I get three values from three variables. How can i check who is the highest number and who is the lowest number?
The numbers are represented like this:
private int _score1;
private int _score2;
private int _score2;
Code:
Public int Highest
{
return the highest number here;
}
public int Lowest
{
return the lowest number here;
}
Can i calculate the highest and the lowest number in my constructor?
The obligatory Linq answer:
Public int Highest(params int[] inputs)
{
return inputs.Max();
}
public int Lowest(params int[] inputs)
{
return inputs.Min();
}
The beauty of this one is that it can take any number of integer inputs. To make it fail-safe you should check for a null/empty inputs array (meaning nothing was passed into the method).
To do this without Linq, you basically just have to mimic the logic performed by the extension method:
Public int Lowest(params int[] inputs)
{
int lowest = inputs[0];
foreach(var input in inputs)
if(input < lowest) lowest = input;
return lowest;
}
Again, to make it foolproof you should check for an empty or null inputs array, because calling Lowest() will throw an ArrayIndexOutOfBoundsException.
This is one way to do it:
public int Highest
{
get { return Math.Max(_score1, Math.Max(_score2, _score3)); }
}
public int Lowest
{
get { return Math.Min(_score1, Math.Min(_score2, _score3)); }
}
int[] numbers = new[] { _score1, _score2, _score3 };
int min = numbers.Min();
int max = numbers.Max();
Highest
return (x > y) ? (x > z ? x : z) : (y > z ? y : z)
Lowest
return (x < y) ? (x < z ? x : z) : (y < z ? y : z)
Here's something you could do:
public class Numbers
{
private int _number1;
private int _number2;
private int _number3;
public readonly int Highest;
public readonly int Lowest;
public Numbers(int num1, int num2, int num3)
{
int high;
int low;
_number1 = num1;
_number2 = num2;
_number3 = num3;
high = num1 > num2 ? num1 : num2;
high = high > num3 ? high : num3;
low = num1 < num2 ? num1 : num2;
low = low < num3 ? low : num3;
Highest = high;
Lowest = low;
}
}
If you want to simply check which is the highest you can do this
private int _highest = _score1;
if (_score2 > _highest)
_highest = _score2
if (_score3 > _highest)
_highest = _score3
Similarly, you can find the lowest like so
private int _lowest = _score1;
if (_score2 < _lowest)
_lowest = _score2
if (_score3 < _lowest)
_lowest = _score3
Using LINQ-to-Objects, you could do something like this.
var numbers = new [] {_score1, _score2, _score3};
numbers.Sort();
var lowest = numbers.First();
var highest = numbers.Last();
For a reference: in some cases you'll be having more than three variables (possibly not knowing how many). If they are stored in an array, here's the way to do it:
int Highest(int[] numbers)
{
int highest = Int32.MinValue();
for (int i = 0; i < numbers.Length; i++)
{
if (numbers[i] > highest)
highest = numbers[i];
}
return highest;
}
int Lowest(int[] numbers)
{
int lowest = Int32.MaxValue();
for (int i = 0; i < numbers.Length; i++)
{
if (numbers[i] < lowest)
lowest = numbers[i];
}
return lowest;
}
This will work for any length of int array.
Find the Largest and smallest number
using System;
namespace LargeSmall;
{
class Program
{
public static void Main()
{
float large, small;
int[] a = new int[50];
Console.WriteLine("Enter the size of Array");
int max = Convert.ToInt32(Console.ReadLine());
Console.WriteLine("Enter the array elements");
for (int i = 0; i < max; i++)
{
string s1 = Console.ReadLine();
a[i] = Int32.Parse(s1);
}
Console.Write("");
large = a[0];
small = a[0];
for (int i = 1; i < max; i++)
{
if (a[i] > large)
large = a[i];
else if (a[i] < small)
small = a[i];
}
Console.WriteLine("Largest element in the array is {0}", large);
Console.WriteLine("Smallest element in the array is {0}", small);
}
}
Here is simple logic for finding Smallest Number
Input : 11, 0 , 3, 33 Output : "0"
namespace PurushLogics
{
class Purush_SmallestNumber
{
static void Main()
{
int count = 0;
Console.WriteLine("Enter Total Number of Integers\n");
count = int.Parse(Console.ReadLine());
int[] numbers = new int[count];
Console.WriteLine("Enter the numbers"); // Input 44, 55, 111, 2 Output = "2"
for (int temp = 0; temp < count; temp++)
{
numbers[temp] = int.Parse(Console.ReadLine());
}
int smallest = numbers[0];
for (int small = 1; small < numbers.Length; small++)
{
if (smallest > numbers[small])
{
smallest = numbers[small];
}
}
Console.WriteLine("Smallest Number is : \"{0}\"",smallest);
Console.ReadKey();
}
}
}
Related
Hi I am trying to make a program where the user can enter up to 25 numbers and then it tells the user the average, smallest and largest number and the mode of the numbers entered but I am not sure on how to do this.
