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Remove two int from three consecutive values from List<int>

I use Winforms and C#. And I have more than 300 integers in a list and this is short example:
List<int> lst = new List<int>();
lst.Add(4);
lst.Add(25);
lst.Add(26);
lst.Add(27);
lst.Add(38);
lst.Add(51);
lst.Add(52);
lst.Add(53);
lst.Add(75);
//Etc.
I need to remove values (25, 26) but keep 27. Then again remove values (51, 52) but keep 53.. etc.
So basically if three consecutive values such as (1, 2, 3) are found, we need to remove (1, 2) and keep value (3).
As far as attempts are concerned, I couldn't scratch my head around any solution so far therefore I would appreciate any help.
Thanks

You could try the following. The key is marking the items to remove by detecting the sequence and picking the first two.
var excludeList = lst.Distinct()
.GroupBy(num =>
Enumerable.Range(num, int.MaxValue - num + 1)
.TakeWhile(lst.Contains)
.Last())
.Where(seq => seq.Count() >= 3)
.SelectMany(seq => seq.OrderBy(num => num).Take(2));
var result = lst.Where(x=> !excludeList.Contains(x));
Output
4
27
38
53
75

You can try this:
static void Test()
{
var list = new List<int>();
list.Add(4);
list.Add(25);
list.Add(26);
list.Add(27);
list.Add(38);
list.Add(51);
list.Add(52);
list.Add(53);
list.Add(75);
var result = new List<int>();
int count = list.Count;
bool passover;
if ( count > 0 )
for ( int index = 0; index < count; )
{
passover = false;
if ( index < count - 3 )
{
int v1 = list[index];
int v2 = list[index + 1];
int v3 = list[index + 2];
if ( v3 == v2 + 1 && v2 == v1 + 1 )
passover = true;
}
if ( passover )
{
result.Add(list[index + 2]);
index += 3;
}
else
{
result.Add(list[index]);
index++;
}
}
foreach ( var item in result )
Console.WriteLine(item);
}
Output:
4
27
38
53
75

Value was either too large or too small for a UInt64.'

My codes are like below.
I just want to try to convert number to binary and summarize binary numbers as if they are decimal ones, e.g. the desired outcome for 3 is 22:
1 -> 1
2 -> 10
3 -> 11
-------
22 == 1 + 10 + 11
But number array is growing and the code blowing :)
static void Main(string[] args)
{
long deger = 1;
for (int i = 0; i < 1000000; i++)
{
deger *= (deger + 1);
int result = solve(deger);
Console.WriteLine(result);
}
}
public static int solve(long a)
{
ulong[] lastValue = new ulong[a];
for (int i = 1; i < a; i++)
{
var binary = Convert.ToString(i, 2);
lastValue[i] = Convert.ToUInt64(binary);// this part gives an error
}
var result = lastValue.Aggregate((t, c) => t + c);
return (int)result;
}

Well, UInt64 is not large enough; you can try either BigInteger or you may sum up strings:
private string MyBinarySum(string left, string right) {
var x = left
.PadLeft(Math.Max(left.Length, right.Length) + 1, '0')
.Reverse()
.Select(c => c - '0')
.ToArray();
var y = right
.PadLeft(Math.Max(left.Length, right.Length) + 1, '0')
.Reverse().Select(c => c - '0')
.ToArray();
StringBuilder sb = new StringBuilder(left.Length);
int shift = 0;
for (int i = 0; i < x.Length; ++i) {
int v = x[i] + y[i] + shift;
shift = v / 2;
v = v % 2;
sb.Append((char)('0' + v));
}
return String.Concat(sb.ToString().TrimEnd('0').Reverse());
}
Then, with a help of Linq
var result = Enumerable
.Range(0, 1000000)
.Select(item => Convert.ToString(item, 2))
.Aggregate((sum, item) => MyBinarySum(sum, item));
Console.Write(result);
Outcome:
111010001101010010010101110011011100000
which is beyond UInt64.MaxValue == 18446744073709551615

Look at UInt64.MaxValue and UInt64.MinValue.
They defined as 18446744073709551615 and 0 corresponding.

