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Figure out max number of consecutive seats

I had an interviewer ask me to write a program in c# to figure out the max number of 4 members families that can sit consecutively in a venue, taking into account that the 4 members must be consecutively seated in one single row, with the following context:
N represents the number of rows availabe.
The Columns are labeled from the letter "A" to "K", purposely ommiting the letter "i" (in other words, {A,B,C,D,E,F,G,H,J,K})
M represents a list of reserved seats
Quick example:
N = 2
M = {"1A","2F","1C"}
Solution = 3
In the representation you can see that, with the reservations and the size given, only three families of 4 can be seated in a consecutive order.
How would you solve this? is it possible to not use for loops? (Linq solutions)
I got mixed up in the for loops when trying to deal with the reservations aray: My idea was to obtain all the reservations that a row has, but then I don't really know how to deal with the letters (Converting directly from letter to number is a no go because the missing "I") and you kinda need the letters to position the reserved sits anyway.
Any approach or insight on how to go about this problem would be nice.
Thanks in advance!

Here is another implementation.
I also tried to explain why certain things have been done.
Good luck.
private static int GetNumberOfAvailablePlacesForAFamilyOfFour(int numberOfRows, string[] reservedSeats)
{
// By just declaring the column names as a string of the characters
// we can query the column index by colulmnNames.IndexOf(char)
string columnNames = "ABCDEFGHJK";
// Here we transform the reserved seats to a matrix
// 1A 2F 1C becomes
// reservedSeatMatrix[0] = [0, 2] -> meaning row 1 and columns A and C, indexes 0 and 2
// reservedSeatMatrix[1] = [5] -> meaning row 2 and column F, index 5
List<List<int>> reservedSeatMatrix = new List<List<int>>();
for (int row = 0; row < numberOfRows; row++)
{
reservedSeatMatrix.Add(new List<int>());
}
foreach (string reservedSeat in reservedSeats)
{
int seatRow = Convert.ToInt32(reservedSeat.Substring(0, reservedSeat.Length - 1));
int seatColumn = columnNames.IndexOf(reservedSeat[reservedSeat.Length - 1]);
reservedSeatMatrix[seatRow - 1].Add(seatColumn);
}
// Then comes the evaluation.
// Which is simple enough to read.
int numberOfAvailablePlacesForAFamilyOfFour = 0;
for (int row = 0; row < numberOfRows; row++)
{
// Reset the number of consecutive seats at the beginning of a new row
int numberOfConsecutiveEmptySeats = 0;
for (int column = 0; column < columnNames.Length; column++)
{
if (reservedSeatMatrix[row].Contains(column))
{
// reset when a reserved seat is reached
numberOfConsecutiveEmptySeats = 0;
continue;
}
numberOfConsecutiveEmptySeats++;
if(numberOfConsecutiveEmptySeats == 4)
{
numberOfAvailablePlacesForAFamilyOfFour++;
numberOfConsecutiveEmptySeats = 0;
}
}
}
return numberOfAvailablePlacesForAFamilyOfFour;
}
static void Main(string[] args)
{
int familyPlans = GetNumberOfAvailablePlacesForAFamilyOfFour(2, new string[] { "1A", "2F", "1C" });
}

