So, this is my problem to solve:
I want to calculate 2^(n) where 0 < n< 10000
I am representing each element of array as a space where 4digit number should be "living" and if extra digit appears, I am replacing it to the next element of this array.
The principle I am using looks like this:
The code I am using is the following:
static string NotEfficient(int power)
{
if (power < 0)
throw new Exception("Power shouldn't be negative");
if (power == 0)
return "1";
if (power == 1)
return "2";
int[] A = new int[3750];
int current4Digit = 0;
//at first 2 is written in first element of array
A[current4Digit] = 2;
int currentPower = 1;
while (currentPower < power)
{
//multiply every 4digit by 2
for (int i = 0; i <= current4Digit; i++)
{
A[i] *= 2;
}
currentPower++;
//checking every 4digit if it
//contains 5 digit and if yes remove and
//put it in next 4digit
for (int i = 0; i <= current4Digit; i++)
{
if (A[i] / 10000 > 0)
{
int more = A[i] / 10000;
A[i] = A[i] % 10000;
A[i + 1] += more;
//if new digit should be opened
if (i + 1 > current4Digit)
{
current4Digit++;
}
}
}
}
//getting data from array to generate answer
string answer = "";
for (int i = current4Digit; i >= 0; i--)
{
answer += A[i].ToString() + ",";
}
return answer;
}
The problem I have is that it doesn't display correctly the number, which contains 0 in reality. for example 2 ^ (50) = 1 125 899 906 842 624 and with my algorithm I get 1 125 899 96 842 624 (0 is missing). This isn't only for 50...
This happens when I have the following situation for example:
How I can make this algorithm better?
Use BigInteger, which is already included in .Net Core or available in the System.Runtime.Numerics Nuget Package.
static string Efficient(int power)
{
var result = BigInteger.Pow(2, power);
return result.ToString(CultureInfo.InvariantCulture);
}
On my machine, NotEfficient takes roughly 80ms, where Efficient takes 0.3ms. You should be able to manipulate that string (if I'm understanding your problem statement correctly):
static string InsertCommas(string value)
{
var sb = new StringBuilder(value);
for (var i = value.Length - 4; i > 0; i -= 4)
{
sb.Insert(i, ',');
}
return sb.ToString();
}
One way to resolve this is to pad your 4-digit numbers with leading zeroes if they are less than four digits by using the PadLeft method:
answer += A[i].ToString().PadLeft(4, '0') + ",";
And then you can use the TrimStart method to remove any leading zeros from the final result:
return answer.TrimStart('0');
Related
I need to validate ICCID, I found only one algo:
int numberStringLength = 18;
int cs = 0;
int dodd;
for (int i = 0; i < numberStringLength; i += 2)
{
dodd = Convert.ToInt32(iccid.Substring(i + 1, 1)) << 1;
cs += Convert.ToInt32(iccid.Substring(i, 1)) + (int)(dodd / 10) + (dodd % 10);
}
cs = (10-(cs % 10)) % 10;
if (cs == Convert.ToInt32(iccid.Substring(numberStringLength, 1)))
{
return true;
}
else
{
return false;
}
but it returns false for 100% right ICCID (89148000005339755555). Where can I get real ICCID algo?
Thanks
According to Wikipedia, ICCIDs use the Luhn algorithm.
Your code that you found is a bit broken, as it assumes that the value has an odd number of digits (an even number of normal digits, plus 1 check digit). It starts parsing the value from the left-most digit, and assumes that this left-most digit ("8" in your example) is not doubled and the next one ("9") is doubled. But this is not correct if the value has an even number of digits. The "8" should be the one that's doubled in your case.
Thankfully, it's very easy to implement the Luhn algorithm ourselves, properly, using that Wikipedia page as reference:
string input = "89148000005339755555";
int sum = 0;
// We'll use index i = 0 means the right-most digit, i = 1 is second-right, etc
for (int i = 0; i < input.Length; i++)
{
// Get the digit at the i'th position from the right
int digit = int.Parse(input[input.Length - i - 1].ToString());
// If it's in an odd position (starting from the right), then double it.
if (i % 2 == 1)
{
digit *= 2;
// If it's now >= 10, subtract 9
if (digit >= 10)
{
digit -= 9;
}
}
sum += digit;
}
// It's a pass if the result is a multiple of 10
bool pass = sum % 10 == 0;
Console.WriteLine(pass ? "Pass" : "Fail");
See it on dotnetfiddle.net.
