Reorder digits in integer using C# - c#

I want to ask how I can reorder the digits in an Int32 so they result in the biggest possible number.
Here is an example which visualizes what I am trying to do:
2927466 -> 9766422
12492771 -> 97742211
I want to perform the ordering of the digits without using the System.Linq namespace and without converting the integer into a string value.
This is what I got so far:
public static int ReorderInt32Digits(int v)
{
int n = Math.Abs(v);
int l = ((int)Math.Log10(n > 0 ? n : 1)) + 1;
int[] d = new int[l];
for (int i = 0; i < l; i++)
{
d[(l - i) - 1] = n % 10;
n /= 10;
}
if (v < 0)
d[0] *= -1;
Array.Sort(d);
Array.Reverse(d);
int h = 0;
for (int i = 0; i < d.Length; i++)
{
int index = d.Length - i - 1;
h += ((int)Math.Pow(10, index)) * d[i];
}
return h;
}
This algorithm works flawlessly but I think it is not very efficient.
I would like to know if there is a way to do the same thing more efficiently and how I could improve my algorithm.

You can use this code:
var digit = 2927466;
String.Join("", digit.ToString().ToCharArray().OrderBy(x => x));
Or
var res = String.Join("", digit.ToString().ToCharArray().OrderByDescending(x => x) );

Not that my answer may or may not be more "efficient", but when I read your code you calculated how many digits there are in your number so you can determine how large to make your array, and then you calculated how to turn your array back into a sorted integer.
It would seem to me that you would want to write your own code that did the sorting part without using built in functionality, which is what my sample does. Plus, I've added the ability to sort in ascending or descending order, which is easy to add in your code too.
UPDATED
The original algorithm sorted the digits, now it sorts the digits so that the end result is the largest or smallest depending on the second parameter passed in. However, when dealing with a negative number the second parameter is treated as opposite.
using System;
public class Program
{
public static void Main()
{
int number1 = 2927466;
int number2 = 12492771;
int number3 = -39284925;
Console.WriteLine(OrderDigits(number1, false));
Console.WriteLine(OrderDigits(number2, true));
Console.WriteLine(OrderDigits(number3, false));
}
private static int OrderDigits(int number, bool asc)
{
// Extract each digit into an array
int[] digits = new int[(int)Math.Floor(Math.Log10(Math.Abs(number)) + 1)];
for (int i = 0; i < digits.Length; i++)
{
digits[i] = number % 10;
number /= 10;
}
// Order the digits
for (int i = 0; i < digits.Length; i++)
{
for (int j = i + 1; j < digits.Length; j++)
{
if ((!asc && digits[j] > digits[i]) ||
(asc && digits[j] < digits[i]))
{
int temp = digits[i];
digits[i] = digits[j];
digits[j] = temp;
}
}
}
// Turn the array of digits back into an integer
int result = 0;
for (int i = digits.Length - 1; i >= 0; i--)
{
result += digits[i] * (int)Math.Pow(10, digits.Length - 1 - i);
}
return result;
}
}
Results:
9766422
11224779
-22345899
See working example here... https://dotnetfiddle.net/RWA4XV

public static int ReorderInt32Digits(int v)
{
var nums = Math.Abs(v).ToString().ToCharArray();
Array.Sort(nums);
bool neg = (v < 0);
if(!neg)
{
Array.Reverse(nums);
}
return int.Parse(new string(nums)) * (neg ? -1 : 1);
}

This code fragment below extracts the digits from variable v. You can modify it to store the digits in an array and sort/reverse.
int v = 2345;
while (v > 0) {
int digit = v % 10;
v = v / 10;
Console.WriteLine(digit);
}
You can use similar logic to reconstruct the number from (sorted) digits: Multiply by 10 and add next digit.

I'm posting this second answer because I think I got the most efficient algorithm of all (thanks for the help Atul) :)
void Main()
{
Console.WriteLine (ReorderInt32Digits2(2927466));
Console.WriteLine (ReorderInt32Digits2(12492771));
Console.WriteLine (ReorderInt32Digits2(-1024));
}
public static int ReorderInt32Digits2(int v)
{
bool neg = (v < 0);
int mult = neg ? -1 : 1;
int result = 0;
var counts = GetDigitCounts(v);
for (int i = 0; i < 10; i++)
{
int idx = neg ? 9 - i : i;
for (int j = 0; j < counts[idx]; j++)
{
result += idx * mult;
mult *= 10;
}
}
return result;
}
// From Atul Sikaria's answer
public static int[] GetDigitCounts(int n)
{
int v = Math.Abs(n);
var result = new int[10];
while (v > 0) {
int digit = v % 10;
v = v / 10;
result[digit]++;
}
return result;
}

Related

Getting Time limit Exceed for last 3 Test Cases in a Program:C# [duplicate]