Why aren't 10 and 100 considered Kaprekar numbers?

I'm trying to do the Modified Kaprekar Numbers problem (https://www.hackerrank.com/challenges/kaprekar-numbers) which describes a Kaprekar number by
Here's an explanation from Wikipedia about the ORIGINAL Kaprekar
Number (spot the difference!): In mathematics, a Kaprekar number for a
given base is a non-negative integer, the representation of whose
square in that base can be split into two parts that add up to the
original number again. For instance, 45 is a Kaprekar number, because
45² = 2025 and 20+25 = 45.
and what I don't understand is why 10 and 100 aren't Kaprekar numbers.
10^2 = 1000 and 10 + 00 = 10
Right?
So my solution
// Returns the number represented by the digits
// in the range arr[i], arr[i + 1], ..., arr[j - 1].
// If there are no elements in range, return 0.
static int NumberInRange(int[] arr, int i, int j)
{
int result = 0;
for(; i < j; ++i)
{
result *= 10;
result += arr[i];
}
return result;
}
// Returns true or false depending on whether k
// is a Kaprekar number.
// Example: IsKaprekar(45) = true because 45^2=2025 and 20+25=45
// Example: IsKaprekar(9) = false because the set of the split
// digits of 7^2=49 are {49,0},{4,9} and
// neither of 49+0 or 4+9 equal 7.
static bool IsKaprekar(int k)
{
int square = k * k;
int[] digits = square.ToString().Select(c => (int)Char.GetNumericValue(c)).ToArray();
for(int i = 0; i < digits.Length; ++i)
{
int right = NumberInRange(digits, 0, i);
int left = NumberInRange(digits, i, digits.Length);
if((right + left) == k)
return true;
}
return false;
}
is saying all the Kaprekar numbers between 1 and 100 are
1 9 10 45 55 99 100
whereas the "right" answer is
1 9 45 55 99

In 100+00 the right is 00, which is wrong because in a kaprekar number the right may start with zero (ex: 025) but cannot be entirely 0.
Therefore you can put a condition in the loop that
if(right==0)
return false;

The reason is because 10 x 10 = 100. Then you substring the right part with a length equals d = 2, that is digit count of original value (10), then the left part would be 1.
So l = 1 and r = 00, l + r = 1, that is not equals to 10.
The same for 100. 100 x 100 = 10000. l = 10, r = 000, so l + r = 10 not equal 100.
Here is my solution in JAVA.
static void kaprekarNumbers(int p, int q) {
long[] result = IntStream.rangeClosed(p, q).mapToLong(Long::valueOf)
.filter(v -> {
int d = String.valueOf(v).length();
Long sq = v * v;
String sqSt = sq.toString();
if (sqSt.length() > 1) {
long r = Long.parseLong(sqSt.substring(sqSt.length() - d));
long l = Long.parseLong(sqSt.substring(0, sqSt.length() - d));
return r + l == v;
} else return v == 1;
}).toArray();
if (result.length > 0) {
for (long l : result) {
System.out.print(l + " ");
}
} else {
System.out.println("INVALID RANGE");
}
}

How about something like this.
static bool IsKaprekar(int k)
{
int t;
for (int digits = new String(k).length(); digits > 0; digits--, t *= 10);
long sq = k * k;
long first = sq / t;
long second = sq % t;
return k == first + second;
}
find a number to divide and mod the square with in order to split it. This number should be a factor of 10 based on the number of digits in the original number.
calculate the square.
split the square.
compare the original to the sum of the splits.