I got asked a question and now I am kicking myself for not being able to come up with the exact/correct result.
Imagine we have a function that splits a string into multiple lines but each line has to have x number of characters before we "split" to the new line:
private string[] GetPagedMessages(string input, int maxCharsPerLine) { ... }
For each line, we need to incorporate, at the end of the line "x/y" which is basically 1/4, 2/4 etc...
Now, the paging mechanism must also be part of the length restriction per line.
I have been overworked and overthinking and tripping up on things and this seems pretty straight forward but for the life of me, I cannot figure it out! What am I not "getting"?
What am I interested in? The calculation and some part of the logic but mainly the calculation of how many lines are required to split the input based on the max chars per line which also needs to include the x/y.
Remember: we can have more than a single digit for the x/y (i.e: not just 1/4 but also 10/17 or 99/200)
Samples:
input = "This is a long message"
maxCharsPerLine = 10
output:
This i 1/4 // << Max 10 chars
s a lo 2/4 // << Max 10 chars
ng mes 3/4 // << Max 10 chars
sage 4/4 // << Max 10 chars
Overall the logic is simple but its just the calculation that is throwing me off.
The idea: First, find how many digits is the number of lines:
(n = input.Length, maxCharsPerLine = 10)
if n <= 9*(10-4) ==> 1 digit
if n <= 9*(10-5) + 90*(10-6) ==> 2 digits
if n <= 9*(10-6) + 90*(10-7) + 900*(10-8) ==> 3 digits
if n <= 9*(10-7) + 90*(10-8) + 900*(10-9) + 9000*(10-10) ==> No solution
Then, subtract the spare number of lines. The solution:
private static int GetNumberOfLines(string input, int maxCharsPerLine)
{
int n = input.Length;
int x = maxCharsPerLine;
for (int i = 4; i < x; i++)
{
int j, sum = 0, d = 9, numberOfLines = 0;
for (j = i; j <= i + i - 4; j++)
{
if (x - j <= 0)
return -1; // No solution
sum += d * (x - j);
numberOfLines += d;
d *= 10;
}
if (n <= sum)
return numberOfLines - (sum - n) / (x - j + 1);
}
return -2; // Invalid
}
Usage:
private static string[] GetPagedMessages(string input, int maxCharsPerLine)
{
int numberOfLines = GetNumberOfLines(input, maxCharsPerLine);
if (numberOfLines < 0)
return null;
string[] result = new string[numberOfLines];
int spaceLeftForLine = maxCharsPerLine - numberOfLines.ToString().Length - 2; // Remove the chars of " x/y" except the incremental 'x'
int inputPosition = 0;
for (int line = 1; line < numberOfLines; line++)
{
int charsInLine = spaceLeftForLine - line.ToString().Length;
result[line - 1] = input.Substring(inputPosition, charsInLine) + $" {line}/{numberOfLines}";
inputPosition += charsInLine;
}
result[numberOfLines-1] = input.Substring(inputPosition) + $" {numberOfLines}/{numberOfLines}";
return result;
}
A naive approach is to start counting the line lengths minus the "pager"'s size, until the line count changes in size ("1/9" is shorter than "1/10", which is shorter than "11/20", and so on):
private static int[] GetLineLengths(string input, int maxCharsPerLine)
{
/* The "pager" (x/y) is at least 4 characters (including the preceding space) and at most ... 8?
* 7/9 (4)
* 1/10 (5)
* 42/69 (6)
* 3/123 (6)
* 42/420 (7)
* 999/999 (8)
*/
int charsRemaining = input.Length;
var lineLengths = new List<int>();
// Start with " 1/2", (1 + 1 + 2) = 4 length
var highestLineNumberLength = 1;
var lineNumber = 0;
do
{
lineNumber++;
var currentLineNumberLength = lineNumber.ToString().Length; // 1 = 1, 99 = 2, ...