Given an array of n integers and a number, d, perform left rotations on the array. Then print the updated array as a single line of space-separated integers.
Sample Input:
5 4
1 2 3 4 5
The first line contains two space-separated integers denoting the respective values of n (the number of integers) and d (the number of left rotations you must perform).
The second line contains n space-separated integers describing the respective elements of the array's initial state.
Sample Output:
5 1 2 3 4
static void Main(String[] args)
{
string[] arr_temp = Console.ReadLine().Split(' ');
int n = Int32.Parse(arr_temp[0]);
int d = Int32.Parse(arr_temp[1]);
string[] arr = Console.ReadLine().Split(' ');
string[] ans = new string[n];
for (int i = 0; i < n; ++i)
{
ans[(i + n - d) % n] = arr[i];
}
for (int j = 0; j < n; ++j)
{
Console.Write(ans[j] + " ");
}
}
How to use less memory to solve this problem?
This will use less memory in most cases as the second array is only as big as the shift.
public static void Main(string[] args)
{
int[] n = { 1, 2, 3, 4, 5 };
LeftShiftArray(n, 4);
Console.WriteLine(String.Join(",", n));
}
public static void LeftShiftArray<T>(T[] arr, int shift)
{
shift = shift % arr.Length;
T[] buffer = new T[shift];
Array.Copy(arr, buffer, shift);
Array.Copy(arr, shift, arr, 0, arr.Length - shift);
Array.Copy(buffer, 0, arr, arr.Length - shift, shift);
}
This problem can get a bit tricky but also has a simple solution if one is familiar with Queues and Stacks.
All I have to do is define a Queue (which will contain the given array) and a Stack.
Next, I just have to Push the Dequeued index to the stack and Enqueue the Popped index in the Queue and finally return the Queue.
Sounds confusing? Check the code below:
static int[] rotLeft(int[] a, int d) {
Queue<int> queue = new Queue<int>(a);
Stack<int> stack = new Stack<int>();
while(d > 0)
{
stack.Push(queue.Dequeue());
queue.Enqueue(stack.Pop());
d--;
}
return queue.ToArray();
}
Do you really need to physically move anything? If not, you could just shift the index instead.
Actually you asked 2 questions:
How to efficiently rotate an array?
and
How to use less memory to solve this problem?
Usually efficiency and low memory usage are mutually exclusive. So I'm going to answer your second question, still providing the most efficient implementation under that memory constraint.
The following method can be used for both left (passing negative count) or right (passing positive count) rotation. It uses O(1) space (single element) and O(n * min(d, n - d)) array element copy operations (O(min(d, n - d)) array block copy operations). In the worst case scenario it performs O(n / 2) block copy operations.
The algorithm is utilizing the fact that
rotate_left(n, d) == rotate_right(n, n - d)
Here it is:
public static class Algorithms
{
public static void Rotate<T>(this T[] array, int count)
{
if (array == null || array.Length < 2) return;
count %= array.Length;
if (count == 0) return;
int left = count < 0 ? -count : array.Length + count;
int right = count > 0 ? count : array.Length - count;
if (left <= right)
{
for (int i = 0; i < left; i++)
{
var temp = array[0];
Array.Copy(array, 1, array, 0, array.Length - 1);
array[array.Length - 1] = temp;
}
}
else
{
for (int i = 0; i < right; i++)
{
var temp = array[array.Length - 1];
Array.Copy(array, 0, array, 1, array.Length - 1);
array[0] = temp;
}
}
}
}
Sample usage like in your example:
var array = Enumerable.Range(1, 5).ToArray(); // { 1, 2, 3, 4, 5 }
array.Rotate(-4); // { 5, 1, 2, 3, 4 }
Isn't using IEnumerables better? Since It won't perform all of those maths, won't allocate that many arrays, etc
public static int[] Rotate(int[] elements, int numberOfRotations)
{
IEnumerable<int> newEnd = elements.Take(numberOfRotations);
IEnumerable<int> newBegin = elements.Skip(numberOfRotations);
return newBegin.Union(newEnd).ToArray();
}
IF you don't actually need to return an array, you can even remove the .ToArray() and return an IEnumerable
Usage:
void Main()
{
int[] n = { 1, 2, 3, 4, 5 };
int d = 4;
int[] rotated = Rotate(n,d);
Console.WriteLine(String.Join(" ", rotated));
}
I have also tried this and below is my approach...
Thank you
public static int[] RotationOfArray(int[] A, int k)
{
if (A == null || A.Length==0)
return null;
int[] result =new int[A.Length];
int arrayLength=A.Length;
int moveBy = k % arrayLength;
for (int i = 0; i < arrayLength; i++)
{
int tmp = i + moveBy;
if (tmp > arrayLength-1)
{
tmp = + (tmp - arrayLength);
}
result[tmp] = A[i];
}
return result;
}
I have tried to used stack and queue in C# to achieve the output as follows:
public int[] rotateArray(int[] A, int rotate)
{
Queue<int> q = new Queue<int>(A);
Stack<int> s;
while (rotate > 0)
{
s = new Stack<int>(q);
int x = s.Pop();
s = new Stack<int>(s);
s.Push(x);
q = new Queue<int>(s);
rotate--;
}
return q.ToArray();
}
I've solve the challange from Hackerrank by following code. Hope it helps.
using System;
using System.Collections.Generic;
using System.IO;
using System.Text;
namespace ConsoleApp1
{
class ArrayLeftRotationSolver
{
TextWriter mTextWriter;
public ArrayLeftRotationSolver()
{
mTextWriter = new StreamWriter(#System.Environment.GetEnvironmentVariable("OUTPUT_PATH"), true);
}
public void Solve()
{
string[] nd = Console.ReadLine().Split(' ');
int n = Convert.ToInt32(nd[0]);
int d = Convert.ToInt32(nd[1]);
int[] a = Array.ConvertAll(Console.ReadLine().Split(' '), aTemp => Convert.ToInt32(aTemp))
;
int[] result = rotLeft(a, d);
mTextWriter.WriteLine(string.Join(" ", result));
mTextWriter.Flush();
mTextWriter.Close();
}
private int[] rotLeft(int[] arr, int shift)
{
int n = arr.Length;
shift %= n;
int[] vec = new int[n];
for (int i = 0; i < n; i++)
{
vec[(n + i - shift) % n] = arr[i];
}
return vec;
}
static void Main(string[] args)
{
ArrayLeftRotationSolver solver = new ArrayLeftRotationSolver();
solver.Solve();
}
}
}
Hope this helps.
public static int[] leftrotation(int[] arr, int d)
{
int[] newarr = new int[arr.Length];
var n = arr.Length;
