array: Store the current index as I.
Else: Store the value of the index of the previous element in the query array, e.g. M[currVal].I = M[query[M[currVal].P-1]].I.
If the value is the last term: Check if [I..curr] is a new best.
Complexity
The complexity of this is O(N), where N is the size of the input array.
N.B.
This code expects that no elements are repeated in the query array. To cater for this, we can use a map M (V → listOf((I,P))). This is O(NhC(Q)), where hC(Q) is the count of the mode for the query array..
Even better would be to use M (V → listOf((linkedList(I), P))). Where repeated elements occur consecutively in the query array, we use a linked list. Updating those values then becomes O(1). The complexity is then O(NhC(D(Q))), where D(Q) is Q with consecutive terms merged.
Implementation
Sample java implementation is available here. This does not work for repeated elements in the query array, nor do error checking, etc.

I don't see how using HashSet and Dictionary will help you in this. Were I faced with this problem, I'd go about it quite differently.
One way to do it (not the most efficient way) is shown below. This code makes the assumption that queryArray contains at least two items.
int FindInArray(int[] a, int start, int value)
{
for (int i = start; i < a.Length; ++i)
{
if (a[i] == value)
return i;
}
return -1;
}
struct Pair
{
int first;
int last;
}
List<Pair> foundPairs = new List<Pair>();
int startPos = 0;
bool found = true;
while (found)
{
found = false;
// find next occurrence of queryArray[0] in inputArray
startPos = FindInArray(inputArray, startPos, queryArray[0]);
if (startPos == -1)
{
// no more occurrences of the first item
break;
}
Pair p = new Pair();
p.first = startPos;
++startPos;
int nextPos = startPos;
// now find occurrences of remaining items
for (int i = 1; i < queryArray.Length; ++i)
{
nextPos = FindInArray(inputArray, nextPos, queryArray[i]);
if (nextPos == -1)
{
break; // didn't find it
}
else
{
p.last = nextPos++;
found = (i == queryArray.Length-1);
}
}
if (found)
{
foundPairs.Add(p);
}
}
// At this point, the foundPairs list contains the (start, end) of all
// sublists that contain the items in order.
// You can then iterate through that list, subtract (last-first), and take
// the item that has the smallest value. That will be the shortest sublist
// that matches the criteria.
With some work, this could be made more efficient. For example, if 'queryArray' contains [1, 2, 3] and inputArray contains [1, 7, 4, 9, 1, 3, 6, 4, 1, 8, 2, 3], the above code will find three matches (starting at positions 0, 4, and 8). Slightly smarter code could determine that when the 1 at position 4 is found, since no 2 was found prior to it, that any sequence starting at the first position would be longer than the sequence starting at position 4, and therefore short-circuit the first sequence and start over at the new position. That complicates the code a bit, though.

You want not a HashSet but a (sorted) tree or array as the value in the dictionary; the dictionary contains mappings from values you find in the input array to the (sorted) list of indices where that value appears.
Then you do the following
Look up the first entry in the query. Pick the lowest index where it appears.
Look up the second entry; pick the lowest entry greater than the index of the first.
Look up the third; pick the lowest greater than the second. (Etc.)
When you reach the last entry in the query, (1 + last index - first index) is the size of the smallest match.
Now pick the second index of the first query, repeat, etc.
Pick the smallest match found from any of the starting indices.
(Note that the "lowest entry greater" is an operation supplied with sorted trees, or can be found via binary search on a sorted array.)
The complexity of this is approximately O(M*n*log n) where M is the length of the query and n is the average number of indices at which a given value appears in the input array. You can modify the strategy by picking that query array value that appears least often for the starting point and going up and down from there; if there are k of those entries (k <= n) then the complexity is O(M*k*log n).