Any Guidance would be appreciated
here is what I have so far
static void Main(string[] args)
{
Console.WriteLine("enter the amount of numbers you would like to use of: ");
int arraylength = Int32.Parse(Console.ReadLine());
int[] AverageArray = new int[25];
//filling the array with user input
for (int i = 0; i < AverageArray.Length; i++)
{
Console.Write("enter the numbers you wish to find the average for: ");
AverageArray[i] = Int32.Parse(Console.ReadLine());
}
//printing out the array
Console.WriteLine("here is the average: ");
for (int i = 0; i < AverageArray.Length; i++)
{
Console.WriteLine(AverageArray[i]);
}
Console.WriteLine(FindAverage(AverageArray));
}
public static double FindAverage(int[] averageNumbers)
{
int arraySum = 0;
for (int i = 0; i < averageNumbers.Length; i++)
arraySum += averageNumbers[i];
return arraySum / averageNumbers.Length;
}
public static double LargestNumber(int[] Nums, int Count)
{
}
public static double SmallestNumber(int[] Nums, int Count)
{
}
public static double Mode(int[] Nums, int Count)
{
}
}
var smallest = AverageArray.Min();
var largest = AverageArray.Max();
var mode = AverageArray.GroupBy(v => v)
.OrderByDescending(g => g.Count())
.First()
.Key;
If you don't know how to use linq
/*Assign Initial Value*/
int highestNum = numberArray[0];
int lowestNum = numberArray[0];
int sum = 0;
foreach (int item in numberArray)
{
/*Get The Highest Number*/
if (item > highestNum)
{
highestNum = item;
}
/*Get The Lowest Number*/
if (item < highestNum)
{
lowestNum = item;
}
//get the sum
sum = item + sum;
}
//display the Highest Num
Console.WriteLine(highestNum);
//display the Lowest Num
Console.WriteLine(lowestNum);
//Display The Average up to 2 decimal Places
Console.WriteLine((sum/numberArray.Length).ToString("0.00"));
public static double LargestNumber(int[] Nums, int Count)
{
double max = Nums[0];
for (int i = 1; i < Count; i++)
if (Nums[i] > max) max = Nums[i];
return min;
}
public static double SmallestNumber(int[] Nums, int Count)
{
double min = Nums[0];
for (int i = 1; i < Count; i++)
if (Nums[i] < min) min = Nums[i];
return min;
}
public static double Mode(int[] Nums, int Count)
{
double mode = Nums[0];
int maxCount = 1;
for (int i = 0; i < Count; i++)
{
int val = Nums[i];
int count = 1;
for (int j = i+1; j < Count; j++)
{
if (Nums[j] == val)
count++;
}
if (count > maxCount)
{
mode = val;
maxCount = count;
}
}
return mode;
}
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;
}
In a program that requires the user to input the number of integers, I cannot find out how to display the minimum of all the values.