if (currentLineNumberLength > highestLineNumberLength)
{
// Pager size changed, reset
highestLineNumberLength = currentLineNumberLength;
lineLengths.Clear();
lineNumber = 0;
charsRemaining = input.Length;
continue;
}
var pagerSize = currentLineNumberLength + highestLineNumberLength + 2;
var lineLength = maxCharsPerLine - pagerSize;
if (lineLength <= 0)
{
throw new ArgumentException($"Can't split input of size {input.Length} into chunks of size {maxCharsPerLine}");
}
lineLengths.Add(lineLength);
charsRemaining -= lineLength;
}
while (charsRemaining > 0);
return lineLengths.ToArray();
}
Usage:
private static string[] GetPagedMessages(string input, int maxCharsPerLine)
{
if (input.Length <= maxCharsPerLine)
{
// Assumption: no pager required for a message that takes one line
return new[] { input };
}
var lineLengths = GetLineLengths(input, maxCharsPerLine);
var result = new string[lineLengths.Length];
// Cut the input and append the pager
var previousIndex = 0;
for (var i = 0; i < lineLengths.Length; i++)
{
var lineLength = Math.Min(lineLengths[i], input.Length - previousIndex); // To cater for final line being shorter
result[i] = input.Substring(previousIndex, lineLength) + " " + (i + 1) + "/" + lineLengths.Length;
previousIndex += lineLength;
}
return result;
}
Prints, for example:
This 1/20
is a 2/20
long 3/20
strin 4/20
g tha 5/20
t wil 6/20
l spa 7/20
n mor 8/20
e tha 9/20
n te 10/20
n li 11/20
nes 12/20
beca 13/20
use 14/20
of i 15/20
ts e 16/20
norm 17/20
ous 18/20
leng 19/20
th 20/20
I need to calculate the similarity between 2 strings. So what exactly do I mean? Let me explain with an example:
The real word: hospital
Mistaken word: haspita
Now my aim is to determine how many characters I need to modify the mistaken word to obtain the real word. In this example, I need to modify 2 letters. So what would be the percent? I take the length of the real word always. So it becomes 2 / 8 = 25% so these 2 given string DSM is 75%.
How can I achieve this with performance being a key consideration?
I just addressed this exact same issue a few weeks ago. Since someone is asking now, I'll share the code. In my exhaustive tests my code is about 10x faster than the C# example on Wikipedia even when no maximum distance is supplied. When a maximum distance is supplied, this performance gain increases to 30x - 100x +. Note a couple key points for performance:
If you need to compare the same words over and over, first convert the words to arrays of integers. The Damerau-Levenshtein algorithm includes many >, <, == comparisons, and ints compare much faster than chars.
It includes a short-circuiting mechanism to quit if the distance exceeds a provided maximum
Use a rotating set of three arrays rather than a massive matrix as in all the implementations I've see elsewhere
Make sure your arrays slice accross the shorter word width.
Code (it works the exact same if you replace int[] with String in the parameter declarations:
/// <summary>
/// Computes the Damerau-Levenshtein Distance between two strings, represented as arrays of
/// integers, where each integer represents the code point of a character in the source string.
/// Includes an optional threshhold which can be used to indicate the maximum allowable distance.
/// </summary>
/// <param name="source">An array of the code points of the first string</param>
/// <param name="target">An array of the code points of the second string</param>
/// <param name="threshold">Maximum allowable distance</param>
/// <returns>Int.MaxValue if threshhold exceeded; otherwise the Damerau-Leveshteim distance between the strings</returns>
public static int DamerauLevenshteinDistance(int[] source, int[] target, int threshold) {
int length1 = source.Length;
int length2 = target.Length;
// Return trivial case - difference in string lengths exceeds threshhold
if (Math.Abs(length1 - length2) > threshold) { return int.MaxValue; }
// Ensure arrays [i] / length1 use shorter length
if (length1 > length2) {
Swap(ref target, ref source);
Swap(ref length1, ref length2);
}
int maxi = length1;
int maxj = length2;
int[] dCurrent = new int[maxi + 1];
int[] dMinus1 = new int[maxi + 1];
int[] dMinus2 = new int[maxi + 1];
int[] dSwap;
for (int i = 0; i <= maxi; i++) { dCurrent[i] = i; }
int jm1 = 0, im1 = 0, im2 = -1;
for (int j = 1; j <= maxj; j++) {
// Rotate
dSwap = dMinus2;
dMinus2 = dMinus1;
dMinus1 = dCurrent;
dCurrent = dSwap;
// Initialize
int minDistance = int.MaxValue;
dCurrent[0] = j;
im1 = 0;
im2 = -1;
for (int i = 1; i <= maxi; i++) {
int cost = source[im1] == target[jm1] ? 0 : 1;
int del = dCurrent[im1] + 1;
int ins = dMinus1[i] + 1;
int sub = dMinus1[im1] + cost;
//Fastest execution for min value of 3 integers
int min = (del > ins) ? (ins > sub ? sub : ins) : (del > sub ? sub : del);
if (i > 1 && j > 1 && source[im2] == target[jm1] && source[im1] == target[j - 2])
min = Math.Min(min, dMinus2[im2] + cost);
dCurrent[i] = min;
if (min < minDistance) { minDistance = min; }
im1++;
im2++;
}
jm1++;
if (minDistance > threshold) { return int.MaxValue; }
}
int result = dCurrent[maxi];
return (result > threshold) ? int.MaxValue : result;
}
Where Swap is:
static void Swap<T>(ref T arg1,ref T arg2) {
T temp = arg1;
arg1 = arg2;
arg2 = temp;
}
What you are looking for is called edit distance or Levenshtein distance. The wikipedia article explains how it is calculated, and has a nice piece of pseudocode at the bottom to help you code this algorithm in C# very easily.
