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How do you do this in C# without using List?

I am new to C#. The following code was a solution I came up to solve a challenge. I am unsure how to do this without using List since my understanding is that you can't push to an array in C# since they are of fixed size.
Is my understanding of what I said so far correct?
Is there a way to do this that doesn't involve creating a new array every time I need to add to an array? If there is no other way, how would I create a new array when the size of the array is unknown before my loop begins?
Return a sorted array of all non-negative numbers less than the given n which are divisible both by 3 and 4. For n = 30, the output should be
threeAndFour(n) = [0, 12, 24].
int[] threeAndFour(int n) {
List<int> l = new List<int>(){ 0 };
for (int i = 12; i < n; ++i)
if (i % 12 == 0)
l.Add(i);
return l.ToArray();
}
EDIT: I have since refactored this code to be..
int[] threeAndFour(int n) {
List<int> l = new List<int>(){ 0 };
for (int i = 12; i < n; i += 12)
l.Add(i);
return l.ToArray();
}

A. Lists is OK
If you want to use a for to find out the numbers, then List is the appropriate data structure for collecting the numbers as you discover them.
B. Use more maths
static int[] threeAndFour(int n) {
var a = new int[(n / 12) + 1];
for (int i = 12; i < n; i += 12) a[i/12] = i;
return a;
}
C. Generator pattern with IEnumerable<int>
I know that this doesn't return an array, but it does avoid a list.
static IEnumerable<int> threeAndFour(int n) {
yield return 0;
for (int i = 12; i < n; i += 12)
yield return i;
}
D. Twist and turn to avoid a list
The code could for twice. First to figure the size or the array, and then to fill it.
int[] threeAndFour(int n) {
// Version: A list is really undesirable, arrays are great.
int size = 1;
for (int i = 12; i < n; i += 12)
size++;
var a = new int[size];
a[0] = 0;
int counter = 1;
for (int i = 12; i < n; i += 12) a[counter++] = i;
}

if (i % 12 == 0)
So you have figured out that the numbers which divides both 3 and 4 are precisely those numbers that divides 12.
Can you figure out how many such numbers there are below a given n? - Can you do so without counting the numbers - if so there is no need for a dynamically growing container, you can just initialize the container to the correct size.
Once you have your array just keep track of the next index to fill.

You could use Linq and Enumerable.Range method for the purpose. For example,
int[] threeAndFour(int n)
{
return Enumerable.Range(0,n).Where(x=>x%12==0).ToArray();
}
Enumerable.Range generates a sequence of integral numbers within a specified range, which is then filtered on the condition (x%12==0) to retrieve the desired result.

Since you know this goes in steps of 12 and you know how many there are before you start, you can do:
Enumerable.Range(0,n/12+1).Select(x => x*12).ToArray();

I am unsure how to do this without using List since my understanding is that you can't push to an array in C# since they are of fixed size.
It is correct that arrays can not grow. List were invented as a wrapper around a array that automagically grows whenever needed. Note that you can give List a integer via the Constructor, wich will tell it the minimum size it should expect. It will allocate at least that much the first time. This can limit growth related overhead.
And dictionaries are just a variation of the list mechanics, with Hash Table key search speed.
There is only 1 other Collection I know of that can grow. However it is rarely mentioned outside of theory and some very specific cases:
Linked Lists. The linked list has a unbeatable growth performance and the lowest issue of running into OutOfMemory Exceptions due to Fragmentation. Unfortunately, their random access times are the worst as a result. Unless you can process those collections exclusively sequentally from the start (or sometimes the end), their performance will be abysmal. Only stacks and queues are likely to use them. There is however still a implementation you could use in .NET: https://learn.microsoft.com/en-us/dotnet/api/system.collections.generic.linkedlist-1
Your code holds some potential too:
for (int i = 12; i < n; ++i)
if (i % 12 == 0)
l.Add(i);
It would way more effective to count up by 12 every itteration - you are only interested in every 12th number after all. You may have to change the loop, but I think a do...while would do. Also the array/minimum List size is easily predicted: Just divide n by 12 and add 1. But I asume that is mostly mock-up code and it is not actually that deterministic.

List generally works pretty well, as I understand your question you have challenged yourself to solve a problem without using the List class. An array (or List) uses a contiguous block of memory to store elements. Arrays are of fixed size. List will dynamically expand to accept new elements but still keeps everything in a single block of memory.
You can use a linked list https://learn.microsoft.com/en-us/dotnet/api/system.collections.generic.linkedlist-1?view=netframework-4.8 to produce a simulation of an array. A linked list allocates additional memory for each element (node) that is used to point to the next (and possibly the previous). This allows you to add elements without large block allocations, but you pay a space cost (increased use of memory) for each element added. The other problem with linked lists are you can't quickly access random elements. To get to element 5, you have to go through elements 0 through 4. There's a reason arrays and array like structures are favored for many tasks, but it's always interesting to try to do common things in a different way.