Sum of Numbers as Distinct Primes

//List Style
using System;
using System.Collections.Generic;
using System.Linq;
public class pr{
static public void Main (){
int n, i, j, k, l, sum,flag = 0;
//int sum = i+j;
//int k = (n-i);
//int l = (n-j);
//System.Console.WriteLine ("Enter a number");
//n = Convert.ToInt32 (Console.ReadLine());
//List <int> primes = new List <int>(); //list to handle the numbers
//HashSet <int> myPrimes = new HashSet <int> (primes);
System.Console.WriteLine ("Enter a number");
n = Convert.ToInt32 (Console.ReadLine());
//myPrimes.Add(n);
//myPrimes.Add(i);
//myPrimes.Add(j);
// var count = string.Join(", ", primes);
//System.Console.WriteLine("The value of n is {0}",myPrimes);
for(i=3; i<n/2; i++){
for(j=3; j<n/2; j++){
if(checkPrime(i) == 1){
if(checkPrime(j) == 1){
if (checkPrime(n-i) == 1){
if (checkPrime(n-j) == 1){
//if(i == j){
//sum = i+j;
System.Console.WriteLine("{0}={1}+{2}\n",n,i,n-i);
//}
}
}
}
}
if (flag == 0 && (n-i) <= 0 && (n-j) <= 0){ //check to avoid dupes
if (n <= 0 && i <= 0 && j <= 0){
Console.Write("{0}\n",n);
}
}
}
}
}
public static int checkPrime(int n){
int i, j, flag = 1;
for (i = 2; i<=(Math.Sqrt(n)); i++){
for (j = 2; j<=(Math.Sqrt(n)); j++){
if (n%i == 0 && n%j == 0 ){ //even number check
i++;
j++;
flag = 0;
}
}
}
return flag;
}
}
So I have been experimenting with this for a while now. I cant seem to print all possible solutions. For example for 24 I am able to print 7+17 but not 2+5+17. There are also some answers being repeated and this might have to do with the fact that I dont have duplicate checks. I tried to push the integers in a list and then use a hashset to only have distinct integers but I got stuck and tried to brute force it. All the numbers to be printed are supposed to be distinct prime integers. I dont understand how to print all distinct numbers and is there an elegant way to print out all the possible.
Thanks for the help!

Don't know if it's elegant enough for you, but I've just mashed a dirty way to make it work:
static void Main()
{
Console.WriteLine("Enter a number");
var numberToSum = Convert.ToInt32(Console.ReadLine());
var primesInRange = GetPrimesUpTo(numberToSum);
var foundSolutions = primesInRange.SubSetsOf().Where(prime => prime.Sum() == numberToSum);
foreach (var solution in foundSolutions.ToList())
{
var formatOperation = solution
.Select(x => x.ToString())
.Aggregate((a, n) => a + " + " + n) + " = " + numberToSum;
Console.WriteLine(formatOperation);
}
Console.ReadLine();
}
public static IEnumerable<int> GetPrimesUpTo(int end)
{
var primes = new HashSet<int>();
for (var i = 2; i <= end; i++)
{
var ok = true;
foreach (var prime in primes)
{
if (prime * prime > i)
break;
if (i % prime == 0)
{
ok = false;
break;
}
}
if (ok)
primes.Add(i);
}
return primes;
}
public static IEnumerable<IEnumerable<T>> SubSetsOf<T>(this IEnumerable<T> source)
{
if (!source.Any())
return Enumerable.Repeat(Enumerable.Empty<T>(), 1);
var element = source.Take(1);
var haveNots = SubSetsOf(source.Skip(1));
var haves = haveNots.Select(set => element.Concat(set));
return haves.Concat(haveNots);
}
I've found your solution quite dirty so I divided the problem to be more understandable. GetPrimesUpTo returns all prime number from 2 to the number you've provided in the input, SubSetsOf returns combination of numbers that summed up equals the input number you've provided and finally the foreach in Main produces formatted output that is easy on the eye. Hope it helps!