static void Main(string[] args)
{
Console.WriteLine("\n Number of values:");
int num = Convert.ToInt32(Console.ReadLine());
int[] number = new int[num];
int i;
for (i = 0; i < number.Length; i++)
{
Console.WriteLine("\n Enter value:");
number[i] = Convert.ToInt32(Console.ReadLine());
}
for (i = 0; i < number.Length; i++)
{
int Min = number[0];
if (number[i + 1] < Min)
{
Min = number[i + 1];
}
}
Console.WriteLine("Smallest is {0}", Min);
}
Declare Min outside for loop
int Min = number[0];
for (i = 1; i < number.Length; i++)
{
if (number[i] < Min)
{
Min = number[i];
}
}
There are methods that do this for you:
int[] number = new int[num];
int min = number.Min();
Use this:
int minNumber = numbers.Min();
You can use the Min method to calculate this.
int min = number.Min();
public static int FindMinimum(int[] values)
{
int result = int.MaxValue;
foreach (int value in values)
result = Math.Min(result, value);
return result;
}
Your current code will overflow the array. You should check index 0 first and then check the rest.
Replace
for (i = 0; i < number.Length; i++)
{
int Min = number[0];
if (number[i + 1] < Min)
{
Min = number[i + 1];
}
}
with
int Min = number[0];
for (i = 1; i < number.Length; i++)
{
if (number[i] < Min)
{
Min = number[i];
}
}
However, you could simply use Enumerable.Min() as int Min = number.Min(x => x)
how about this?!
int[] Numbers = new int[5] { 3, 5, 7, 9, 11, 15 };
var q = (from Num in Numbers
select Num).Min();
Have a look at LINQ samples from MSDN: http://code.msdn.microsoft.com/101-LINQ-Samples-3fb9811b
you can use linq to object.
use Min method
sort your data in descending and take the first one
Declare Min as global static for example - public static int Min
Change the loop condition to for (i = 0; i < number.Length -1 ; i++) you exceed by one index range
Put Console.ReadLine on the end.
Or you don't have to store all numbers
static void Main(string[] args)
{
int min = int.MaxValue;
Console.WriteLine("\n Number of values:");
int num = Convert.ToInt32(Console.ReadLine());
var enteredValue = 0;
for (var i = 0; i < num; i++)
{
Console.WriteLine("\n Enter value:");
enteredValue = Convert.ToInt32(Console.ReadLine());
if (min>enteredValue) min = enteredValue;
}
Console.WriteLine("Smallest is {0}", min);
Console.ReadLine();
}
Tried to googled it but with no luck.
How can I find the second maximum number in an array with the smallest complexity?
code OR idea will be much help.
I can loop through an array and look for the maximum number
after that, I have the maximum number and then loop the array again to find the second the same way.
But for sure it is not efficient.
You could sort the array and choose the item at the second index, but the following O(n) loop will be much faster.
int[] myArray = new int[] { 0, 1, 2, 3, 13, 8, 5 };
int largest = int.MinValue;
int second = int.MinValue;
foreach (int i in myArray)
{
if (i > largest)
{
second = largest;
largest = i;
}
else if (i > second)
second = i;
}
System.Console.WriteLine(second);
OR
Try this (using LINQ):
int secondHighest = (from number in test
orderby number descending
select number).Distinct().Skip(1).First()
How to get the second highest number in an array in Visual C#?
Answer in C# :
static void Main(string[] args)
{
//let us define array and vars
var arr = new int[]{ 100, -3, 95,100,95, 177,-5,-4,177,101 };
int biggest =0, secondBiggest=0;
for (int i = 0; i < arr.Length; ++i)
{
int arrItem = arr[i];
if(arrItem > biggest)
{
secondBiggest = biggest; //always store the prev value of biggest
//to second biggest...