C# why does binarysearch have to be made on sorted arrays and lists?

C# why does binarysearch have to be made on sorted arrays and lists?
Is there any other method that does not require me to sort the list?
It kinda messes with my program in a way that I cannot sort the list for it to work as I want to.

A binary search works by dividing the list of candidates in half using equality. Imagine the following set:
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
We can also represent this as a binary tree, to make it easier to visualise:
Source
Now, say we want to find the number 3. We can do it like so:
Is 3 smaller than 8? Yes. OK, now we're looking at everything between 1 and 7.
Is 3 smaller than 4? Yes. OK, now we're looking at everything between 1 and 3.
Is 3 smaller than 2? No. OK, now we're looking at 3.
We found it!
Now, if your list isn't sorted, how will we divide the list in half? The simple answer is: we can't. If we swap 3 and 15 in the example above, it would work like this:
Is 3 smaller than 8? Yes. OK, now we're looking at everything between 1 and 7.
Is 3 smaller than 4? Yes. OK, now we're looking at everything between 1 and 3 (except we swapped it with 15).
Is 3 smaller than 2? No. OK, now we're looking at 15.
Huh? There's no more items to check but we didn't find it. I guess it's not in the list.
The solution is to use an appropriate data type instead. For fast lookups of key/value pairs, I'll use a Dictionary. For fast checks if something already exists, I'll use a HashSet. For general storage I'll use a List or an array.
Dictionary example:
var values = new Dictionary<int, string>();
values[1] = "hello";
values[2] = "goodbye";
var value2 = values[2]; // this lookup will be fast because Dictionaries are internally optimised inside and partition keys' hash codes into buckets.
HashSet example:
var mySet = new HashSet<int>();
mySet.Add(1);
mySet.Add(2);
if (mySet.Contains(2)) // this lookup is fast for the same reason as a dictionary.
{
// do something
}
List exmaple:
var list = new List<int>();
list.Add(1);
list.Add(2);
if (list.Contains(2)) // this isn't fast because it has to visit each item in the list, but it works OK for small sets or places where performance isn't so important
{
}
var idx2 = list.IndexOf(2);
If you have multiple values with the same key, you could store a list in a Dictionary like this:
var values = new Dictionary<int, List<string>>();
if (!values.ContainsKey(key))
{
values[key] = new List<string>();
}
values[key].Add("value1");
values[key].Add("value2");

There is no way you use binary search on unordered collections. Sorting collection is the main concept of the binary search. The key is that on every move u take the middle index between l and r. On first step they are 0 and size - 1, after every step one of them becomes middle index between them. If x > arr[m] then l becomes m + 1, otherwise r becomes m - 1. Basically, on every step you take half of the array you had and, of course, it remains sorted. This code is recursive, if you don't know what recursion is(which is very important in programming), you can review and learn here.
// C# implementation of recursive Binary Search
using System;
class GFG {
// Returns index of x if it is present in
// arr[l..r], else return -1
static int binarySearch(int[] arr, int l,
int r, int x)
{
if (r >= l) {
int mid = l + (r - l) / 2;
// If the element is present at the
// middle itself
if (arr[mid] == x)
return mid;
// If element is smaller than mid, then
// it can only be present in left subarray
if (arr[mid] > x)
return binarySearch(arr, l, mid - 1, x);
// Else the element can only be present
// in right subarray
return binarySearch(arr, mid + 1, r, x);
}
// We reach here when element is not present
// in array
return -1;
}
// Driver method to test above
public static void Main()
{
int[] arr = { 2, 3, 4, 10, 40 };
int n = arr.Length;
int x = 10;
int result = binarySearch(arr, 0, n - 1, x);
if (result == -1)
Console.WriteLine("Element not present");
else
Console.WriteLine("Element found at index "
+ result);
}
}
Output:
Element is present at index 3

Sure there is.
var list = new List<int>();
list.Add(42);
list.Add(1);
list.Add(54);
var index = list.IndexOf(1); //TADA!!!!
EDIT: Ok, I hoped the irony was obvious. But strictly speaking, if your array is not sorted, you are pretty much stuck with the linear search, readily available by means of IndexOf() or IEnumerable.First().