Good luck on your interview
As always, you will be asked how could you improve that? So you'd consider complexity stuff like O(N), O(wtf).
Underlying implementation would always need for or foreach. Just importantly, never do unnecessary in a loop. For example, if there's only 3 seats left in a row, you don't need to keep hunting on that row because it is not possible to find any.
This might help a bit:
var n = 2;
var m = new string[] { "1A", "2F", "1C" };
// We use 2 dimension bool array here. If it is memory constraint, we can use BitArray.
var seats = new bool[n, 10];
// If you just need the count, you don't need a list. This is for returning more information.
var results = new List<object>();
// Set reservations.
foreach (var r in m)
{
var row = r[0] - '1';
// If it's after 'H', then calculate index based on 'J'.
// 8 is index of J.
var col = r[1] > 'H' ? (8 + r[1] - 'J') : r[1] - 'A';
seats[row, col] = true;
}
// Now you should all reserved seats marked as true.
// This is O(N*M) where N is number of rows, M is number of columns.
for (int row = 0; row < n; row++)
{
int start = -1;
int length = 0;
for (int col = 0; col < 10; col++)
{
if (start < 0)
{
if (!seats[row, col])
{
// If there's no consecutive seats has started, and current seat is available, let's start!
start = col;
length = 1;
}
}
else
{
// If have started, check if we could have 4 seats.
if (!seats[row, col])
{
length++;
if (length == 4)
{
results.Add(new { row, start });
start = -1;
length = 0;
}
}
else
{
// // We won't be able to reach 4 seats, so reset
start = -1;
length = 0;
}
}
if (start < 0 && col > 6)
{
// We are on column H now (only have 3 seats left), and we do not have a consecutive sequence started yet,
// we won't be able to make it, so break and continue next row.
break;
}
}
}
var solution = results.Count;
LINQ, for and foreach are similar things. It is possible you could wrap the above into a custom iterator like:
class ConsecutiveEnumerator : IEnumerable
{
public IEnumerator GetEnumerator()
{
}
}
Then you could start using LINQ.

If you represent your matrix in simple for developers format, it will be easier. You can accomplish it either by dictionary or perform not so complex mapping by hand. In any case this will calculate count of free consecutive seats:
public static void Main(string[] args)
{
var count = 0;//total count
var N = 2; //rows
var M = 10; //columns
var familySize = 4;
var matrix = new []{Tuple.Create(0,0),Tuple.Create(1,5), Tuple.Create(0,2)}.OrderBy(x=> x.Item1).ThenBy(x=> x.Item2).GroupBy(x=> x.Item1, x=> x.Item2);
foreach(var row in matrix)
{
var prevColumn = -1;
var currColumn = 0;
var free = 0;
var div = 0;
//Instead of enumerating entire matrix, we just calculate intervals in between reserved seats.
//Then we divide them by family size to know how many families can be contained within
foreach(var column in row)
{
currColumn = column;
free = (currColumn - prevColumn - 1)/familySize;
count += free;
prevColumn = currColumn;
}
currColumn = M;
free = (currColumn - prevColumn - 1)/familySize;
count += free;
}
Console.WriteLine("Result: {0}", count);
}

Calculate all possible permutations/combinations, then check if the result is equal to a value

Best way I can explain it is using an example:
You are visiting a shop with $2000, your goal is to have $0 at the end of your trip.
You do not know how many items are going to be available, nor how much they cost.
Say that there are currently 3 items costing $1000, $750, $500.
(The point is to calculate all possible solutions, not the most efficient one.)
You can spend $2000, this means:
You can buy the $1000 item 0, 1 or 2 times.
You can buy the $750 item 0, 1 or 2 times.
You can buy the $500 item 0, 1, 2, 3 or 4 times.
At the end I need to be able to have all solutions, in this case it will be
2*$1000
1*$1000 and 2*$500
2*$750 and 1*$500
4*$500
Side note: you can't have a duplicate solution (like this)
1*$1000 and 2*$500
2*$500 and 1*$1000
This is what I tried:
You first call this function using
goalmoney = convert.ToInt32(goalMoneyTextBox.Text);
totalmoney = Convert.ToInt32(totalMoneyTextBox.Text);
int[] list = new int[usingListBox.Items.Count];
Calculate(0, currentmoney, list);
The function:
public void Calculate(int level, int money, int[] list)
{
string item = usingListBox.Items[level].ToString();
int cost = ItemDict[item];
for (int i = 0; i <= (totalmoney / cost); i++)
{
int[] templist = list;
int tempmoney = money - (cost * i);
templist[level] = i;
if (tempmoney == goalmoney)
{
resultsFound++;
}
if (level < usingListBox.Items.Count - 1 && tempmoney != goalmoney) Calculate(level + 1, tempmoney, templist);
}
}