Providing that you have collection of primes and IsPrime method
private static int[] primes = new[] {
2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37 };
private static bool IsPrime(int value) {
return primes.Contains(value);
}
You can implement recoursive solution
private List<List<int>> ToListOfPrimes(int value, List<int> parts = null) {
if (null == parts)
parts = new List<int>();
List<List<int>> result = new List<List<int>>();
if (value == 0) {
result.Add(parts.ToList());
return result;
}
int minPrime = parts.Count <= 0 ? 0 : parts[parts.Count - 1];
if (value <= minPrime)
return result;
// not that efficient: binary search will be a better choice here
for (int i = 0; i < primes.Length; ++i) {
int p = primes[i];
if (p <= minPrime)
continue;
else if (p > value)
break;
var list = parts.ToList();
list.Add(p);
var outcome = ToListOfPrimes(value - p, list);
foreach (var solution in outcome)
result.Add(solution);
}
return result;
}
Test
var result = ToListOfPrimes(28);
string report = String.Join(Environment.NewLine, result
.Select(line => String.Join(", ", line)));
Console.Write(report);
Outcome (28)
2, 3, 5, 7, 11
2, 3, 23
2, 7, 19
3, 5, 7, 13
5, 23
11, 17
For 24
2, 3, 19
2, 5, 17
5, 19
7, 17
11, 13

If you really want to implement it in other languages just throw your solution to rubbish bin. You should be more explicit about what is happening during execution. Nested for loops with multiple if statements are not explicit at all. What's even worse in the sample - you'll need to add new for loop every time you want more numbers in the sum. I do believe it's hard to understand it for a novice, but I find recursion the only way to go here.
See for yourself:
it's hard to say why output of your program is wrong, because of the logic
Variables should be named meaningfully so you know what they store instead of blind guessing.
Your checkPrime method returns int even though you return 0 or 1 so it should really return bool type
Use debugger and a piece of paper to understand how recursion works either in my previous answer or the one provided by Dmitry Bychenko

Finding an integer sum in an array of 1 000 000

Given a large list of integers (more than 1 000 000 values) find how many ways there are of selecting two of them that add up to 0.... Is the question
What I have done is create a positive random integer list:
Random pos = new Random();
int POSNO = pos.Next(1, 1000000);
lstPOS.Items.Add(POSNO);
lblPLus.Text = lstPOS.Items.Count.ToString();
POSCount++;
And created a negative list:
Random neg = new Random();
int NEGNO = neg.Next(100000, 1000000);
lstNEG.Items.Add("-" + NEGNO);
lblNegative.Text = lstNEG.Items.Count.ToString();
NegCount++;
To do the sum checking I am using:
foreach (var item in lstPOS.Items)
{
int POSItem = Convert.ToInt32(item.ToString());
foreach (var negItem in lstNEG.Items)
{
int NEGItem = Convert.ToInt32(negItem.ToString());
int Total = POSItem - NEGItem;
if (Total == 0)
{
lstADD.Items.Add(POSItem + "-" + NEGItem + "=" + Total);
lblAddition.Text = lstADD.Items.Count.ToString();
}
}
}
I know this is not the fastest route. I have considered using an array. Do you have any suggestions?

Let's see; your array is something like this:
int[] data = new int[] {
6, -2, 3, 2, 0, 0, 5, 7, 0, -2
};
you can add up to zero in two different ways:
a + (-a) // positive + negative
0 + 0 // any two zeros
in the sample above there're five pairs:
-2 + 2 (two pairs): [1] + [3] and [3] + [9]
0 + 0 (three pairs): [4] + [5], [4] + [8] and [5] + [8]
So you have to track positive/negative pairs and zeros. The implementation
Dictionary<int, int> positives = new Dictionary<int, int>();
Dictionary<int, int> negatives = new Dictionary<int, int>();
int zeros = 0;
foreach(var item in data) {
int v;
if (item < 0)
if (negatives.TryGetValue(item, out v))
negatives[item] = negatives[item] + 1;
else
negatives[item] = 1;
else if (item > 0)
if (positives.TryGetValue(item, out v))
positives[item] = positives[item] + 1;
else
positives[item] = 1;
else
zeros += 1;
}
// zeros: binomal coefficent: (2, zeros)
int result = zeros * (zeros - 1) / 2;
// positive/negative pairs
foreach (var p in positives) {
int n;
if (negatives.TryGetValue(-p.Key, out n))
result += n * p.Value;
}
// Test (5)
Console.Write(result);
Note, that there's no sorting, and dictionaries (i.e. hash tables) are used for positives and negatives so the execution time will be linear, O(n); the dark side of the implementation is that two additional structures (i.e. additional memory) required. In your case (millions integers only - Megabytes) you have that memory.
Edit: terser, but less readable Linq solution:
var dict = data
.GroupBy(item => item)
.ToDictionary(chunk => chunk.Key, chunk => chunk.Count());
int result = dict.ContainsKey(0) ? dict[0] * (dict[0] - 1) / 2 : 0;
result += dict
.Sum(pair => pair.Key > 0 && dict.ContainsKey(-pair.Key) ? pair.Value * dict[-pair.Key] : 0);