Find smallest number in given range in an array

Hi i have an array of size N. The array values will always have either 1, 2, 3 integer values only. Now i need to find the lowest number between a given range of array indices. So for e.g. array = 2 1 3 1 2 3 1 3 3 2. the lowest value for ranges like [2-4] = 1, [4-5] = 2, [7-8] = 3, etc.
Below is my code :
static void Main(String[] args) {
string[] width_temp = Console.ReadLine().Split(' ');
int[] width = Array.ConvertAll(width_temp,Int32.Parse); // Main Array
string[] tokens_i = Console.ReadLine().Split(' ');
int i = Convert.ToInt32(tokens_i[0]);
int j = Convert.ToInt32(tokens_i[1]);
int vehicle = width[i];
for (int beg = i+1; beg <= j; beg++) {
if (vehicle > width[beg]) {
vehicle = width[beg];
}
}
Console.WriteLine("{0}", vehicle);
}
The above code works fine. But my concern is about efficiency. In above I am just taking one set of array range, but in actual there will be n number of ranges and I would have to return the lowest for each range. Now the problem is if there is a range like [0-N], N is array size, then I would end up comparing all the items for lowest. So I was wondering if there is a way around to optimize the code for efficiency???

I think it is a RMQ (Range Minimum Query) and there is several implementation which may fit your scenario.
Here is a nice TopCoder Tutorial cover a lot of them, I recommend two of them:
Using the notation in the tutorial, define <P, T> as <Preprocess Complexity, Query Complexity>, there is two famous and common implementation / data structure which can handle RMQ: Square Rooting Array & Segment Tree.
Segment Tree is famous yet hard to implement, it can solve RMQ in <O(n), O(lg n)> though, which has better complexity than Square Rooting Array (<O(n), O(sqrt(n))>)
Square Rooting Array (<O(n), O(sqrt(n))>)
Note That It is not a official name of the technique nor any data structure, indeed I do not know if there is any official naming of this technique since I learnt it...but here we go
For query time, it is definitely not the best you can got to solve RMQ, but it has an advantage: Easy Implementation! (Compared to Segment Tree...)
Here is the high level concept of how it works:
Let N be the length of the array, we split the array into sqrt(N) groups, each contain sqrt(N) elements.
Now we use O(N) time to find the minimum value of each groups, store them into another array call M
So using the above array, M[0] = min(A[0..2]), M[1] = min(A[3..5]), M[2] = min(A[6..8]), M[3] = min(A[9..9])
(The image from TopCoder Tutorial is storing the index of the minimum element)
Now let's see how to query:
For any range [p..q], we can always split this range into 3 parts at most.
Two parts for the left boundaries which is some left over elements that cannot be form a whole group.
One part is the elements in between, which forms some groups.
Using the same example, RMQ(2,7) can be split into 3 parts:
Left Boundary (left over elements): A[2]
Right Boundary (left over elements): A[6], A[7]
In between elements (elements across whole group): A[3],A[4],A[5]
Notice that for those in between elements, we have already preprocessed their minimum using M, so we do not need to look at each element, we can look and compare M instead, there is at most O(sqrt(N)) of them (it is the length of M afterall)
For boundary parts, as they cannot form a whole group by definition, means there is at most O(sqrt(N)) of them (it is the length of one whole group afterall)
So combining two boundary parts, with one part of in between elements, we only need to compare O(3*sqrt(N)) = O(sqrt(N)) elements
You can refer to the tutorial for more details (even for some pseudo codes).

You could do this using Linq extension methods.
List<int> numbers = new List<int> {2, 1, 3, 1, 2, 3, 1, 3, 3, 2};
int minindex =1, maxindex =3, minimum=-1;
if(minindex <= maxindex && maxindex>=0 && maxindex >=0 && maxindex < numbers.Count())
{
minimum = Enumerable.Range(minindex, maxindex-minindex+1) // max inclusive, remove +1 if you want to exclude
.Select(x=> numbers[x]) // Get the elements between given indices
.Min(); // Get the minimum among.
}
Check this Demo

This seems a fun little problem. My first point would be that scanning a fixed array tends to be pretty fast (millions per second), so you'd need a vast amount of data to warrant a more complex solution.
The obvious first thing, is to break from the loop when you have found a 1, as you've found your lowest value then.
If you want something more advanced.
Create a new array of int. Create a pre load function that populates each item of this array with the next index where it gets lower.
Create a loop that uses the new array to skip.
Here is what I mean. Take the following arrays.
int[] intialArray = new int[] { 3, 3, 3, 3, 2, 2, 2, 1 };
int[] searchArray = new int[] { 4, 4, 4, 4, 7, 7, 7, 7 };
So the idea is to find the lowest between positions 0-7.
Start at initialArray[0] and get value 3.
Read searchArray[0] and get the value 4. The 4 is the next index where the number is lower.
Read initialArray[4] and get the value 2.
etc.
So basically you'd need to put some effort to build the searcharray, but onces it's complete you would scan each range much faster.