Your problem can be reduced to a well known mathematical problem labeled Frobenius equation which is closely related to the well known Coin problem. Suppose you have N items, where i-th item costs c[i] and you need to spent exactly S$. So you need to find all non negative integer solutions (or decide whether there are no solutions at all) of equation
c[1]*n[1] + c[2]*n[2] + ... + c[N]*n[N] = S
where all n[i] are unknown variables and each n[i] is the number of bought items of i-th type.
This equation can be solved in a various ways. The following function allSolutions (I suppose it can be additionally simplified) finds all solutions of a given equation:
public static List<int[]> allSolutions(int[] system, int total) {
ArrayList<int[]> all = new ArrayList<>();
int[] solution = new int[system.length];//initialized by zeros
int pointer = system.length - 1, temp;
out:
while (true) {
do { //the following loop can be optimized by calculation of remainder
++solution[pointer];
} while ((temp = total(system, solution)) < total);
if (temp == total && pointer != 0)
all.add(solution.clone());
do {
if (pointer == 0) {
if (temp == total) //not lose the last solution!
all.add(solution.clone());
break out;
}
for (int i = pointer; i < system.length; ++i)
solution[i] = 0;
++solution[--pointer];
} while ((temp = total(system, solution)) > total);
pointer = system.length - 1;
if (temp == total)
all.add(solution.clone());
}
return all;
}
public static int total(int[] system, int[] solution) {
int total = 0;
for (int i = 0; i < system.length; ++i)
total += system[i] * solution[i];
return total;
}
In the above code system is array of coefficients c[i] and total is S. There is an obvious restriction: system should have no any zero elements (this lead to infinite number of solutions). A slight modification of the above code avoids this restriction.

Assuming you have class Product which exposes a property called Price, this is a way to do it:
public List<List<Product>> GetAffordableCombinations(double availableMoney, List<Product> availableProducts)
{
List<Product> sortedProducts = availableProducts.OrderByDescending(p => p.Price).ToList();
//we have to cycle through the list multiple times while keeping track of the current
//position in each subsequent cycle. we're using a list of integers to save these positions
List<int> layerPointer = new List<int>();
layerPointer.Add(0);
int currentLayer = 0;
List<List<Product>> affordableCombinations = new List<List<Product>>();
List<Product> tempList = new List<Product>();
//when we went through all product on the top layer, we're done
while (layerPointer[0] < sortedProducts.Count)
{
//take the product in the current position on the current layer
var currentProduct = sortedProducts[layerPointer[currentLayer]];
var currentSum = tempList.Sum(p => p.Price);
if ((currentSum + currentProduct.Price) <= availableMoney)
{
//if the sum doesn't exeed our maximum we add that prod to a temp list
tempList.Add(currentProduct);
//then we advance to the next layer
currentLayer++;
//if it doesn't exist, we create it and set the 'start product' on that layer
//to the current product of the current layer
if (currentLayer >= layerPointer.Count)
layerPointer.Add(layerPointer[currentLayer - 1]);
}
else
{
//if the sum would exeed our maximum we move to the next prod on the current layer
layerPointer[currentLayer]++;
if (layerPointer[currentLayer] >= sortedProducts.Count)
{
//if we've reached the end of the list on the current layer,
//there are no more cheaper products to add, and this cycle is complete
//so we add the list we have so far to the possible combinations
affordableCombinations.Add(tempList);
tempList = new List<Product>();
//move to the next product on the top layer
layerPointer[0]++;
currentLayer = 0;
//set the current products on each subsequent layer to the current of the top layer
for (int i = 1; i < layerPointer.Count; i++)
{
layerPointer[i] = layerPointer[0];
}
}
}
}
return affordableCombinations;
}

Find the contiguous sequence with the largest product in an integer array

I have come up with the code below but that doesn't satisfy all cases, e.g.:
Array consisting all 0's
Array having negative values(it's bit tricky since it's about finding product as two negative ints give positive value)
public static int LargestProduct(int[] arr)
{
//returning arr[0] if it has only one element
if (arr.Length == 1) return arr[0];
int product = 1;
int maxProduct = Int32.MinValue;
for (int i = 0; i < arr.Length; i++)
{
//this block store the largest product so far when it finds 0
if (arr[i] == 0)
{
if (maxProduct < product)
{
maxProduct = product;
}
product = 1;
}
else
{
product *= arr[i];
}
}
if (maxProduct > product)
return maxProduct;
else
return product;
}
How can I incorporate the above cases/correct the code. Please suggest.

