Related
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();
}
}
}
I am trying to randomize a sett of predetermine elements in a 2d array.
using System;
namespace array
{
public class tiles
{
static void Main(string[] args)
{
Random random = new Random();
int[,] tilearr = { { 0, 1, 2 }, { 3, 4, 5 }, { 6, 7, 8 } };
for (int i = 0; i < 3; i++)
{
for (int j = 0; j < 3; j++)
{
Console.Write(tilearr[i, j] + " ");
}
Console.WriteLine();
}
Console.ReadLine();
}
}
}
My problem is if I do something like tilearr[i, j] = random.Next(0, 8); it randomizes the array but doesn't care if there are any duplicates of the same element.
2 6 7
1 1 3
2 7 0
what I am after is more like this:
2 4 8 1 3 0
0 3 1 5 6 8
6 7 5, 2 4 7
A simple and to the point solution would be to have a list of available numbers and then go position by position and randomly select the numbers out of the list.
Like this:
var numbers = new List<int> { 0, 1, 2, 3, 4, 5, 6, 7, 8 };
for(int x = 0; x < 3; ++x) {
for(int y = 0; y < 3; ++y) {
// select a random number from the list ...
int rand = random.Next(0, numbers.Count - 1);
tilearr[x, y] = numbers[rand];
// ... and remove it from the list
numbers.RemoveAt(rand);
}
}
As user Wai Ha Lee stated in the comments a shuffle will achieve what you are looking for. I would recommend the Fisher Yates Shuffle.
public static void Shuffle<T>(Random random, T[,] array)
{
int lengthRow = array.GetLength(1);
for (int i = array.Length - 1; i > 0; i--)
{
int i0 = i / lengthRow;
int i1 = i % lengthRow;
int j = random.Next(i + 1);
int j0 = j / lengthRow;
int j1 = j % lengthRow;
T temp = array[i0, i1];
array[i0, i1] = array[j0, j1];
array[j0, j1] = temp;
}
}
I retrieved this implementation from this answer.
This should be implemented in your code like this,
using System;
namespace array
{
public class tiles
{
static void Main(string[] args)
{
Random random = new Random();
int[,] tilearr = { { 0, 1, 2 }, { 3, 4, 5 }, { 6, 7, 8 } };
Shuffle<int>(random, tilearr);
for (int i = 0; i < 3; i++)
{
for (int j = 0; j < 3; j++)
{
Console.Write(tilearr[i, j] + " ");
}
Console.WriteLine();
}
Console.ReadLine();
}
}
}
Make sure to place the shuffle generic function within your class.
HERE is an example of my implementation on dotnetfiddle.net.
If you generate the array from scratch, it's easier to randomize a one dimensional array, and then load it into a 2D array.
static int[,] GenerateArray(int size)
{
Random r = new Random();
var arr = new int[size, size];
var values = Enumerable.Range(0, size * size).OrderBy(x => r.Next()).ToArray();
for (int i = 0; i < size; i++)
for (int j = 0; j < size; j++)
arr[i, j] = values[i * size + j];
return arr;
}
One way to Randomize would be flatten the 2d array, shuffle it and then recreate based on original dimension. If you want to use Linq/Extension methods, you could do the following
Random random = new Random();
int[,] tilearr = {{ 0, 1, 2 }, { 3, 4, 5 }, { 6, 7, 8 }};
var result = tilearr.OfType<int>()
.OrderBy(x=> random.Next())
.ChunkBy(tilearr.GetLength(1))
.To2DArray();
Where ChunkBy and To2DArray are defined as
public static class Extensions
{
public static IEnumerable<IEnumerable<T>> ChunkBy<T>(this IEnumerable<T> source, int chunkSize)
{
return source
.Select((x, i) => new { Index = i, Value = x })
.GroupBy(x => x.Index / chunkSize)
.Select(x => x.Select(v => v.Value));
}
public static T[,] To2DArray<T>(this IEnumerable<IEnumerable<T>> source)
{
var data = source
.Select(x => x.ToArray())
.ToArray();
var res = new T[data.Length, data.Max(x => x.Length)];
for (var i = 0; i < data.Length; ++i)
{
for (var j = 0; j < data[i].Length; ++j)
{
res[i,j] = data[i][j];
}
}
return res;
}
}
Sample Demo
I want to move all duplicate numbers at the end of array like this.