bool isswapped = false;
for (int i = 0; i < n; i++)
{
int index = Math.Abs((i) -d);
if(index == 0)
{
isswapped = true;
}
if (!isswapped)
{
int finalindex = (n) - index;
newarr[finalindex] = arr[i];
}
else
{
newarr[index] = arr[i];
}
}
return newarr;
}
Take the Item at position 0 and add it at the end. remove the item at position 0. repeat n times.
List<int> iList = new List<int>();
private void shift(int n)
{
for (int i = 0; i < n; i++)
{
iList.Add(iList[0]);
iList.RemoveAt(0);
}
}
An old question, but I thought I'd add another possible solution using just one intermediate array (really, 2 if you include the LINQ Take expression). This code rotates to right rather than left, but may be useful nonetheless.
public static Int32[] ArrayRightRotation(Int32[] A, Int32 k)
{
if (A == null)
{
return A;
}
if (!A.Any())
{
return A;
}
if (k % A.Length == 0)
{
return A;
}
if (A.Length == 1)
{
return A;
}
if (A.Distinct().Count() == 1)
{
return A;
}
for (var i = 0; i < k; i++)
{
var intermediateArray = new List<Int32> {A.Last()};
intermediateArray.AddRange(A.Take(A.Length - 1).ToList());
A = intermediateArray.ToArray();
}
return A;
}
O(1) space, O(n) time solution
I think in theory this is as optimal as it gets, since it makes a.Length in-place swaps and 1 temp variable swap per inner loop.
However I suspect O(d) space solutions would be faster in real life due to less code branching (fewer CPU command pipeline resets) and cache locality (mostly sequential access vs in d element steps).
static int[] RotateInplaceLeft(int[] a, int d)
{
var swapCount = 0;
//get canonical/actual d
d = d % a.Length;
if(d < 0) d += a.Length;
if(d == 0) return a;
for (var i = 0; swapCount < a.Length; i++) //we're done after a.Length swaps
{
var dstIdx = i; //we need this becasue of ~this: https://youtu.be/lJ3CD9M3nEQ?t=251
var first = a[i]; //save first element in this group
for (var j = 0; j < a.Length; j++)
{
var srcIdx = (dstIdx + d) % a.Length;
if(srcIdx == i)// circled around
{
a[dstIdx] = first;
swapCount++;
break; //hence we're done with this group
}
a[dstIdx] = a[srcIdx];
dstIdx = srcIdx;
swapCount++;
}
}
return a;
}
If you take a look at constrains you will see that d <= n (number of rotations <= number of elements in array). Because of that this can be solved in 1 line.
static int[] rotLeft(int[] a, int d)
{
return a.Skip(d).Concat(a.Take(d)).ToArray();
}
// using the same same array, and only one temp variable
// shifting everything several times by one
// works, simple, but slow
public static int[] ArrayRotateLeftCyclical(int[] a, int shift)
{
var length = a.Length;
for (int j = 0; j < shift; j++)
{
int t = a[0];
for (int i = 0; i < length; i++)
{
if (i == length - 1)
a[i] = t;
else
a[i] = a[i + 1];
}
}
return a;
}
Let's say if I have a array of integer 'Arr'. To rotate the array 'n' you can do as follows:
static int[] leftRotation(int[] Arr, int n)
{
int tempVariable = 0;
Queue<int> TempQueue = new Queue<int>(a);
for(int i=1;i<=d;i++)
{
tempVariable = TempQueue.Dequeue();
TempQueue.Enqueue(t);
}
return TempQueue.ToArray();`
}
Let me know if any comments. Thanks!
This is my attempt. It is easy, but for some reason it timed out on big chunks of data:
int arrayLength = arr.Length;
int tmpCell = 0;
for (int rotation = 1; rotation <= d; rotation++)
{
for (int i = 0; i < arrayLength; i++)
{
if (arr[i] < arrayElementMinValue || arr[i] > arrayElementMaxValue)
{
throw new ArgumentException($"Array element needs to be between {arrayElementMinValue} and {arrayElementMaxValue}");
}
if (i == 0)
{
tmpCell = arr[0];
arr[0] = arr[1];
}
else if (i == arrayLength - 1)
{
arr[arrayLength - 1] = tmpCell;
}
else
{
arr[i] = arr[i + 1];
}
}
}
what about this?
public static void RotateArrayAndPrint(int[] n, int rotate)
{
for (int i = 1; i <= n.Length; i++)
{
var arrIndex = (i + rotate) > n.Length ? n.Length - (i + rotate) : (i + rotate);
arrIndex = arrIndex < 0 ? arrIndex * -1 : arrIndex;
var output = n[arrIndex-1];
Console.Write(output + " ");
}
}
It's very straight forward answer.
Main thing is how you choose the start index.
public static List<int> rotateLeft(int d, List<int> arr) {
int n = arr.Count;
List<int> t = new List<int>();
int h = d;
for (int j = 0; j < n; j++)
{
if ((j + d) % n == 0)
{
h = 0;
}
t.Add(arr[h]);
h++;
}
return t;
}
using this code, I have successfully submitted to hacker rank problem,
// fast and beautiful method
// reusing the same array
// using small temp array to store replaced values when unavoidable
// a - array, s - shift
public static int[] ArrayRotateLeftWithSmallTempArray(int[] a, int s)
{
var l = a.Length;
var t = new int[s]; // temp array with size s = shift
for (int i = 0; i < l; i++)
{
// save cells which will be replaced by shift
if (i < s)
t[i] = a[i];
if (i + s < l)
a[i] = a[i + s];
else
a[i] = t[i + s - l];
}
return a;
}
https://github.com/sam-klok/ArraysRotation
public static void Rotate(int[] arr, int steps)
{
for (int i = 0; i < steps; i++)
{
int previousValue = arr[arr.Length - 1];
for (int j = 0; j < arr.Length; j++)
{
int currentValue = arr[j];
arr[j] = previousValue;
previousValue = currentValue;
}
}
}
Here is an in-place Rotate implementation of a trick posted by גלעד ברקן in another question. The trick is:
Example, k = 3:
1234567
First reverse in place each of the two sections delineated by n-k:
4321 765
Now reverse the whole array:
5671234
My implementation, based on the Array.Reverse method:
/// <summary>
/// Rotate left for negative k. Rotate right for positive k.
/// </summary>
public static void Rotate<T>(T[] array, int k)
{
ArgumentNullException.ThrowIfNull(array);
k = k % array.Length;
if (k < 0) k += array.Length;
if (k == 0) return;
Debug.Assert(k > 0);
Debug.Assert(k < array.Length);
Array.Reverse(array, 0, array.Length - k);
Array.Reverse(array, array.Length - k, k);
Array.Reverse(array);
}
Live demo.
Output:
Array: 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12
Rotate(5)
Array: 8, 9, 10, 11, 12, 1, 2, 3, 4, 5, 6, 7
Rotate(-2)
Array: 10, 11, 12, 1, 2, 3, 4, 5, 6, 7, 8, 9