After you got all the positions(indexes) in the inputArray:
2 --> position {0,2} // note: I change them to 0-based array
1 --> position {5,6} // I suppose it's {5,6} to make it more complex, in your code it's only {5}
7 --> position {7}
I use a recursion to get all possible paths. [0->5->7] [0->6->7] [2->5->7] [2->6->7]. The total is 2*2*1=4 possible paths. Obviously the one who has Min(Last-First) is the shortest path(smallest window), those numbers in the middle of the path don't matter. Here comes the code.
struct Pair
{
public int Number; // the number in queryArray
public int[] Indexes; // the positions of the number
}
static List<int[]> results = new List<int[]>(); //store all possible paths
static Stack<int> currResult = new Stack<int>(); // the container of current path
static int[] inputArray, queryArray;
static Pair[] pairs;
After the data structures, here is the Main.
inputArray = new int[] { 2, 7, 1, 5, 2, 8, 0, 1, 4, 7 }; //my test case
queryArray = new int[] { 2, 1, 7 };
pairs = (from n in queryArray
select new Pair { Number = n, Indexes = inputArray.FindAllIndexes(i => i == n) }).ToArray();
Go(0);
FindAllIndexes is an extension method to help find all the indexes.
public static int[] FindAllIndexes<T>(this IEnumerable<T> source, Func<T,bool> predicate)
{
//do necessary check here, then
Queue<int> indexes = new Queue<int>();
for (int i = 0;i<source.Count();i++)
if (predicate(source.ElementAt(i))) indexes.Enqueue(i);
return indexes.ToArray();
}
The recursion method:
static void Go(int depth)
{
if (depth == pairs.Length)
{
results.Add(currResult.Reverse().ToArray());
}
else
{
var indexes = pairs[depth].Indexes;
for (int i = 0; i < indexes.Length; i++)
{
if (depth == 0 || indexes[i] > currResult.Last())
{
currResult.Push(indexes[i]);
Go(depth + 1);
currResult.Pop();
}
}
}
}
At last, a loop of results can find the Min(Last-First) result(shortest window).

Algorithm:
get all indexes into the inputArray
of all queryArray values
order them ascending by index
using each index (x) as a starting
point find the first higher index
(y) such that the segment
inputArray[x-y] contains all
queryArray values
keep only those segments that have the queryArray items in order
order the segments by their lengths,
ascending
c# implementation:
First get all indexes into the inputArray of all queryArray values and order them ascending by index.
public static int[] SmallestWindow(int[] inputArray, int[] queryArray)
{
var indexed = queryArray
.SelectMany(x => inputArray
.Select((y, i) => new
{
Value = y,
Index = i
})
.Where(y => y.Value == x))
.OrderBy(x => x.Index)
.ToList();
Next, using each index (x) as a starting point find the first higher index (y) such that the segment inputArray[x-y] contains all queryArray values.
var segments = indexed
.Select(x =>
{
var unique = new HashSet<int>();
return new
{
Item = x,
Followers = indexed
.Where(y => y.Index >= x.Index)
.TakeWhile(y => unique.Count != queryArray.Length)
.Select(y =>
{
unique.Add(y.Value);
return y;
})
.ToList(),
IsComplete = unique.Count == queryArray.Length
};
})
.Where(x => x.IsComplete);
Now keep only those segments that have the queryArray items in order.
var queryIndexed = segments
.Select(x => x.Followers.Select(y => new
{
QIndex = Array.IndexOf(queryArray, y.Value),
y.Index,
y.Value
}).ToArray());
var queryOrdered = queryIndexed
.Where(item =>
{
var qindex = item.Select(x => x.QIndex).ToList();
bool changed;
do
{
changed = false;
for (int i = 1; i < qindex.Count; i++)
{
if (qindex[i] <= qindex[i - 1])
{
qindex.RemoveAt(i);
changed = true;
}
}
} while (changed);
return qindex.Count == queryArray.Length;
});
Finally, order the segments by their lengths, ascending. The first segment in the result is the smallest window into inputArray that contains all queryArray values in the order of queryArray.
var result = queryOrdered
.Select(x => new[]
{
x.First().Index,
x.Last().Index
})
.OrderBy(x => x[1] - x[0]);
var best = result.FirstOrDefault();
return best;
}
test it with
public void Test()
{
var inputArray = new[] { 2, 1, 5, 6, 8, 1, 8, 6, 2, 9, 2, 9, 1, 2 };
var queryArray = new[] { 6, 1, 2 };
var result = SmallestWindow(inputArray, queryArray);
if (result == null)
{
Console.WriteLine("no matching window");
}
else
{
Console.WriteLine("Smallest window is indexes " + result[0] + " to " + result[1]);
}
}
output:
Smallest window is indexes 3 to 8