{4,1,5,4,3,1,6,5}
{4,1,5,3,6,4,1,5}
also i want to know number of dups. that i will use to resize array.
here is the code i tried but this code is not compatible when i insert more than 2 dups at starting.
static void RemoveRepeated(ref int[] array)
{
int count = 0; bool flag;
for (int i = 0; i < array.Length; i++)
{
flag = true;
for (int j = i+1; j < array.Length-1; j++)
{
if (array[i] == array[j] )
{
int temp = array[j];
array[j] = array[j + 1];
array[j + 1] = temp;
if (flag)
{
count++;
flag = false;
}
}
}
}
Array.Resize(ref array,array.Length-count);
}
Thanks in advance :)
A good solution will be to use a fitting data-structure. It will not fix your algorithm but replace it. Here a HashSet<T> is perfect. A HashSet<T> remove itself all duplicate. Check the msdn for more informations.
Demo on .NETFiddle
using System;
using System.Linq;
using System.Collections.Generic;
public class Program
{
public static void Main()
{
var array = new int[]{ 4,1,5,4,3,1,6,5 };
RemoveRepeated(ref array);
foreach (var item in array)
Console.WriteLine(item);
}
static void RemoveRepeated(ref int[] array)
{
array = new HashSet<int>(array).ToArray();
}
}
By the way you don't really need ref here. I would remove it and change void RemoveRepeated(ref int[] array) to int[] RemoveRepeated(int[] array). See ref parameter or return value?
What you are doing is equivalent to leaving only the unique elements in their original order. Here is simpler way to do this:
static void RemoveRepeated(ref int[] array)
{
HashSet<int> seen = new HashSet<int>();
List<int> newArray = new List<int>(array.Length);
foreach(int x in array)
{
if(!seen.Contains(x))
{
seen.Add(x);
newArray.Add(x);
}
}
array = newArray.ToArray();
}
Don't use an array here. You can try to (just one example) group a List<int> and Count() all groups that have more than one element. When you have the count, you can use Distinct() to get only the distinct elements.
In my opinion, resizing an array like this is always a very bad idea.
Edit:
Well, like the other answers already stated, a HashSet is a even better way of doing it.
class Program
{
static void Main(string[] args)
{
int[] arr = new int[] { 4, 1, 5, 4, 3, 1, 6, 5 };
RemoveRepeated(ref arr);
foreach (int i in arr)
{
Console.WriteLine(i);
}
Console.ReadKey();
}
private static void RemoveRepeated(ref int[] array)
{
int count =0;
for (int i = 0; i < array.Length; i++)
{
for (int j = i + 1; j < array.Length; j++)
{
if (array[i] == array[j])
{
int temp = array[j];
count++;
for (int k = j; k < array.Length-1; k++)
{
array[k] = array[k + 1];
}
array[array.Length - 1] = temp;
j = array.Length;
}
}
}
Array.Resize(ref array, array.Length - count);
}
}
Try this:
class Program
{
static void Main(string[] args)
{
int[] ints = {4,1,5,4,3,1,6,5};
var query = ints.GroupBy(x => x).OrderBy(x => x.Count()).Select(x => x);
// print ordered array showing dupes are last
query.ToList().ForEach(x => Console.WriteLine(x.Key.ToString()));
// Get number of dupes
int dupeCount = query.Where(x => x.Count() > 1).Count();
// put unique items into a new array
var newArray = query.Where(x => x.Count() == 1).Select(x => x.Key).ToArray();
// put distinct items into an array
var distinctArray = ints.Distinct().ToArray();
Console.ReadKey();
}
}
Edit: Added distinct elements
If you want to move the duplicates to the end of array (but preserve order) you can do this:
static void RemoveRepeated(ref int[] array)
{
var lookup = array.ToLookup(x => x);
var maxdupes = lookup.Select(x => x.Count()).Max();
var reordered =
Enumerable
.Range(0, maxdupes)
.SelectMany(x => lookup.SelectMany(y => y.Skip(x).Take(1)))
.ToArray();
Array.Copy(reordered, array, reordered.Length);
}
This would turn { 4, 1, 5, 4, 3, 1, 6, 5 } into { 4, 1, 5, 3, 6, 4, 1, 5 }
You can easily get the number of duplicates by doing this:
lookup
.Select(x => new
{
Value = x.Key,
Count = x.Count(),
});
This returns:
4 2
1 2
5 2
3 1
6 1
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();
}
}
}
How to get the most common value in an Int array using C#
eg: Array has the following values: 1, 1, 1, 2
Ans should be 1
var query = (from item in array
group item by item into g
orderby g.Count() descending
select new { Item = g.Key, Count = g.Count() }).First();
For just the value and not the count, you can do
var query = (from item in array
group item by item into g
orderby g.Count() descending
select g.Key).First();
Lambda version on the second:
var query = array.GroupBy(item => item).OrderByDescending(g => g.Count()).Select(g => g.Key).First();
Some old fashioned efficient looping:
var cnt = new Dictionary<int, int>();
foreach (int value in theArray) {
if (cnt.ContainsKey(value)) {
cnt[value]++;
} else {
cnt.Add(value, 1);
}
}
int mostCommonValue = 0;
int highestCount = 0;
foreach (KeyValuePair<int, int> pair in cnt) {
if (pair.Value > highestCount) {
mostCommonValue = pair.Key;
highestCount = pair.Value;
}
}
Now mostCommonValue contains the most common value, and highestCount contains how many times it occured.
I know this post is old, but someone asked me the inverse of this question today.
LINQ Grouping
sourceArray.GroupBy(value => value).OrderByDescending(group => group.Count()).First().First();
Temp Collection, similar to Guffa's:
var counts = new Dictionary<int, int>();
foreach (var i in sourceArray)
{
if (!counts.ContainsKey(i)) { counts.Add(i, 0); }
counts[i]++;
}
return counts.OrderByDescending(kv => kv.Value).First().Key;
public static int get_occure(int[] a)
{
int[] arr = a;
int c = 1, maxcount = 1, maxvalue = 0;
int result = 0;
for (int i = 0; i < arr.Length; i++)
{
maxvalue = arr[i];
for (int j = 0; j <arr.Length; j++)
{
if (maxvalue == arr[j] && j != i)
{
c++;
if (c > maxcount)
{
maxcount = c;
result = arr[i];
}
}
else
{
c=1;
}
}
}
return result;
}
Maybe O(n log n), but fast:
sort the array a[n]
// assuming n > 0
int iBest = -1; // index of first number in most popular subset
int nBest = -1; // popularity of most popular number
// for each subset of numbers
for(int i = 0; i < n; ){
int ii = i; // ii = index of first number in subset
int nn = 0; // nn = count of numbers in subset
// for each number in subset, count it
for (; i < n && a[i]==a[ii]; i++, nn++ ){}
// if the subset has more numbers than the best so far
// remember it as the new best
if (nBest < nn){nBest = nn; iBest = ii;}
}
// print the most popular value and how popular it is
print a[iBest], nBest
Yet another solution with linq:
static int[] GetMostCommonIntegers(int[] nums)
{
return nums
.ToLookup(n => n)
.ToLookup(l => l.Count(), l => l.Key)
.OrderBy(l => l.Key)
.Last()
.ToArray();
}
This solution can handle case when several numbers have the same number of occurences:
[1,4,5,7,1] => [1]
[1,1,2,2,3,4,5] => [1,2]
[6,6,6,2,2,1] => [6]