How to get integers to an array (from integer i to integer j). c#

I want to store integers(in an array or anything) that in range of int "i" and int "j".
eg:-Think, "int i = 1" and "int j = 10".I want to store integers from 1 and 10.
So that (1,2,3,4,5,6,7,8,9,10)
Because I want to answer to HackerRank "Beautiful Days at the Movies".
link below.
https://www.hackerrank.com/challenges/beautiful-days-at-the-movies/problem?isFullScreen=false
here is my code and it a garbage.
static int beautifulDays(int i, int j, int k) {
var total = 0;
for(var a = i; a <= j; a++ )
{
if (a != 0)
{
int ri = Reverse(i);
int rj = Reverse(j);
var ra = Reverse(a);
if((ra/k) % 1 == 0)
{
total++;
}
if((rj/k) % 1 == 0)
{
total++;
}
if((ri/k) % 1 == 0)
{
total++;
}
}
return total;
}
return total;
}
public static int Reverse(int inval)
{
int result = 0;
do
{
result = (result * 10) + (inval % 10);
inval = inval / 10;
}
while(inval > 0);
return result;
}
simply, can you give me the answer of HackerRank "Beautiful Days at the Movies".
link below.
https://www.hackerrank.com/challenges/beautiful-days-at-the-movies/problem?isFullScreen=false
Using Java you can easily stream a range of numbers with IntStream, then map the reverse function for each value, then filter those that fulfils the condition and count. With streams you don't need to store, you can get straight to the answer.
IntUnaryOperator reverse = (opperand) -> {
int reversed = 0;
int num = opperand;
while (num != 0) {
int digit = num % 10;
reversed = reversed * 10 + digit;
num /= 10;
}
return Math.abs(opperand - reversed);
};
return (int) IntStream.rangeClosed(i, j).map(reverse)
.filter(v -> v % k == 0).count();