Thank you everyone for your inputs. I have changed my code a bit and find it working. Though it might not be very efficient but I'm happy to solve using my head :). Please give your feedback
Here is my Pair class with having number and position as variable
public class Pair
{
public int Number;
public List<int> Position;
}
Here is a method which will return the list of all Pairs.
public static Pair[] GetIndex(int[] inputArray, int[] query)
{
Pair[] pairList = new Pair[query.Length];
int pairIndex = 0;
foreach (int i in query)
{
Pair pair = new Pair();
int index = 0;
pair.Position = new List<int>();
foreach (int j in inputArray)
{
if (i == j)
{
pair.Position.Add(index);
}
index++;
}
pair.Number = i;
pairList[pairIndex] = pair;
pairIndex++;
}
return pairList;
}
Here is the line of code in Main method
Pair[] pairs = NewCollection.GetIndex(array, intQuery);
List<int> minWindow = new List<int>();
for (int i = 0; i <pairs.Length - 1; i++)
{
List<int> first = pairs[i].Position;
List<int> second = pairs[i + 1].Position;
int? temp = null;
int? temp1 = null;
foreach(int m in first)
{
foreach (int n in second)
{
if (n > m)
{
temp = m;
temp1 = n;
}
}
}
if (temp.HasValue && temp1.HasValue)
{
if (!minWindow.Contains((int)temp))
minWindow.Add((int)temp);
if (!minWindow.Contains((int)temp1))
minWindow.Add((int)temp1);
}
else
{
Console.WriteLine(" Bad Query array");
minWindow.Clear();
break;
}
}
if(minWindow.Count > 0)
{
Console.WriteLine("Minimum Window is :");
foreach(int i in minWindow)
{
Console.WriteLine(i + " ");
}
}

It is worth noting that this problem is related to the longest common subsequence problem, so coming up with algorithms that run in better than O(n^2) time in the general case with duplicates would be challenging.

Just in case someone is interested in C++ implementation with O(nlog(k))
void findMinWindow(const vector<int>& input, const vector<int>& query) {
map<int, int> qtree;
for(vector<int>::const_iterator itr=query.begin(); itr!=query.end(); itr++) {
qtree[*itr] = 0;
}
int first_ptr=0;
int begin_ptr=0;
int index1 = 0;
int queptr = 0;
int flip = 0;
while(true) {
//check if value is in query
if(qtree.find(input[index1]) != qtree.end()) {
int x = qtree[input[index1]];
if(0 == x) {
flip++;
}
qtree[input[index1]] = ++x;
}
//remove all nodes that are not required and
//yet satisfy the all query condition.
while(query.size() == flip) {
//done nothing more
if(queptr == input.size()) {
break;
}
//check if queptr is pointing to node in the query
if(qtree.find(input[queptr]) != qtree.end()) {
int y = qtree[input[queptr]];
//more nodes and the queue is pointing to deleteable node
//condense the nodes
if(y > 1) {
qtree[input[queptr]] = --y;
queptr++;
} else {
//cant condense more just keep that memory
if((!first_ptr && !begin_ptr) ||
((first_ptr-begin_ptr)>(index1-queptr))) {
first_ptr=index1;
begin_ptr=queptr;
}
break;
}
} else {
queptr++;
}
}
index1++;
if(index1==input.size()) {
break;
}
}
cout<<"["<<begin_ptr<<" - "<<first_ptr<<"]"<<endl;
}
here the main for calling it.
#include <iostream>
#include <vector>
#include <map>
using namespace std;
int main() {
vector<int> input;
input.push_back(2);
input.push_back(5);
input.push_back(2);
input.push_back(8);
input.push_back(0);
input.push_back(1);
input.push_back(4);
input.push_back(7);
vector<int> query1;
query1.push_back(2);
query1.push_back(8);
query1.push_back(0);
vector<int> query2;
query2.push_back(2);
query2.push_back(1);
query2.push_back(7);
vector<int> query3;
query3.push_back(1);
query3.push_back(4);
findMinWindow(input, query1);
findMinWindow(input, query2);
findMinWindow(input, query3);
}