biggest = arrItem;
}
else if (arrItem > secondBiggest && arrItem < biggest) //if in our
//iteration we will find a number that is bigger than secondBiggest and smaller than biggest
secondBiggest = arrItem;
}
Console.WriteLine($"Biggest Number:{biggest}, SecondBiggestNumber:
{secondBiggest}");
Console.ReadLine(); //make program wait
}
Output : Biggest Number:177, SecondBiggestNumber:101
public static int F(int[] array)
{
array = array.OrderByDescending(c => c).Distinct().ToArray();
switch (array.Count())
{
case 0:
return -1;
case 1:
return array[0];
}
return array[1];
}
static void Main(string[] args)
{
int[] myArray = new int[] { 0, 11, 2, 15, 16, 8, 16 ,8,15};
int Smallest = myArray.Min();
int Largest = myArray.Max();
foreach (int i in myArray)
{
if(i>Smallest && i<Largest)
{
Smallest=i;
}
}
System.Console.WriteLine(Smallest);
Console.ReadLine();
}
This will work even if you have reputation of items in an array
int[] arr = {-10, -3, -3, -6};
int h = int.MinValue, m = int.MinValue;
foreach (var t in arr)
{
if (t == h || t == m)
continue;
if (t > h)
{
m = h;
h = t;
}
else if(t > m )
{
m = t;
}
}
Console.WriteLine("High: {0} 2nd High: {1}", h, m);
//or,
m = arr.OrderByDescending(i => i).Distinct().Skip(1).First();
Console.WriteLine("High: {0} 2nd High: {1}", h, m);
/* we can use recursion */
var counter = 0;
findSecondMax = (arr)=> {
let max = Math.max(...arr);
counter++;
return counter == 1 ? findSecondMax(arr.slice(0,arr.indexOf(max)).concat(arr.slice(arr.indexOf(max)+1))) : max;
}
console.log(findSecondMax([1,5,2,3,0]))
static void Main(string[] args){
int[] arr = new int[5];
int i, j,k;
Console.WriteLine("Enter Array");
for (i = 0; i < 5; i++) {
Console.Write("element - {0} : ", i);
arr[i] = Convert.ToInt32(Console.ReadLine());
}
Console.Write("\nElements in array are: ");
j=arr[0];
k=j;
for (i = 1; i < 5; i++) {
if (j < arr[i])
{
if(j>k)
{
k=j;
}
j=arr[i];
}
}
Console.WriteLine("First Greatest element: "+ j);
Console.WriteLine("Second Greatest element: "+ k);
Console.Write("\n");
}
int max = 0;
int secondmax = 0;
int[] arr = { 2, 11, 15, 1, 7, 99, 6, 85, 4 };
for (int r = 0; r < arr.Length; r++)
{
if (max < arr[r])
{
max = arr[r];
}
}
for (int r = 0; r < arr.Length; r++)
{
if (secondmax < arr[r] && arr[r] < max)
{
secondmax = arr[r];
}
}
Console.WriteLine(max);
Console.WriteLine(secondmax);
Console.Read();
Python 36>=
def sec_max(array: list) -> int:
_max_: int = max(array)
second: int = 0
for element in array:
if second < element < _max_:
second = element
else:
continue
return second
Using below code we can find out second highest number, even array contains multiple max numbers
// int[] myArray = { 25, 25, 5, 20, 50, 23, 10 };
public static int GetSecondHighestNumberForUniqueNumbers(int[] numbers)
{
int highestNumber = 0, Seconhight = 0;
List<int> numberList = new List<int>();
for (int i = 0; i < numbers.Length; i++)
{
//For loop should move forward only for unique items
if (numberList.Contains(numbers[i]))
continue;
else
numberList.Add(numbers[i]);
//find higest number
if (highestNumber < numbers[i])
{
Seconhight = highestNumber;
highestNumber = numbers[i];
} //find second highest number
else if (Seconhight < numbers[i])
{
Seconhight = numbers[i];
}
}
It's not like that your structure is a tree...It's just a simple array, right?
The best solution is to sort the array. And depending on descending or ascending, display the second or the 2nd last element respectively.
The other alternative is to use some inbuilt methods, to get the initial max. Pop that element, and then search for the max again. Don't know C#, so can't give the direct code.