Integer to Integer Array C# [duplicate]

This question already has answers here:
Is there an easy way to turn an int into an array of ints of each digit?
(11 answers)
Closed 1 year ago.
I had to split an int "123456" each value of it to an Int[] and i have already a Solution but i dont know is there any better way :
My solution was :
public static int[] intToArray(int num){
String holder = num.ToString();
int[] numbers = new int[Holder.ToString().Length];
for(int i=0;i<numbers.length;i++){
numbers[i] = Convert.toInt32(holder.CharAt(i));
}
return numbers;
}
A simple solution using LINQ
int[] result = yourInt.ToString().Select(o=> Convert.ToInt32(o) - 48 ).ToArray()
I believe this will be better than converting back and forth. As opposed to JBSnorro´s answer I reverse after converting to an array and therefore avoid IEnumerable´s which I think will contribute to a little bit faster code. This method work for non negative numbers, so 0 will return new int[1] { 0 }.
If it should work for negative numbers, you could do a n = Math.Abs(n) but I don't think that makes sense.
Furthermore, if it should be more performant, I could create the final array to begin with by making a binary-search like combination of if-statements to determine the number of digits.
public static int[] digitArr(int n)
{
if (n == 0) return new int[1] { 0 };
var digits = new List<int>();
for (; n != 0; n /= 10)
digits.Add(n % 10);
var arr = digits.ToArray();
Array.Reverse(arr);
return arr;
}
Update 2018:
public static int numDigits(int n) {
if (n < 0) {
n = (n == Int32.MinValue) ? Int32.MaxValue : -n;
}
if (n < 10) return 1;
if (n < 100) return 2;
if (n < 1000) return 3;
if (n < 10000) return 4;
if (n < 100000) return 5;
if (n < 1000000) return 6;
if (n < 10000000) return 7;
if (n < 100000000) return 8;
if (n < 1000000000) return 9;
return 10;
}
public static int[] digitArr2(int n)
{
var result = new int[numDigits(n)];
for (int i = result.Length - 1; i >= 0; i--) {
result[i] = n % 10;
n /= 10;
}
return result;
}
int[] outarry = Array.ConvertAll(num.ToString().ToArray(), x=>(int)x);
but if you want to convert it to 1,2,3,4,5:
int[] outarry = Array.ConvertAll(num.ToString().ToArray(), x=>(int)x - 48);
I'd do it like this:
var result = new List<int>();
while (num != 0) {
result.Insert(0, num % 10);
num = num / 10;
}
return result.ToArray();
Slightly less performant but possibly more elegant is:
return num.ToString().Select(c => Convert.ToInt32(c.ToString())).ToArray();
Note that these both return 1,2,3,4,5,6 rather than 49,50,51,52,53,54 (i.e. the byte codes for the characters '1','2','3','4','5','6') as your code does. I assume this is the actual intent?
Using conversion from int to string and back probably isn't that fast. I would use the following
public static int[] ToDigitArray(int i)
{
List<int> result = new List<int>();
while (i != 0)
{
result.Add(i % 10);
i /= 10;
}
return result.Reverse().ToArray();
}
I do have to note that this only works for strictly positive integers.
EDIT:
I came up with an alternative. If performance really is an issue, this will probably be faster, although you can only be sure by checking it yourself for your specific usage and application.
public static int[] ToDigitArray(int n)
{
int[] result = new int[GetDigitArrayLength(n)];
for (int i = 0; i < result.Length; i++)
{
result[result.Length - i - 1] = n % 10;
n /= 10;
}
return result;
}
private static int GetDigitArrayLength(int n)
{
if (n == 0)
return 1;
return 1 + (int)Math.Log10(n);
}
This works when n is nonnegative.
Thanks to ASCII character table. The simple answer using LINQ above yields answer + 48.
Either
int[] result = youtInt.ToString().Select(o => Convert.ToInt32(o) - 48).ToArray();
or
int[] result = youtInt.ToString().Select(o => int.Parse(o.ToString())).ToArray();
can be used
You can do that without converting it to a string and back:
public static int[] intToArray(int num) {
List<int> numbers = new List<int>();
do {
numbers.Insert(0, num % 10);
num /= 10;
} while (num > 0);
return numbers.ToArray();
}
It only works for positive values, of course, but your original code also have that limitation.
I would convert it in the below manner
if (num == 0) return new int[1] { 0 };
var digits = new List<int>();
while (num > 0)
{
digits.Add(num % 10);
num /= 10;
}
var arr = digits.ToArray().Reverse().ToArray();
string DecimalToBase(int iDec, int numbase)
{
string strBin = "";
int[] result = new int[32];
int MaxBit = 32;
for(; iDec > 0; iDec/=numbase)
{
int rem = iDec % numbase;
result[--MaxBit] = rem;
}
for (int i=0;i<result.Length;i++)
if ((int)result.GetValue(i) >= base10)
strBin += cHexa[(int)result.GetValue(i)%base10];
else
strBin += result.GetValue(i);
strBin = strBin.TrimStart(new char[] {'0'});
return strBin;
}