Merging two lists into one and sorting the items

Is there a way to merge(union without dupes) two given lists into one and store the items in sorted way by using ONE for loop?
Also, i am looking for a solution which does not makes use of API methods ( like, union, sort etc).
Sample Code.
private static void MergeAndOrder()
{
var listOne = new List<int> {3, 4, 1, 2, 7, 6, 9, 11};
var listTwo = new List<int> {1, 7, 8, 3, 5, 10, 15, 12};
//Without Using C# helper methods...
//ToDo.............................
//Using C# APi.
var expectedResult = listOne.Union(listTwo).ToList();
expectedResult.Sort();//Output: 1,2,3,4,5,6,7,8,9,10,11,12,15
//I need the same result without using API methods, and that too by iterating over items only once.
}
PS: I have been asked this question in an interview, but couldn't find answer as yet.

Why can't you use the api methods? Re-inventing the wheel is dumb. Also, it's the .ToList() call that's killing you. Never call .ToList() or .ToArray() until you absolutely have to, because they break your lazy evaluation.
Do it like this and you'll enumerate the lists with the minimum amount necessary:
var expectedResult = listOne.Union(listTwo).OrderBy(i => i);
This will do the union in one loop using a hashset, and lazy execution means the base-pass for the sort will piggyback on the union. But I don't think it's possible finish the sort in a single iteration, because sorting is not a O(n) operation.

Without the precondition that both lists are sorted before the merge + sort operation, you can't do this in O(n) time (or "using one loop").
Add that precondition and the problem is very easy.
Keep two iterators, one for each list. On each loop, compare the element from each list and choose the smaller. Increment that list's iterator. If the element you are about to insert in the final list is already the last element in that list, skip the insert.
In pseudocode:
List a = { 1, 3, 5, 7, 9 }
List b = { 2, 4, 6, 8, 10 }
List result = { }
int i=0, j=0, lastIndex=0
while(i < a.length || j < b.length)
// If we're done with a, just gobble up b (but don't add duplicates)
if(i >= a.length)
if(result[lastIndex] != b[j])
result[++lastIndex] = b[j]
j++
continue
// If we're done with b, just gobble up a (but don't add duplicates)
if(j >= b.length)
if(result[lastIndex] != a[i])
result[++lastIndex] = a[i]
i++
continue
int smallestVal
// Choose the smaller of a or b
if(a[i] < b[j])
smallestVal = a[i++]
else
smallestVal = b[j++]
// Don't insert duplicates
if(result[lastIndex] != smallestVal)
result[++lastIndex] = smallestVal
end while

private static void MergeTwoSortedArray(int[] first, int[] second)
{
//throw new NotImplementedException();
int[] result = new int[first.Length + second.Length];
int i=0 , j=0 , k=0;
while(i < first.Length && j <second.Length)
{
if(first[i] < second[j])
{
result[k++] = first[i++];
}
else
{
result[k++] = second[j++];
}
}
if (i < first.Length)
{
for (int a = i; a < first.Length; a++)
result[k] = first[a];
}
if (j < second.Length)
{
for (int a = j; a < second.Length; a++)
result[k++] = second[a];
}
foreach (int a in result)
Console.Write(a + " ");
Console.WriteLine();
}

Using iterators and streaming interface the task is not that complicated:
class MergeTwoSortedLists
{
static void Main(string[] args) {
var list1 = new List<int?>() {
1,3,5,9,11
};
var list2 = new List<int?>() {
2,5,6,11,15,17,19,29
};
foreach (var c in SortedAndMerged(list1.GetEnumerator(), list2.GetEnumerator())) {
Console.Write(c+" ");
}
Console.ReadKey();
}
private static IEnumerable<int> SortedAndMerged(IEnumerator<int?> e1, IEnumerator<int?> e2) {
e2.MoveNext();
e1.MoveNext();
do {
while (e1.Current < e2.Current) {
if (e1.Current != null) yield return e1.Current.Value;
e1.MoveNext();
}
if (e2.Current != null) yield return e2.Current.Value;
e2.MoveNext();
} while (!(e1.Current == null && e2.Current == null));
}
}