Here's an implementation from the first site linked below:
private static int CalcLevenshteinDistance(string a, string b)
{
if (String.IsNullOrEmpty(a) && String.IsNullOrEmpty(b)) {
return 0;
}
if (String.IsNullOrEmpty(a)) {
return b.Length;
}
if (String.IsNullOrEmpty(b)) {
return a.Length;
}
int lengthA = a.Length;
int lengthB = b.Length;
var distances = new int[lengthA + 1, lengthB + 1];
for (int i = 0; i <= lengthA; distances[i, 0] = i++);
for (int j = 0; j <= lengthB; distances[0, j] = j++);
for (int i = 1; i <= lengthA; i++)
for (int j = 1; j <= lengthB; j++)
{
int cost = b[j - 1] == a[i - 1] ? 0 : 1;
distances[i, j] = Math.Min
(
Math.Min(distances[i - 1, j] + 1, distances[i, j - 1] + 1),
distances[i - 1, j - 1] + cost
);
}
return distances[lengthA, lengthB];
}
There is a big number of string similarity distance algorithms that can be used. Some listed here (but not exhaustively listed are):
Levenstein
Needleman Wunch
Smith Waterman
Smith Waterman Gotoh
Jaro, Jaro Winkler
Jaccard Similarity
Euclidean Distance
Dice Similarity
Cosine Similarity
Monge Elkan
A library that contains implementation to all of these is called SimMetrics
which has both java and c# implementations.
I have found that Levenshtein and Jaro Winkler are great for small differences betwen strings such as:
Spelling mistakes; or
รถ instead of o in a persons name.
However when comparing something like article titles where significant chunks of the text would be the same but with "noise" around the edges, Smith-Waterman-Gotoh has been fantastic:
compare these 2 titles (that are the same but worded differently from different sources):
An endonuclease from Escherichia coli that introduces single polynucleotide chain scissions in ultraviolet-irradiated DNA
Endonuclease III: An Endonuclease from Escherichia coli That Introduces Single Polynucleotide Chain Scissions in Ultraviolet-Irradiated DNA
This site that provides algorithm comparison of the strings shows:
Levenshtein: 81
Smith-Waterman Gotoh 94
Jaro Winkler 78
Jaro Winkler and Levenshtein are not as competent as Smith Waterman Gotoh in detecting the similarity. If we compare two titles that are not the same article, but have some matching text:
Fat metabolism in higher plants. The function of acyl thioesterases in the metabolism of acyl-coenzymes A and acyl-acyl carrier proteins
Fat metabolism in higher plants. The determination of acyl-acyl carrier protein and acyl coenzyme A in a complex lipid mixture
Jaro Winkler gives a false positive, but Smith Waterman Gotoh does not:
Levenshtein: 54
Smith-Waterman Gotoh 49
Jaro Winkler 89
As Anastasiosyal pointed out, SimMetrics has the java code for these algorithms. I had success using the SmithWatermanGotoh java code from SimMetrics.