int BaseToDecimal(string sBase, int numbase)
{
int dec = 0;
int b;
int iProduct=1;
string sHexa = "";
if (numbase > base10)
for (int i=0;i<cHexa.Length;i++)
sHexa += cHexa.GetValue(i).ToString();
for(int i=sBase.Length-1; i>=0; i--,iProduct *= numbase)
{
string sValue = sBase[i].ToString();
if (sValue.IndexOfAny(cHexa) >=0)
b=iHexaNumeric[sHexa.IndexOf(sBase[i])];
else
b= (int) sBase[i] - asciiDiff;
dec += (b * iProduct);
}
return dec;
}
i had similar requirement .. i took from many good ideas, and added a couple missing pieces .. where many folks weren’t handling zero or negative values. this is what i came up with:
public static int[] DigitsFromInteger(int n)
{
int _n = Math.Abs(n);
int length = ((int)Math.Log10(_n > 0 ? _n : 1)) + 1;
int[] digits = new int[length];
for (int i = 0; i < length; i++)
{
digits[(length - i) - 1] = _n % 10 * ((i == (length - 1) && n < 0) ? -1 : 1);
_n /= 10;
}
return digits;
}
i think this is pretty clean .. although, it is true we're doing a conditional check and several extraneous calculations with each iteration .. while i think they’re nominal in this case, you could optimize a step further this way:
public static int[] DigitsFromInteger(int n)
{
int _n = Math.Abs(n);
int length = ((int)Math.Log10(_n > 0 ? _n : 1)) + 1;
int[] digits = new int[length];
for (int i = 0; i < length; i++)
{
//digits[(length - i) - 1] = _n % 10 * ((i == (length - 1) && n < 0) ? -1 : 1);
digits[(length - i) - 1] = _n % 10;
_n /= 10;
}
if (n < 0)
digits[0] *= -1;
return digits;
}
A slightly more concise way to do MarkXA's one-line version:
int[] result = n.ToString().Select(c => (int)Char.GetNumericValue(c)).ToArray();
GetNumericValue returns the visible number in your char as a double, so if your string is "12345", it will return the doubles 1,2,3,4,5, which can each be cast to an int. Note that using Convert.ToInt32 on a char in C# returns the ASCII code, so you would get 49,50,51,52,53. This can understandably lead to a mistake.
Here is a Good Solution for Convert Your Integer into Array i.e:
int a= 5478 into int[]
There is no issue if You Have a String and You want to convert a String into integer Array for example
string str=4561; //Convert into
array[0]=4;
array[1]=5;
array[2]=6;
array[3]=7;
Note: The Number of zero (0) in devider are Equal to the Length of input and Set Your Array Length According to Your input length
Now Check the Coding:
string str=4587;
int value = Convert.ToInt32(str);
int[] arr = new int[4];
int devider = 10000;
for (int i = 0; i < str.Length; i++)
{
int m = 0;
devider /= 10;
arr[i] = value / devider;
m = value / devider;
value -= (m * devider);
}
private static int[] ConvertIntToArray(int variable)
{
string converter = "" + variable;
int[] convertedArray = new int[converter.Length];
for (int i=0; i < convertedArray.Length;i++) //it can be also converter.Length
{
convertedArray[i] = int.Parse(converter.Substring(i, 1));
}
return convertedArray;
}
We get int via using method. Then, convert it to string immediately (123456->"123456"). We have a string called converter and carry to int value. Our string have a string.Length, especially same length of int so, we create an array called convertedArray that we have the length, that is converter(string) length. Then, we get in the loop where we are convert the string to int one by one via using string.Substring(i,1), and assign the value convertedArray[i]. Then, return the convertedArray.At the main or any method you can easily call the method.
public static int[] intToArray(int num)
{
num = Math.Abs(num);
int length = num.ToString().Length;
int[] arr = new int[length];
do
{
arr[--length] = num % 10;
num /= 10;
} while (num != 0);
return arr;
}
Dividing by system base (decimal in this case) removes the right most digit, and we get that digit by remainder operator. We keep repeating until we end up with a zero. Each time a digit is removed it will be stored in an array starting from the end of the array and backward to avoid the need of revering the array at the end. The Math.Abs() function is to handle the negative input, also the array is instantiated with the same size as input length.
Integer or Long to Integer Array C#
Convert it to char array and subtract 48.
public static int[] IntToArray(int value, int length)
{
char[] charArray = new char[length];
charArray = value.ToString().ToCharArray();
int[] intArray = new int[length];
for (int i = 0; i < intArray.Length; i++)
{
intArray[i] = charArray[i] - 48;
}
return intArray;
}
public static int[] LongToIntArray(long value, int length)
{
char[] charArray = new char[length];
charArray = value.ToString().ToCharArray();
int[] intArray = new int[length];
for (int i = 0; i < intArray.Length; i++)
{
intArray[i] = charArray[i] - 48;
}
return intArray;
}