Try this:
public static IEnumerable<T> MergeWith<T>(IEnumerable<T> collection1, IEnumerable<T> collection2,
IComparer<T> comparer)
{
using (var enumerator1 = collection1.GetEnumerator())
using (var enumerator2 = collection2.GetEnumerator())
{
var isMoveNext1 = enumerator1.MoveNext();
var isMoveNext2 = enumerator2.MoveNext();
do
{
while (comparer.Compare(enumerator1.Current, enumerator2.Current) < 0 || !isMoveNext2)
{
if (isMoveNext1)
yield return enumerator1.Current;
else
break;
isMoveNext1 = enumerator1.MoveNext();
}
if (isMoveNext2)
yield return enumerator2.Current;
isMoveNext2 = enumerator2.MoveNext();
} while (isMoveNext1 || isMoveNext2);
}
}

You could write a loop that merges and de-dups the lists and uses a binary-search approach to insert new values into the destination list.

var listOne = new List<int> { 3, 4, 1, 2, 7, 6, 9, 11 };
var listTwo = new List<int> { 1, 7, 8, 3, 5, 10, 15, 12 };
var result = listOne.ToList();
foreach (var n in listTwo)
{
if (result.IndexOf(n) == -1)
result.Add(n);
}

The closest solution I see would be to allocate an array knowing that integers are bounded to some value.
int[] values = new int[ Integer.MAX ]; // initialize with 0
int size1 = list1.size();
int size2 = list2.size();
for( int pos = 0; pos < size1 + size2 ; pos++ )
{
int val = pos > size1 ? list2[ pos-size1 ] : list1[ pos ] ;
values[ val ]++;
}
Then you can argue that you have the sorted array in a "special" form :-) To get a clean sorted array, you need to traverse the values array, skip all position with 0 count, and build the final list.

This will only work for lists of integers, but happily that is what you have!
List<int> sortedList = new List<int>();
foreach (int x in listOne)
{
sortedList<x> = x;
}
foreach (int x in listTwo)
{
sortedList<x> = x;
}
This is using the values in each list as the index position at which to store the value. Any duplicate values will overwrite the previous entry at that index position. It meets the requirement of only one iteration over the values.
It does of course mean that there will be 'empty' positions in the list.
I suspect the job position has been filled by now though.... :-)

Find the first occurrence/starting index of the sub-array in C#

Given two arrays as parameters (x and y) and find the starting index where the first occurrence of y in x. I am wondering what the simplest or the fastest implementation would be.
Example:
when x = {1,2,4,2,3,4,5,6}
y = {2,3}
result
starting index should be 3
Update: Since my code is wrong I removed it from the question.

Simplest to write?
return (from i in Enumerable.Range(0, 1 + x.Length - y.Length)
where x.Skip(i).Take(y.Length).SequenceEqual(y)
select (int?)i).FirstOrDefault().GetValueOrDefault(-1);
Not quite as efficient, of course... a bit more like it:
private static bool IsSubArrayEqual(int[] x, int[] y, int start) {
for (int i = 0; i < y.Length; i++) {
if (x[start++] != y[i]) return false;
}
return true;
}
public static int StartingIndex(this int[] x, int[] y) {
int max = 1 + x.Length - y.Length;
for(int i = 0 ; i < max ; i++) {
if(IsSubArrayEqual(x,y,i)) return i;
}
return -1;
}

Here is a simple (yet fairly efficient) implementation that finds all occurances of the array, not just the first one:
static class ArrayExtensions {
public static IEnumerable<int> StartingIndex(this int[] x, int[] y) {
IEnumerable<int> index = Enumerable.Range(0, x.Length - y.Length + 1);
for (int i = 0; i < y.Length; i++) {
index = index.Where(n => x[n + i] == y[i]).ToArray();
}
return index;
}
}
Example:
int[] x = { 1, 2, 3, 4, 1, 2, 3, 4, 1, 2, 3, 4 };
int[] y = { 2, 3 };
foreach (int i in x.StartingIndex(y)) {
Console.WriteLine(i);
}
Output:
1
5
9
The method first loops through the x array to find all occurances of the first item in the y array, and place the index of those in the index array. Then it goes on to reduce the matches by checking which of those also match the second item in the y array. When all items in the y array is checked, the index array contains only the full matches.
Edit:
An alternative implementation would be to remove the ToArray call from the statement in the loop, making it just:
index = index.Where(n => x[n + i] == y[i]);
This would totally change how the method works. Instead of looping through the items level by level, it would return an enumerator with nested expressions, deferring the search to the time when the enumerator was iterated. That means that you could get only the first match if you wanted:
int index = x.StartingIndex(y).First();
This would not find all matches and then return the first, it would just search until the first was found and then return it.