I am basing my answer on the assumption that if you have more than 1 element in the array, you would want to multiply at least 2 contiguous integers for checking the output, i.e. in array of {-1, 15}, the output that you want is -15 and not 15).
The problem that we need to solve is to look at all possible multiplication combinations and find out the max product out of them.
The total number of products in an array of n integers would be nC2 i.e. if there are 2 elements, then the total multiplication combinations would be 1, for 3, it would be 3, for 4, it would be 6 and so on.
For each number that we have in the incoming array, it has to multiply with all the multiplications that we did with the last element and keep the max product till now and if we do it for all the elements, at the end we would be left with the maximum product.
This should work for negatives and zeros.
public static long LargestProduct(int[] arr)
{
if (arr.Length == 1)
return arr[0];
int lastNumber = 1;
List<long> latestProducts = new List<long>();
long maxProduct = Int64.MinValue;
for (int i = 0; i < arr.Length; i++)
{
var item = arr[i];
var latest = lastNumber * item;
var temp = new long[latestProducts.Count];
latestProducts.CopyTo(temp);
latestProducts.Clear();
foreach (var p in temp)
{
var product = p * item;
if (product > maxProduct)
maxProduct = product;
latestProducts.Add(product);
}
if (i != 0)
{
if (latest > maxProduct)
maxProduct = latest;
latestProducts.Add(latest);
}
lastNumber = item;
}
return maxProduct;
}
If you want the maximum product to also incorporate the single element present in the array i.e. {-1, 15} should written 15, then you can compare the max product with the element of the array being processed and that should give you the max product if the single element is the max number.
This can be achieved by adding the following code inside the for loop at the end.
if (item > maxProduct)
maxProduct = item;

Your basic problem is 2 parts. Break them down and solving it becomes easier.
1) Find all contiguous subsets.
Since your source sequence can have negative values, you are not all that equipped to make any value judgments until you're found each subset, as a negative can later be "cancelled" by another. So let the first phase be to only find the subsets.
An example of how you might do this is the following code
// will contain all contiguous subsets
var sequences = new List<Tuple<bool, List<int>>>();
// build subsets
foreach (int item in source)
{
var deadCopies = new List<Tuple<bool, List<int>>>();
foreach (var record in sequences.Where(r => r.Item1 && !r.Item2.Contains(0)))
{
// make a copy that is "dead"
var deadCopy = new Tuple<bool, List<int>>(false, record.Item2.ToList());
deadCopies.Add(deadCopy);
record.Item2.Add(item);
}
sequences.Add(new Tuple<bool, List<int>>(true, new List<int> { item }));
sequences.AddRange(deadCopies);
}
In the above code, I'm building all my contiguous subsets, while taking the liberty of not adding anything to a given subset that already has a 0 value. You can omit that particular behavior if you wish.
2) Calculate each subset's product and compare that to a max value.
Once you have found all of your qualifying subsets, the next part is easy.
// find subset with highest product
int maxProduct = int.MinValue;
IEnumerable<int> maxSequence = Enumerable.Empty<int>();
foreach (var record in sequences)
{
int product = record.Item2.Aggregate((a, b) => a * b);
if (product > maxProduct)
{
maxProduct = product;
maxSequence = record.Item2;
}
}
Add whatever logic you wish to restrict the length of the original source or the subset candidates or product values. For example, if you wish to enforce minimum length requirements on either, or if a subset product of 0 is allowed if a non-zero product is available.
Also, I make no claims as to the performance of the code, it is merely to illustrate breaking the problem down into its parts.

I think you should have 2 products at the same time - they will differ in signs.
About case, when all values are zero - you can check at the end if maxProduct is still Int32.MinValue (if Int32.MinValue is really not possible)
My variant:
int maxProduct = Int32.MinValue;
int? productWithPositiveStart = null;
int? productWithNegativeStart = null;
for (int i = 0; i < arr.Length; i++)
{
if (arr[i] == 0)
{
productWithPositiveStart = null;
productWithNegativeStart = null;
}
else
{
if (arr[i] > 0 && productWithPositiveStart == null)
{
productWithPositiveStart = arr[i];
}
else if (productWithPositiveStart != null)
{
productWithPositiveStart *= arr[i];
maxProduct = Math.max(maxProduct, productWithPositiveStart);
}
if (arr[i] < 0 && productWithNegativeStart == null)
{
productWithNegativeStart = arr[i];
}
else if (productWithNegativeStart != null)
{
productWithNegativeStart *= arr[i];
maxProduct = Math.max(maxProduct, productWithNegativeStart);
}
maxProduct = Math.max(arr[i], maxProduct);
}
}
if (maxProduct == Int32.MinValue)
{
maxProduct = 0;
}