Fastest way without sorting!.
First of all you know that the sum of two integers are only 0 when they have equal absolute value but one is negative and the other is positive. So you dont need to sort. what you need is to Intersect positive list with negative list (by comparing absolute value). the result is numbers that ended up 0 sum.
Intersect has time complexity of O(n+m) where n is size of first list and m is size of second one.
private static void Main(string[] args)
{
Random random = new Random();
int[] positive = Enumerable.Range(0, 1000000).Select(n => random.Next(1, 1000000)).ToArray();
int[] negative = Enumerable.Range(0, 1000000).Select(n => random.Next(-1000000, -1)).ToArray();
var zeroSum = positive.Intersect(negative, new AbsoluteEqual());
foreach (var i in zeroSum)
{
Console.WriteLine("{0} - {1} = 0", i, i);
}
}
You also need to use this IEqualityComparer.
public class AbsoluteEqual : IEqualityComparer<int>
{
public bool Equals(int x, int y)
{
return (x < 0 ? -x : x) == (y < 0 ? -y : y);
}
public int GetHashCode(int obj)
{
return obj < 0 ? (-obj).GetHashCode() : obj.GetHashCode();
}
}

You tried to avoid check two numbers that are close (1, 2 are close, 3, 4 are close), but you didn't avoid check like (-100000, 1), (-1, 100000). Time complexity is O(n^2).
To avoid that you need to sort them first, then search from two direction.
var random = new Random();
var input = Enumerable.Range(1, 100).Select(_ => random.Next(200) - 100).ToArray();
Array.Sort(input); // This causes most computation. Time Complexity is O(n*log(n));
var expectedSum = 0;
var i = 0;
var j = input.Length - 1;
while (i < j) // This has liner time complexity O(n);
{
var result = input[i] + input[j];
if(expectedSum == result)
{
var anchori = i;
while (i < input.Length && input[i] == input[anchori] )
{
i++;
}
var anchorj = j;
while (j >= 0 && input[j] == input[anchorj])
{
j--;
}
// Exclude (self, self) combination
Func<int, int, int> combination = (n, k) =>
{
var mink = k * 2 < n ? k : n - k;
return mink == 0 ? 1
: Enumerable.Range(0, mink).Aggregate(1, (x, y) => x * (n - y))
/ Enumerable.Range(1, mink).Aggregate((x, y) => x * y);
};
var c = i < j ? (i - anchori) * (anchorj - j) : combination(i - anchori, 2);
for (int _ = 0; _ < c; _++)
{
// C# 6.0 String.Format
Console.WriteLine($"{input[anchori]}, {input[anchorj]}");
}
}
else if(result < expectedSum) {
i++;
}
else if(result > expectedSum) {
j--;
}
}

Here is another solution using (huh) LINQ. Hope the code is self explanatory
First some data
var random = new Random();
var data = new int[1000000];
for (int i = 0; i < data.Length; i++) data[i] = random.Next(-100000, 100000);
And now the solution
var result = data
.Where(value => value != int.MinValue)
.GroupBy(value => Math.Abs(value), (key, values) =>
{
if (key == 0)
{
var zeroCount = values.Count();
return zeroCount * (zeroCount - 1) / 2;
}
else
{
int positiveCount = 0, negativeCount = 0;
foreach (var value in values)
if (value > 0) positiveCount++; else negativeCount++;
return positiveCount * negativeCount;
}
})
.Sum();
Theoretically the above should have O(N) time and O(M) space complexity, where M is the count of the unique absolute values in the list.