C# fibonacci function returning errors

I am practising a C# console application, and I am trying to get the function to verify if the number appears in a fibonacci series or not but I'm getting errors.
What I did was:
class Program
{
static void Main(string[] args)
{
System.Console.WriteLine(isFibonacci(20));
}
static int isFibonacci(int n)
{
int[] fib = new int[100];
fib[0] = 1;
fib[1] = 1;
for (int i = 2; i <= 100; i++)
{
fib[i] = fib[i - 1] + fib[i - 2];
if (n == fib[i])
{
return 1;
}
}
return 0;
}
}
Can anybody tell me what am I doing wrong here?
Here's a fun solution using an infinite iterator block:
IEnumerable<int> Fibonacci()
{
int n1 = 0;
int n2 = 1;
yield return 1;
while (true)
{
int n = n1 + n2;
n1 = n2;
n2 = n;
yield return n;
}
}
bool isFibonacci(int n)
{
foreach (int f in Fibonacci())
{
if (f > n) return false;
if (f == n) return true;
}
}
I actually really like this kind of Fibonacci implementation vs the tradition recursive solution, because it keeps the work used to complete a term available to complete the next. The traditional recursive solution duplicates some work, because it needs two recursive calls each term.
The problem lies in <= the following statement:
for (int i = 2; i <= 100; i++)
more to the point the =. There is no fib[100] (C# zero counts) so when you check on i=100 you get an exception.
the proper statement should be
for (int i = 2; i < 100; i++)
or even better
for (int i = 2; i < fib.Length; i++)
And here is a solution that beats all of yours!
Because, why iteration when you have smart mathematicians doing closed-form solutions for you? :)
static bool IsFibonacci(int number)
{
//Uses a closed form solution for the fibonacci number calculation.
//http://en.wikipedia.org/wiki/Fibonacci_number#Closed-form_expression
double fi = (1 + Math.Sqrt(5)) / 2.0; //Golden ratio
int n = (int) Math.Floor(Math.Log(number * Math.Sqrt(5) + 0.5, fi)); //Find's the index (n) of the given number in the fibonacci sequence
int actualFibonacciNumber = (int)Math.Floor(Math.Pow(fi, n) / Math.Sqrt(5) + 0.5); //Finds the actual number corresponding to given index (n)
return actualFibonacciNumber == number;
}
Well, for starters your array is only 10 long and you're filling it with ~100 items (out-of-range-exception) - but there are better ways to do this...
for example, using this post:
long val = ...
bool isFib = Fibonacci().TakeWhile(x => x <= val).Last() == val;
int[] fib = new int[10];
for (int i = 2; i <= *100*; i++)
You're going out of the bounds of your array because your loop conditional is too large. A more traditional approach would be to bound the loop by the size of the array:
for (int i = 2; i < fib.Length; i++)
And make your array bigger, but as Marc said, there are better ways to do this, and I would advise you spend some time reading the wikipedia article on Fibonacci numbers.
One thing you can do is check for an early exit. Since you're trying to determine if a given number is in the Fibonacci sequence, you can do bounds checking to exit early.
Example:
static bool isFibonacci(int n)
{
int[] fib = new int[100];
fib[0] = 1;
fib[1] = 1;
for (int i = 2; i <= fib.Length; i++)
{
fib[i] = fib[i - 1] + fib[i - 2];
if (n == fib[i])
{
return true;
}
else if (n < fib[i])
{
return false; //your number has been surpassed in the fib seq
}
}
return false;
}
public static int FibNo(int n) {
int result = 0; int No = 0; int N1 = 1;
if (n< 0)
{ throw new ArguementException("number must be a positive value"); }
if (n <= 1)
{ result = n; return result; }
for(int x=1; x < n; x++)
{ result = No + N1; No = N1; N1=result; }
return result;
}