Here is my implementation of Damerau Levenshtein Distance, which returns not only similarity coefficient, but also returns error locations in corrected word (this feature can be used in text editors). Also my implementation supports different weights of errors (substitution, deletion, insertion, transposition).
public static List<Mistake> OptimalStringAlignmentDistance(
string word, string correctedWord,
bool transposition = true,
int substitutionCost = 1,
int insertionCost = 1,
int deletionCost = 1,
int transpositionCost = 1)
{
int w_length = word.Length;
int cw_length = correctedWord.Length;
var d = new KeyValuePair<int, CharMistakeType>[w_length + 1, cw_length + 1];
var result = new List<Mistake>(Math.Max(w_length, cw_length));
if (w_length == 0)
{
for (int i = 0; i < cw_length; i++)
result.Add(new Mistake(i, CharMistakeType.Insertion));
return result;
}
for (int i = 0; i <= w_length; i++)
d[i, 0] = new KeyValuePair<int, CharMistakeType>(i, CharMistakeType.None);
for (int j = 0; j <= cw_length; j++)
d[0, j] = new KeyValuePair<int, CharMistakeType>(j, CharMistakeType.None);
for (int i = 1; i <= w_length; i++)
{
for (int j = 1; j <= cw_length; j++)
{
bool equal = correctedWord[j - 1] == word[i - 1];
int delCost = d[i - 1, j].Key + deletionCost;
int insCost = d[i, j - 1].Key + insertionCost;
int subCost = d[i - 1, j - 1].Key;
if (!equal)
subCost += substitutionCost;
int transCost = int.MaxValue;
if (transposition && i > 1 && j > 1 && word[i - 1] == correctedWord[j - 2] && word[i - 2] == correctedWord[j - 1])
{
transCost = d[i - 2, j - 2].Key;
if (!equal)
transCost += transpositionCost;
}
int min = delCost;
CharMistakeType mistakeType = CharMistakeType.Deletion;
if (insCost < min)
{
min = insCost;
mistakeType = CharMistakeType.Insertion;
}
if (subCost < min)
{
min = subCost;
mistakeType = equal ? CharMistakeType.None : CharMistakeType.Substitution;
}
if (transCost < min)
{
min = transCost;
mistakeType = CharMistakeType.Transposition;
}
d[i, j] = new KeyValuePair<int, CharMistakeType>(min, mistakeType);
}
}
int w_ind = w_length;
int cw_ind = cw_length;
while (w_ind >= 0 && cw_ind >= 0)
{
switch (d[w_ind, cw_ind].Value)
{
case CharMistakeType.None:
w_ind--;
cw_ind--;
break;
case CharMistakeType.Substitution:
result.Add(new Mistake(cw_ind - 1, CharMistakeType.Substitution));
w_ind--;
cw_ind--;
break;
case CharMistakeType.Deletion:
result.Add(new Mistake(cw_ind, CharMistakeType.Deletion));
w_ind--;
break;
case CharMistakeType.Insertion:
result.Add(new Mistake(cw_ind - 1, CharMistakeType.Insertion));
cw_ind--;
break;
case CharMistakeType.Transposition:
result.Add(new Mistake(cw_ind - 2, CharMistakeType.Transposition));
w_ind -= 2;
cw_ind -= 2;
break;
}
}
if (d[w_length, cw_length].Key > result.Count)
{
int delMistakesCount = d[w_length, cw_length].Key - result.Count;
for (int i = 0; i < delMistakesCount; i++)
result.Add(new Mistake(0, CharMistakeType.Deletion));
}
result.Reverse();
return result;
}
public struct Mistake
{
public int Position;
public CharMistakeType Type;
public Mistake(int position, CharMistakeType type)
{
Position = position;
Type = type;
}
public override string ToString()
{
return Position + ", " + Type;
}
}
public enum CharMistakeType
{
None,
Substitution,
Insertion,
Deletion,
Transposition
}
This code is a part of my project: Yandex-Linguistics.NET.
I wrote some tests and it's seems to me that method is working.
But comments and remarks are welcome.
Here is an alternative approach:
A typical method for finding similarity is Levenshtein distance, and there is no doubt a library with code available.
Unfortunately, this requires comparing to every string. You might be able to write a specialized version of the code to short-circuit the calculation if the distance is greater than some threshold, you would still have to do all the comparisons.
Another idea is to use some variant of trigrams or n-grams. These are sequences of n characters (or n words or n genomic sequences or n whatever). Keep a mapping of trigrams to strings and choose the ones that have the biggest overlap. A typical choice of n is "3", hence the name.
For instance, English would have these trigrams:
Eng
ngl
gli
lis
ish
And England would have:
Eng
ngl
gla
lan
and
Well, 2 out of 7 (or 4 out of 10) match. If this works for you, and you can index the trigram/string table and get a faster search.