Form your looping like the following:
int[] inputArray = { 2, 1, 3, 1, 2, 3, 1, 3, 3, 2 };
int minIndex = 2;
int maxIndex = 5;
int minVal = 3;
for (int i = minIndex; i <= maxIndex; i++)
{
if (inputArray[i] <= minVal)
minVal = inputArray[i];
}
Console.WriteLine("Minimum value in the Given range is ={0}", minVal);

Determining how close an array is to the target array

I'm playing a little experiment to increase my knowledge and I have reached a part where I feel i could really optimize it, but am not quite sure how to do this.
I have many arrays of numbers. (for simplicity, lets say each array has 4 numbers: 1, 2, 3, and 4)
The target is to have all of the numbers in ascending order (ie,
1-2-3-4), but the numbers are all scrambled in the different arrays.
A higher weight is placed upon larger numbers.
I need to sort all of these arrays in order of how close they are to
the target.
Ie, 4-3-2-1 would be the worst possible case.
Some example cases:
3-4-2-1 is better than 4-3-2-1
2-3-4-1 is better than 1-4-3-2 (even though two numbers match (1 and 3).
the biggest number is closer to its spot.)
So the big numbers always take precedence over the smaller numbers. Here is my attempt:
var tmp = from m in moves
let mx = m.Max()
let ranking = m.IndexOf(s => s == mx)
orderby ranking descending
select m;
return tmp.ToArray();
P.S IndexOf in the above example, is an extension I wrote to take an array and expression, and return the index of the element that satisfies the expression. It is needed because the situation is really a little more complicated, i'm simplifying it with my example.
The problem with my attempt here though, is that it would only sort by the biggest number, and forget all of the other numbers. it SHOULD rank by biggest number first, then by second largest, then by third.
Also, since it will be doing this operation over and over again for several minutes, it should be as efficient as possible.

You could implement a bubble sort, and count the number of times you have to move data around. The number of data moves will be large on arrays that are far away from the sorted ideal.
int GetUnorderedness<T>(T[] data) where T : IComparable<T>
{
data = (T[])data.Clone(); // don't modify the input data,
// we weren't asked to actually sort.
int swapCount = 0;
bool isSorted;
do
{
isSorted = true;
for(int i = 1; i < data.Length; i++)
{
if(data[i-1].CompareTo(data[i]) > 0)
{
T temp = data[i];
data[i] = data[i-1];
data[i-1] = temp;
swapCount++;
isSorted = false;
}
}
} while(!isSorted);
}
From your sample data, this will give slightly different results than you specified.
Some example cases:
3-4-2-1 is better than 4-3-2-1
2-3-4-1 is better than 1-4-3-2
3-4-2-1 will take 5 swaps to sort, 4-3-2-1 will take 6, so that works.
2-3-4-1 will take 3, 1-4-3-2 will also take 3, so this doesn't match up with your expected results.
This algorithm doesn't treat the largest number as the most important, which it seems you want; all numbers are treated equally. From your description, you'd consider 2-1-3-4 as much better than 1-2-4-3, because the first one has both the largest and second largest numbers in their proper place. This algorithm would consider those two equal, because each requires only 1 swap to sort the array.
This algorithm does have the advantage that it's not just a comparison algorithm, each input has a discrete output, so you only need to run the algorithm once for each input array.

I hope this helps
var i = 0;
var temp = (from m in moves select m).ToArray();
do
{
temp = (from m in temp
orderby m[i] descending
select m).ToArray();
}
while (++i < moves[0].Length);

Optimizing this C# algorithm (K Difference)