Sum of digits in C#

What's the fastest and easiest to read implementation of calculating the sum of digits?
I.e. Given the number: 17463 = 1 + 7 + 4 + 6 + 3 = 21
You could do it arithmetically, without using a string:
sum = 0;
while (n != 0) {
sum += n % 10;
n /= 10;
}
I use
int result = 17463.ToString().Sum(c => c - '0');
It uses only 1 line of code.
For integer numbers, Greg Hewgill has most of the answer, but forgets to account for the n < 0. The sum of the digits of -1234 should still be 10, not -10.
n = Math.Abs(n);
sum = 0;
while (n != 0) {
sum += n % 10;
n /= 10;
}
It the number is a floating point number, a different approach should be taken, and chaowman's solution will completely fail when it hits the decimal point.
public static int SumDigits(int value)
{
int sum = 0;
while (value != 0)
{
int rem;
value = Math.DivRem(value, 10, out rem);
sum += rem;
}
return sum;
}
int num = 12346;
int sum = 0;
for (int n = num; n > 0; sum += n % 10, n /= 10) ;
I like the chaowman's response, but would do one change
int result = 17463.ToString().Sum(c => Convert.ToInt32(c));
I'm not even sure the c - '0', syntax would work? (substracting two characters should give a character as a result I think?)
I think it's the most readable version (using of the word sum in combination with the lambda expression showing that you'll do it for every char). But indeed, I don't think it will be the fastest.
I thought I'd just post this for completion's sake:
If you need a recursive sum of digits, e.g: 17463 -> 1 + 7 + 4 + 6 + 3 = 21 -> 2 + 1 = 3
then the best solution would be
int result = input % 9;
return (result == 0 && input > 0) ? 9 : result;
int n = 17463; int sum = 0;
for (int i = n; i > 0; i = i / 10)
{
sum = sum + i % 10;
}
Console.WriteLine(sum);
Console.ReadLine();
I would suggest that the easiest to read implementation would be something like:
public int sum(int number)
{
int ret = 0;
foreach (char c in Math.Abs(number).ToString())
ret += c - '0';
return ret;
}
This works, and is quite easy to read. BTW: Convert.ToInt32('3') gives 51, not 3. Convert.ToInt32('3' - '0') gives 3.
I would assume that the fastest implementation is Greg Hewgill's arithmetric solution.
private static int getDigitSum(int ds)
{
int dssum = 0;
while (ds > 0)
{
dssum += ds % 10;
ds /= 10;
if (dssum > 9)
{
dssum -= 9;
}
}
return dssum;
}
This is to provide the sum of digits between 0-9
public static int SumDigits1(int n)
{
int sum = 0;
int rem;
while (n != 0)
{
n = Math.DivRem(n, 10, out rem);
sum += rem;
}
return sum;
}
public static int SumDigits2(int n)
{
int sum = 0;
int rem;
for (sum = 0; n != 0; sum += rem)
n = Math.DivRem(n, 10, out rem);
return sum;
}
public static int SumDigits3(int n)
{
int sum = 0;
while (n != 0)
{
sum += n % 10;
n /= 10;
}
return sum;
}
Complete code in: https://dotnetfiddle.net/lwKHyA
int j, k = 1234;
for(j=0;j+=k%10,k/=10;);
A while back, I had to find the digit sum of something. I used Muhammad Hasan Khan's code, however it kept returning the right number as a recurring decimal, i.e. when the digit sum was 4, i'd get 4.44444444444444 etc.
Hence I edited it, getting the digit sum correct each time with this code:
double a, n, sumD;
for (n = a; n > 0; sumD += n % 10, n /= 10);
int sumI = (int)Math.Floor(sumD);
where a is the number whose digit sum you want, n is a double used for this process, sumD is the digit sum in double and sumI is the digit sum in integer, so the correct digit sum.
static int SumOfDigits(int num)
{
string stringNum = num.ToString();
int sum = 0;
for (int i = 0; i < stringNum.Length; i++)
{
sum+= int.Parse(Convert.ToString(stringNum[i]));
}
return sum;
}
If one wants to perform specific operations like add odd numbers/even numbers only, add numbers with odd index/even index only, then following code suits best. In this example, I have added odd numbers from the input number.
using System;
public class Program
{
public static void Main()
{
Console.WriteLine("Please Input number");
Console.WriteLine(GetSum(Console.ReadLine()));
}
public static int GetSum(string num){
int summ = 0;
for(int i=0; i < num.Length; i++){
int currentNum;
if(int.TryParse(num[i].ToString(),out currentNum)){
if(currentNum % 2 == 1){
summ += currentNum;
}
}
}
return summ;
}
}
The simplest and easiest way would be using loops to find sum of digits.
int sum = 0;
int n = 1234;
while(n > 0)
{
sum += n%10;
n /= 10;
}
#include <stdio.h>
int main (void) {
int sum = 0;
int n;
printf("Enter ir num ");
scanf("%i", &n);
while (n > 0) {
sum += n % 10;
n /= 10;
}
printf("Sum of digits is %i\n", sum);
return 0;
}
Surprised nobody considered the Substring method. Don't know whether its more efficient or not. For anyone who knows how to use this method, its quite intuitive for cases like this.
string number = "17463";
int sum = 0;
String singleDigit = "";
for (int i = 0; i < number.Length; i++)
{
singleDigit = number.Substring(i, 1);
sum = sum + int.Parse(singleDigit);
}
Console.WriteLine(sum);
Console.ReadLine();

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