You can also combine this with Levenshtein to reduce the set of comparison to those that have some minimum number of n-grams in common.
Here's a VB.net implementation:
Public Shared Function LevenshteinDistance(ByVal v1 As String, ByVal v2 As String) As Integer
Dim cost(v1.Length, v2.Length) As Integer
If v1.Length = 0 Then
Return v2.Length 'if string 1 is empty, the number of edits will be the insertion of all characters in string 2
ElseIf v2.Length = 0 Then
Return v1.Length 'if string 2 is empty, the number of edits will be the insertion of all characters in string 1
Else
'setup the base costs for inserting the correct characters
For v1Count As Integer = 0 To v1.Length
cost(v1Count, 0) = v1Count
Next v1Count
For v2Count As Integer = 0 To v2.Length
cost(0, v2Count) = v2Count
Next v2Count
'now work out the cheapest route to having the correct characters
For v1Count As Integer = 1 To v1.Length
For v2Count As Integer = 1 To v2.Length
'the first min term is the cost of editing the character in place (which will be the cost-to-date or the cost-to-date + 1 (depending on whether a change is required)
'the second min term is the cost of inserting the correct character into string 1 (cost-to-date + 1),
'the third min term is the cost of inserting the correct character into string 2 (cost-to-date + 1) and
cost(v1Count, v2Count) = Math.Min(
cost(v1Count - 1, v2Count - 1) + If(v1.Chars(v1Count - 1) = v2.Chars(v2Count - 1), 0, 1),
Math.Min(
cost(v1Count - 1, v2Count) + 1,
cost(v1Count, v2Count - 1) + 1
)
)
Next v2Count
Next v1Count
'the final result is the cheapest cost to get the two strings to match, which is the bottom right cell in the matrix
'in the event of strings being equal, this will be the result of zipping diagonally down the matrix (which will be square as the strings are the same length)
Return cost(v1.Length, v2.Length)
End If
End Function
At first user gives a number (n) to program, for example 5.
the program must find the smallest number that can be divided to n (5).
and this number can only consist of digits 0 and 9 not any other digits.
for example if user gives 5 to program.
numbers that can be divided to 5 are:
5, 10, 15, 20, 25, 30, ..., 85, 90, 95, ...
but 90 here is the smallest number that can be divided to 5 and also consist of digits (0 , 9). so answer for 5 must be 90.
and answer for 9 is 9, because it can be divided to 9 and consist of digit (9).
my code
string a = txtNumber.Text;
Int64 x = Convert.ToInt64(a);
Int64 i ,j=1,y=x;
bool t = false;
for (i = x + 1; t == false; i++)
{
if (i % 9 == 0 && i % 10 == 0 && i % x == 0)
{
j = i;
for (; (i /= 10) != 0; )
{
i /= 10;
if (i == 0)
t = true;
continue;
}
}
}
lblAnswer.Text = Convert.ToString(j);
If you're happy to go purely functional then this works:
Func<IEnumerable<long>> generate = () =>
{
Func<long, IEnumerable<long>> extend =
x => new [] { x * 10, x * 10 + 9 };
Func<IEnumerable<long>, IEnumerable<long>> generate2 = null;
generate2 = ns =>
{
var clean = ns.Where(n => n > 0).ToArray();
return clean.Any()
? clean.Concat(generate2(clean.SelectMany(extend)))
: Enumerable.Empty<long>();
};
return generate2(new[] { 9L, });
};
Func<long, long?> f = n =>
generate()
.Where(x => x % n == 0L)
.Cast<long?>()
.FirstOrDefault();
So rather than iterate through all possible values and test for 0 & 9 and divisibility, this just generates only numbers with 0 & 9 and then only tests for visibility. It is much faster this way.
I can call it like this:
var result = f(5L); // 90L
result = f(23L); //990909L
result = f(123L); //99999L
result = f(12321L); //90900999009L
result = f(123212L); //99909990090000900L
result = f(117238L); //990990990099990990L
result = f(1172438L); //null == No answer
These results are super fast. f(117238L) returns a result on my computer in 138ms.