At a high level, your current algorithm splits the array upon a 0 and returns the largest contiguous product of these sub-arrays. Any further iterations will be on the process of finding the largest contiguous product of a sub-array where no elements are 0.
To take into account negative numbers, we obviously first need to test if the product of one of these sub-arrays is negative, and take some special action if it is.
The negative result comes from an odd number of negative values, so we need to remove one of these negative values to make the result positive again. To do this we remove all elements up the the first negative number, or the last negative number and all elements after that, whichever results in the highest product.
To take into account an array of all 0's, simply use 0 as your starting maxProduct. If the array is a single negative value, you're special case handling of a single element will mean that is returned. After that, there will always be a positive sub-sequence product, or else the whole array is 0 and it should return 0 anyway.

it can be done in O(N). it is based on the simple idea: calculate the minimum (minCurrent) and maximum (maxCurrent) till i. This can be easily changed to fit for the condition like: {0,0,-2,0} or {-2,-3, -8} or {0,0}
a[] = {6, -3, 2, 0, 3, -2, -4, -2, 4, 5}
steps of the algorithm given below for the above array a :
private static int getMaxProduct(int[] a) {
if (a.length == 0) {
throw new IllegalArgumentException();
}
int minCurrent = 1, maxCurrent = 1, max = Integer.MIN_VALUE;
for (int current : a) {
if (current > 0) {
maxCurrent = maxCurrent * current;
minCurrent = Math.min(minCurrent * current, 1);
} else if (current == 0) {
maxCurrent = 1;
minCurrent = 1;
} else {
int x = maxCurrent;
maxCurrent = Math.max(minCurrent * current, 1);
minCurrent = x * current;
}
if (max < maxCurrent) {
max = maxCurrent;
}
}
//System.out.println(minCurrent);
return max;
}

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);
}

C#: Cleanest way to divide a string array into N instances N items long

I know how to do this in an ugly way, but am wondering if there is a more elegant and succinct method.
I have a string array of e-mail addresses. Assume the string array is of arbitrary length -- it could have a few items or it could have a great many items. I want to build another string consisting of say, 50 email addresses from the string array, until the end of the array, and invoke a send operation after each 50, using the string of 50 addresses in the Send() method.
The question more generally is what's the cleanest/clearest way to do this kind of thing. I have a solution that's a legacy of my VBScript learnings, but I'm betting there's a better way in C#.

You want elegant and succinct, I'll give you elegant and succinct:
var fifties = from index in Enumerable.Range(0, addresses.Length)
group addresses[index] by index/50;
foreach(var fifty in fifties)
Send(string.Join(";", fifty.ToArray());
Why mess around with all that awful looping code when you don't have to? You want to group things by fifties, then group them by fifties.
That's what the group operator is for!
UPDATE: commenter MoreCoffee asks how this works. Let's suppose we wanted to group by threes, because that's easier to type.
var threes = from index in Enumerable.Range(0, addresses.Length)
group addresses[index] by index/3;
Let's suppose that there are nine addresses, indexed zero through eight
What does this query mean?
The Enumerable.Range is a range of nine numbers starting at zero, so 0, 1, 2, 3, 4, 5, 6, 7, 8.
Range variable index takes on each of these values in turn.
We then go over each corresponding addresses[index] and assign it to a group.
What group do we assign it to? To group index/3. Integer arithmetic rounds towards zero in C#, so indexes 0, 1 and 2 become 0 when divided by 3. Indexes 3, 4, 5 become 1 when divided by 3. Indexes 6, 7, 8 become 2.
So we assign addresses[0], addresses[1] and addresses[2] to group 0, addresses[3], addresses[4] and addresses[5] to group 1, and so on.
The result of the query is a sequence of three groups, and each group is a sequence of three items.
Does that make sense?
Remember also that the result of the query expression is a query which represents this operation. It does not perform the operation until the foreach loop executes.