The simplest way is probably this:
public static class ArrayExtensions
{
private static bool isMatch(int[] x, int[] y, int index)
{
for (int j = 0; j < y.Length; ++j)
if (x[j + index] != y[j]) return false;
return true;
}
public static int IndexOf(this int[] x, int[] y)
{
for (int i = 0; i < x.Length - y.Length + 1; ++i)
if (isMatch(x, y, i)) return i;
return -1;
}
}
But it's definitely not the fastest way.

This is based off of Mark Gravell's answer but I made it generic and added some simple bounds checking to keep exceptions from being thrown
private static bool IsSubArrayEqual<T>(T[] source, T[] compare, int start) where T:IEquatable<T>
{
if (compare.Length > source.Length - start)
{
//If the compare string is shorter than the test area it is not a match.
return false;
}
for (int i = 0; i < compare.Length; i++)
{
if (source[start++].Equals(compare[i]) == false) return false;
}
return true;
}
Could be improved further by implementing Boyer-Moore but for short patterns it works fine.

"Simplest" and "fastest" are opposites in this case, and besides, in order to describe fast algorithms we need to know lots of things about how the source array and the search array are related to each other.
This is essentially the same problem as finding a substring inside a string. Suppose you are looking for "fox" in "the quick brown fox jumps over the lazy dog". The naive string matching algorithm is extremely good in this case. If you are searching for "bananananananananananananananana" inside a million-character string that is of the form "banananananabanananabananabananabanananananbananana..." then the naive substring matching algorithm is terrible -- far faster results can be obtained by using more complex and sophisticated string matching algorithms. Basically, the naive algorithm is O(nm) where n and m are the lengths of the source and search strings. There are O(n+m) algorithms but they are far more complex.
Can you tell us more about the data you're searching? How big is it, how redundant is it, how long are the search arrays, and what is the likelihood of a bad match?

I find something along the following lines more intuitive, but that may be a matter of taste.
public static class ArrayExtensions
{
public static int StartingIndex(this int[] x, int[] y)
{
var xIndex = 0;
while(xIndex < x.length)
{
var found = xIndex;
var yIndex = 0;
while(yIndex < y.length && xIndex < x.length && x[xIndex] == y[yIndex])
{
xIndex++;
yIndex++;
}
if(yIndex == y.length-1)
{
return found;
}
xIndex = found + 1;
}
return -1;
}
}
This code also addresses an issue I believe your implementation may have in cases like x = {3, 3, 7}, y = {3, 7}. I think what would happen with your code is that it matches the first number, then resets itself on the second, but starts matching again on the third, rather than stepping back to the index just after where it started matching. May be missing something, but it's definitely something to consider and should be easily fixable in your code.

//this is the best in C#
//bool contains(array,subarray)
// when find (subarray[0])
// while subarray[next] IS OK
// subarray.end then Return True
public static bool ContainSubArray<T>(T[] findIn, out int found_index,
params T[]toFind)
{
found_index = -1;
if (toFind.Length < findIn.Length)
{
int index = 0;
Func<int, bool> NextOk = (i) =>
{
if(index < findIn.Length-1)
return findIn[++index].Equals(toFind[i]);
return false;
};
//----------
int n=0;
for (; index < findIn.Length; index++)
{
if (findIn[index].Equals(toFind[0]))
{
found_index=index;n=1;
while (n < toFind.Length && NextOk(n))
n++;
}
if (n == toFind.Length)
{
return true;
}
}
}
return false;
}

using System;
using System.Linq;
public class Test
{
public static void Main()
{
int[] x = {1,2,4,2,3,4,5,6};
int[] y = {2,3};
int? index = null;
for(int i=0; i<x.Length; ++i)
{
if (y.SequenceEqual(x.Skip(i).Take(y.Length)))
{
index = i;
break;
}
}
Console.WriteLine($"{index}");
}
}
Output
3