You can try this way :
string a = txtNumber.Text;
Int64 x = Convert.ToInt64(a);
int counter;
for (counter = 1; !isValid(x * counter); counter++)
{
}
lblAnswer.Text = Convert.ToString(counter*x);
code above works by searching multiple of x incrementally until result that satisfy criteria : "consist of only 0 and or 9 digits" found. By searching only multiple of x, it is guaranteed to be divisible by x. So the rest is checking validity of result candidate, in this case using following isValid() function :
private static bool isValid(int number)
{
var lastDigit = number%10;
//last digit is invalid, return false
if (lastDigit != 0 & lastDigit != 9) return false;
//last digit is valid, but there is other digit(s)
if(number/10 >= 1)
{
//check validity of digit(s) before the last
return isValid(number/10);
}
//last digit is valid, and there is no other digit. return true
return true;
}
About strange empty for loop in snippet above, it is just syntactic sugar, to make the code a bit shorter. It is equal to following while loop :
counter = 1;
while(!isValid(input*counter))
{
counter++;
}
Use this simple code
int inputNumber = 5/*Or every other number, you can get this number from input.*/;
int result=1;
for (int i = 1; !IsOk(result,inputNumber); i++)
{
result = i*inputNumber;
}
Print(result);
IsOk method is here:
bool IsOk(int result, int inputNumber)
{
if(result%inputNumber!=0)
return false;
if(result.ToString().Replace("9",string.Empty).Replace("0",string.Empty).Length!=0)
return false;
return true;
}
My first solution has very bad performance, because of converting a number to string and looking for characters '9' and '0'.
New solution:
My new solution has very good performance and is a technical approach since of using Breadth-first search(BFS).
Algorithm of this solution:
For every input number, test 9, if it is answer print it, else add 2 child numbers (90 & 99) to queue, and continue till finding answer.
int inputNumber = 5;/*Or every other number, you can get this number from input.*/
long result;
var q = new Queue<long>();
q.Enqueue(9);
while (true)
{
result = q.Dequeue();
if (result%inputNumber == 0)
{
Print(result);
break;
}
q.Enqueue(result*10);
q.Enqueue(result*10 + 9);
}
Trace of number creation:
9
90,99
900,909,990,999
9000,9009,9090,9099,9900,9909,9990,9999
.
.
.
I wrote this code for console, and i used goto command however it is not prefered but i could not write it with only for.
using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.Threading.Tasks;
namespace main
{
class Program
{
static void Main(string[] args)
{
Console.WriteLine("Enter your number");
Int64 x = Convert.ToInt64(Console.ReadLine());
Int64 y, j, i, k, z = x;
x = x + 5;
loop:
x++;
for (i = 0, y = x; y != 0; i++)
y /= 10;
for (j = x, k = i; k != 0; j /= 10, k--)
{
if (j % 10 != 9)
if (j % 10 != 0)
goto loop;
}
if (x % z != 0)
goto loop;
Console.WriteLine("answer:{0}",x);
Console.ReadKey();
}
}
}
My problem is similar to that question but slightly different. I have a registration number having the form :
14AAA999 : 14 is the current year, each A represents a caracter and each 9 represents a digit.
I know how to extract a number from a string but I would like to increment also the letters. Here is what I would like to achieve :
The first guy registered on the year 2014 would have this registration number : 14AAA000, the second 14AAA001, ..., until 14AAA999 and then it would become 14AAB000, and so on until 14ZZZ999. I'm looking for a good way to increment the letters for each overflow.
Thanks for your help !
In your case you have a Number that is two separate bases. For the first 3 digits you are base 10 in the next base 26 where A = 0. you can simply create a number convert that converts from your representation to to other. This is slightly different than ISE solution.
class foo
{
public static string ToRep(int Year,int Num)
{
StringBuilder SB = new StringBuilder(Year.ToString());
SB.Append(":");
int DecPart = Num % 1000;
int Base26Part = Num / 1000;
for(int x = 0 ; x < 3 ; x++)
{
char NewChar =(char)( 'A'+ Base26Part % 26);
Base26Part /= 26;
SB.Insert(3,NewChar);
}
SB.Append(DecPart.ToString("000"));
return SB.ToString();
}
}
Pseudo-code:
char[] minValues { "14AAA000" }
char[] maxValues { "99ZZZ999" }
int index = input.Length - 1;
do
{
if (input[index] < maxValues[index])
{
input[index]++;
for (int j = index + 1; j < input.Length)
input[j] := minValues[j];
break;
}
else
index--;
} while (true);