Seems similar to this question: Split a collection into n parts with LINQ?
A modified version of Hasan Khan's answer there should do the trick:
public static IEnumerable<IEnumerable<T>> Chunk<T>(
this IEnumerable<T> list, int chunkSize)
{
int i = 0;
var chunks = from name in list
group name by i++ / chunkSize into part
select part.AsEnumerable();
return chunks;
}
Usage example:
var addresses = new[] { "a#example.com", "b#example.org", ...... };
foreach (var chunk in Chunk(addresses, 50))
{
SendEmail(chunk.ToArray(), "Buy V14gr4");
}

It sounds like the input consists of separate email address strings in a large array, not several email address in one string, right? And in the output, each batch is a single combined string.
string[] allAddresses = GetLongArrayOfAddresses();
const int batchSize = 50;
for (int n = 0; n < allAddresses.Length; n += batchSize)
{
string batch = string.Join(";", allAddresses, n,
Math.Min(batchSize, allAddresses.Length - n));
// use batch somehow
}

Assuming you are using .NET 3.5 and C# 3, something like this should work nicely:
string[] s = new string[] {"1", "2", "3", "4"....};
for (int i = 0; i < s.Count(); i = i + 50)
{
string s = string.Join(";", s.Skip(i).Take(50).ToArray());
DoSomething(s);
}

I would just loop through the array and using StringBuilder to create the list (I'm assuming it's separated by ; like you would for email). Just send when you hit mod 50 or the end.
void Foo(string[] addresses)
{
StringBuilder sb = new StringBuilder();
for (int i = 0; i < addresses.Length; i++)
{
sb.Append(addresses[i]);
if ((i + 1) % 50 == 0 || i == addresses.Length - 1)
{
Send(sb.ToString());
sb = new StringBuilder();
}
else
{
sb.Append("; ");
}
}
}
void Send(string addresses)
{
}

I think we need to have a little bit more context on what exactly this list looks like to give a definitive answer. For now I'm assuming that it's a semicolon delimeted list of email addresses. If so you can do the following to get a chunked up list.
public IEnumerable<string> DivideEmailList(string list) {
var last = 0;
var cur = list.IndexOf(';');
while ( cur >= 0 ) {
yield return list.SubString(last, cur-last);
last = cur + 1;
cur = list.IndexOf(';', last);
}
}
public IEnumerable<List<string>> ChunkEmails(string list) {
using ( var e = DivideEmailList(list).GetEnumerator() ) {
var list = new List<string>();
while ( e.MoveNext() ) {
list.Add(e.Current);
if ( list.Count == 50 ) {
yield return list;
list = new List<string>();
}
}
if ( list.Count != 0 ) {
yield return list;
}
}
}

I think this is simple and fast enough.The example below divides the long sentence into 15 parts,but you can pass batch size as parameter to make it dynamic.Here I simply divide using "/n".
private static string Concatenated(string longsentence)
{
const int batchSize = 15;
string concatanated = "";
int chanks = longsentence.Length / batchSize;
int currentIndex = 0;
while (chanks > 0)
{
var sub = longsentence.Substring(currentIndex, batchSize);
concatanated += sub + "/n";
chanks -= 1;
currentIndex += batchSize;
}
if (currentIndex < longsentence.Length)
{
int start = currentIndex;
var finalsub = longsentence.Substring(start);
concatanated += finalsub;
}
return concatanated;
}
This show result of split operation.
var parts = Concatenated(longsentence).Split(new string[] { "/n" }, StringSplitOptions.None);

Extensions methods based on Eric's answer:
public static IEnumerable<IEnumerable<T>> SplitIntoChunks<T>(this T[] source, int chunkSize)
{
var chunks = from index in Enumerable.Range(0, source.Length)
group source[index] by index / chunkSize;
return chunks;
}
public static T[][] SplitIntoArrayChunks<T>(this T[] source, int chunkSize)
{
var chunks = from index in Enumerable.Range(0, source.Length)
group source[index] by index / chunkSize;
return chunks.Select(e => e.ToArray()).ToArray();
}