This is the problem I'm solving (it's a sample problem, not a real problem):
Given N numbers , [N<=10^5] we need to count the total pairs of
numbers that have a difference of K. [K>0 and K<1e9]
Input Format: 1st line contains N & K (integers). 2nd line contains N
numbers of the set. All the N numbers are assured to be distinct.
Output Format: One integer saying the no of pairs of numbers that have
a diff K.
Sample Input #00:
5 2
1 5 3 4 2
Sample Output #00:
3
Sample Input #01:
10 1
363374326 364147530 61825163 1073065718 1281246024 1399469912 428047635 491595254 879792181 1069262793
Sample Output #01:
0
I already have a solution (and I haven't been able to optimize it as well as I had hoped). Currently my solution gets a score of 12/15 when it is run, and I'm wondering why I can't get 15/15 (my solution to another problem wasn't nearly as efficient, but got all of the points). Apparently, the code is run using "Mono 2.10.1, C# 4".
So can anyone think of a better way to optimize this further? The VS profiler says to avoid calling String.Split and Int32.Parse. The calls to Int32.Parse can't be avoided, although I guess I could optimize tokenizing the array.
My current solution:
using System;
using System.Collections.Generic;
using System.Text;
using System.Linq;
namespace KDifference
{
class Solution
{
static void Main(string[] args)
{
char[] space = { ' ' };
string[] NK = Console.ReadLine().Split(space);
int N = Int32.Parse(NK[0]), K = Int32.Parse(NK[1]);
int[] nums = Console.ReadLine().Split(space, N).Select(x => Int32.Parse(x)).OrderBy(x => x).ToArray();
int KHits = 0;
for (int i = nums.Length - 1, j, k; i >= 1; i--)
{
for (j = 0; j < i; j++)
{
k = nums[i] - nums[j];
if (k == K)
{
KHits++;
}
else if (k < K)
{
break;
}
}
}
Console.Write(KHits);
}
}
}

Your algorithm is still O(n^2), even with the sorting and the early-out. And even if you eliminated the O(n^2) bit, the sort is still O(n lg n). You can use an O(n) algorithm to solve this problem. Here's one way to do it:
Suppose the set you have is S1 = { 1, 7, 4, 6, 3 } and the difference is 2.
Construct the set S2 = { 1 + 2, 7 + 2, 4 + 2, 6 + 2, 3 + 2 } = { 3, 9, 6, 8, 5 }.
The answer you seek is the cardinality of the intersection of S1 and S2. The intersection is {6, 3}, which has two elements, so the answer is 2.
You can implement this solution in a single line of code, provided that you have sequence of integers sequence, and integer difference:
int result = sequence.Intersect(from item in sequence select item + difference).Count();
The Intersect method will build an efficient hash table for you that is O(n) to determine the intersection.

Try this (note, untested):
Sort the array
Start two indexes at 0
If difference between the numbers at those two positions is equal to K, increase count, and increase one of the two indexes (if numbers aren't duplicated, increase both)
If difference is larger than K, increase index #1
If difference is less than K, increase index #2, if that would place it outside the array, you're done
Otherwise, go back to 3 and keep going
Basically, try to keep the two indexes apart by K value difference.
You should write up a series of unit-tests for your algorithm, and try to come up with edge cases.

This would allow you to do it in a single pass. Using hash sets is beneficial if there are many values to parse/check. You might also want to use a bloom filter in combination with hash sets to reduce lookups.
Initialize. Let A and B be two empty hash sets. Let c be zero.
Parse loop. Parse the next value v. If there are no more values the algorithm is done and the result is in c.
Back check. If v exists in A then increment c and jump back to 2.
Low match. If v - K > 0 then:
insert v - K into A
if v - K exists in B then increment c (and optionally remove v - K from B).
High match. If v + K < 1e9 then:
insert v + K into A
if v + K exists in B then increment c (and optionally remove v + K from B).
Remember. Insert v into B.
Jump back to 2.

// php solution for this k difference
function getEqualSumSubstring($l,$s) {
$s = str_replace(' ','',$s);
$l = str_replace(' ','',$l);
for($i=0;$i<strlen($s);$i++)
{
$array1[] = $s[$i];
}
for($i=0;$i<strlen($s);$i++)
{
$array2[] = $s[$i] + $l[1];
}
return count(array_intersect($array1,$array2));
}
echo getEqualSumSubstring("5 2","1 3 5 4 2");

Actually that's trivially to solve with a hashmap:
First put each number into a hashmap: dict((x, x) for x in numbers) in "pythony" pseudo code ;)
Now you just iterate through every number in the hashmap and check if number + K is in the hashmap. If yes, increase count by one.
The obvious improvement to the naive solution is to ONLY check for the higher (or lower) bound, otherwise you get the double results and have to divide by 2 afterwards - useless.
This is O(N) for creating the hashmap when reading the values in and O(N) when iterating through, i.e. O(N) and about 8loc in python (and it is correct, I just solved it ;-) )

Following Eric's answer, paste the implementation of Interscet method below, it is O(n):
private static IEnumerable<TSource> IntersectIterator<TSource>(IEnumerable<TSource> first, IEnumerable<TSource> second, IEqualityComparer<TSource> comparer)
{
Set<TSource> set = new Set<TSource>(comparer);
foreach (TSource current in second)
{
set.Add(current);
}
foreach (TSource current2 in first)
{
if (set.Remove(current2))
{
yield return current2;
}
}
yield break;
}