You'd want to sort the numbers, then just take the second largest. Here's a snippet without any consideration of efficiency:
var numbers = new int[] { 3, 5, 1, 5, 4 };
var result=numbers.OrderByDescending(x=>x).Distinct().Skip(1).First();
This isn't too bad:
int[] myArray = new int[] { 0, 1, 2, 3, 13, 8, 5 };
var secondMax =
myArray.Skip(2).Aggregate(
myArray.Take(2).OrderByDescending(x => x).AsEnumerable(),
(a, x) => a.Concat(new [] { x }).OrderByDescending(y => y).Take(2))
.Skip(1)
.First();
It's fairly low on complexity as it only every sorts a maximum of three elements
namespace ConsoleApplication1
{
class Program
{
static void Main(string[] args)
{
int size;
Console.WriteLine("Enter the size of array");
size = Convert.ToInt32(Console.ReadLine());
Console.WriteLine("Enter the element of array");
int[] arr = new int[size];
for (int i = 0; i < size; i++)
{
arr[i] = Convert.ToInt32(Console.ReadLine());
}
int length = arr.Length;
Program program = new Program();
program.SeconadLargestValue(arr, length);
}
private void SeconadLargestValue(int[] arr, int length)
{
int maxValue = 0;
int secondMaxValue = 0;
for (int i = 0; i < length; i++)
{
if (arr[i] > maxValue)
{
secondMaxValue = maxValue;
maxValue = arr[i];
}
else if(arr[i] > secondMaxValue)
{
secondMaxValue = arr[i];
}
}
Console.WriteLine("First Largest number :"+maxValue);
Console.WriteLine("Second Largest number :"+secondMaxValue);
Console.ReadLine();
}
}
}
My solution below.
class Program
{
static void Main(string[] args)
{
Program pg = new Program();
Console.WriteLine("*****************************Program to Find 2nd Highest and 2nd lowest from set of values.**************************");
Console.WriteLine("Please enter the comma seperated numbers : ");
string[] val = Console.ReadLine().Split(',');
int[] inval = Array.ConvertAll(val, int.Parse); // Converts Array from one type to other in single line or Following line
// val.Select(int.Parse)
Array.Sort(inval);
Console.WriteLine("2nd Highest is : {0} \n 2nd Lowest is : {1}", pg.Return2ndHighest(inval), pg.Return2ndLowest(inval));
Console.ReadLine();
}
//Method to get the 2nd lowest and 2nd highest from list of integers ex 1000,20,-10,40,100,200,400
public int Return2ndHighest(int[] values)
{
if (values.Length >= 2)
return values[values.Length - 2];
else
return values[0];
}
public int Return2ndLowest(int[] values)
{
if (values.Length > 2)
return values[1];
else
return values[0];
}
}
I am giving solution that's in JavaScript, it takes o(n/2) complexity to find the highest and second highest number.
here is the working Fiddler Link
var num=[1020215,2000,35,2,54546,456,2,2345,24,545,132,5469,25653,0,2315648978523];
var j=num.length-1;
var firstHighest=0,seoncdHighest=0;
num[0] >num[num.length-1]?(firstHighest=num[0],seoncdHighest=num[num.length-1]):(firstHighest=num[num.length-1], seoncdHighest=num[0]);
j--;
for(var i=1;i<=num.length/2;i++,j--)
{
if(num[i] < num[j] )
{
if(firstHighest < num[j]){
seoncdHighest=firstHighest;
firstHighest= num[j];
}
else if(seoncdHighest < num[j] ) {
seoncdHighest= num[j];
}
}
else {
if(firstHighest < num[i])
{
seoncdHighest=firstHighest;
firstHighest= num[i];
}
else if(seoncdHighest < num[i] ) {
seoncdHighest= num[i];
}
}
}
Sort the array and take the second to last value?
var result = (from elements in inputElements
orderby elements descending
select elements).Distinct().Skip(1).Take(1);
return result.FirstOrDefault();
namespace FindSecondLargestNumber
{
class Program
{
static void Main(string[] args)
{
int max=0;
int smax=0;
int i;
int[] a = new int[20];
Console.WriteLine("enter the size of the array");
int n = int.Parse(Console.ReadLine());
Console.WriteLine("elements");
for (i = 0; i < n; i++)
{
a[i] = int.Parse(Console.ReadLine());
}
for (i = 0; i < n; i++)
{
if ( a[i]>max)
{
smax = max;
max= a[i];
}
else if(a[i]>smax)
{
smax=a[i];
}
}
Console.WriteLine("max:" + max);
Console.WriteLine("second max:"+smax);
Console.ReadLine();
}
}
}
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();