How to find highest ,second highest number, Lowest Second Lowest number in given Array

Question

Can anyone Please tell me how to How to find highest ,second highest number, Lowest Second Lowest number in given Array

var numbers = new[] {855,3,64,6,24,75,3,6,24,45};

Any pointer and suggestion would really helpful . Thanks

Answer 1

Shortest way to find out largest, second largest, smallest & second smallest number, using conventional method:

int LargestNumber = 0;
int SecondLargestNumber = 0;
int SmallestNumber = 0;
int SecondSmallestNumber = 0;
int[] array = new int[] { 250, 5, 17, 50, 98, 352, 2, 8, 67, 150, 200, 1 };

for (int i = 0; i < array.Length; i++)
{
    if (array[i] > LargestNumber)
    {
        SecondLargestNumber = LargestNumber;
        LargestNumber = array[i];
    }
    else if (array[i] > SecondLargestNumber)
    {
        SecondLargestNumber = array[i];
    }

    if (i == 0)
    {
        SmallestNumber = array[0];
        SecondSmallestNumber = array[1];
    }

    if (array[i] < SmallestNumber)
    {
        SecondSmallestNumber = SmallestNumber;
        SmallestNumber = array[i];
    }
    else if (array[i] < SecondSmallestNumber)
    {
        SecondSmallestNumber = array[i];
    }
}

Console.WriteLine("Largest Number : " + LargestNumber);
Console.WriteLine("Second Largest Number : " + SecondLargestNumber);
Console.WriteLine("Smallest Number : " + SmallestNumber);
Console.WriteLine("Second Smallest Number : " + SecondSmallestNumber);

Answer 2

Here is an easy solution assuming you array has at least two members.

For the lowest number in the array.

              
 Array.Sort(new);

            string newNumber = string.Join(" ", new[0]);
            return newNumber;

For second lowest number in the array.

                Array.Sort(new);
                string newNumber = string.Join(" ", new[1]);
                return newNumber;

for the highest number in the array.

                    Array.Sort(new);
                    string newNumber = string.Join(" ", new[new.Length - 1]);
                    return newNumber;

For the second highest number in the array.

                        Array.Sort(new);
                        string newNumber = string.Join(" ", new[new.Length - 1]);
                        return newNumber;

Answer 3

This algorithm is applicable even with duplicate record.

int[] intArrayInput= new int[] { 855, 3, 64, 6, 24, 75, 3, 6, 24, 45, 855 };

Using Sort:

Array.Sort(intArrayInput);
int intMax = intArrayInput[intInput.Length - 1];
int intLow = intArrayInput[0];

And using LINQ:

int intSecondMax = intArrayInput.OrderByDescending(x => x).Distinct().Skip(1).First();
int intSecondLow = intArrayInput.OrderBy(x => x).Distinct().Skip(1).First();

Answer 4

Using Linq Concepts

var a = new int[] { 855, 3, 64, 6, 24, 75, 3, 6, 24, 45 };

var Max = a.Max(z => z);
var Min = a.Min( z => z);
var SMax = a.OrderByDescending(z=>z).Skip(1).First();
var SMin = a.OrderBy(z => z).Skip(1).First();

Answer 5

I have first arranged them using Selection Sort algorithm in ascending order, then i displayed values.

static void Main(string[] args)
    {
        int[] Num =new int[] { 3, 4, 5, 6, 7, 0,99,105,55 };

        int temp;
        for(int a=0;a<Num.Length-1;a++)
        {
            for(int b=a+1;b<Num.Length;b++)
            {
                if(Num[a]>Num[b])
                {
                    temp = Num[a];
                    Num[a] = Num[b];
                    Num[b] = temp;
                }
            }

        }


        Console.WriteLine("Max value ="+Num[Num.Length-1]+"\nSecond largest Max value="+ Num[Num.Length - 2]+"\nMin value =" + Num[0] + "\nSecond smallest Min value=" + Num[1]);
        Console.WriteLine("\nPress to close");
        Console.ReadLine();
    }

Answer 6

Why two loops when can be done in

        int[] myArray = new int[] { 2, 4, 3, 6, 9 };

        int max1 = 0;
        int max2 = 0;

        for (int i = 0; i < myArray.Length; i++)
        {
            if (myArray[i] > max1)
            {
                max2 = max1;
                max1 = myArray[i];

            }
            else
            {
                max2 = myArray[i];
            }
        }

        Console.WriteLine("first" + max1.ToString());
        Console.WriteLine("Second" + max2.ToString());
        Console.ReadKey();

Answer 7

int[] myUnSortArray = { 1, 5, 8, 3, 10, 6, 19, 5, 4, 4 };

int[] SortedArray = (from number in myUnSortArray
                     orderby number ascending
                     select number).ToArray();

int highestValue = SortedArray.Max();
int SecondHighest = SortedArray.Last(m => m < highestValue);

Answer 8

        int[] i = new int[] { 4, 8, 1, 9, 2, 7, 3 };
        Array.Sort(i);
        Console.WriteLine("Highest number :" + i[i.Length-1]);
        Console.WriteLine("Second highest number :"+i[i.Length-2]);
        Console.WriteLine("Lowest number :" + i[i.Length-i.Length]);
        Console.WriteLine("Second Lowest number :" + i[i.Length -i.Length+1]);

        Output : Highest number : 9

                 Second highest number : 8

                 Lowest number : 1

                 Second Lowest number : 2

Answer 9

You can also try this -

using System;
using System.Collections.Generic;
using System.Linq;
using System.Web;
using System.Web.UI;
using System.Web.UI.WebControls;

public partial class secondHighestLowest : System.Web.UI.Page
{
    int[] arr = new int[10] { 45, 3, 64, 6, 24, 75, 3, 6, 24, 45 };

    protected void Page_Load(object sender, EventArgs e)
    {
        secondHighestLowestNumber();
        secoundLowestNumber();
    }

    private void secondHighestLowestNumber()
    {
        int firstHighestNumber = arr[0];
        int secondHighestNumber = arr[0];
        for(int i = 0; i<arr.Length; i++)
        {
            if (arr[i]>firstHighestNumber)
            {
                firstHighestNumber = arr[i];
            }
        }

        for (int x = 0; x < arr.Length; x++)
        {
            if (arr[x]>secondHighestNumber && firstHighestNumber!=arr[x])
            {
                secondHighestNumber = arr[x];
            }
        }

        Response.Write("secondHighestNumber---- " + secondHighestNumber + "</br>");
    }

    private void secoundLowestNumber()
    {
        int firstLowestNumber = arr[0];
        int secondLowestNumber = arr[0];
        for (int i = 0; i < arr.Length; i++)
        {
            if (arr[i] < firstLowestNumber)
            {
                firstLowestNumber = arr[i];
            }
        }

        for (int x = 0; x < arr.Length; x++)
        {
            if (arr[x] < secondLowestNumber && firstLowestNumber != arr[x])
            {
                secondLowestNumber = arr[x];
            }
        }

        Response.Write("secondLowestNumber---- " + secondLowestNumber + "</br>");
    }
}

Hope this is helpful :)

Answer 10

Assuming you have at least 2 items in the array you can use OrderBy() and ElementAt():

var numbers = new[] { 855, 3, 64, 6, 24, 75, 3, 6, 24, 45 };
var secondLowest = numbers.OrderBy(num => num).ElementAt(1);
var secondHighest = numbers.OrderBy(num => num).Reverse().ElementAt(1);

Getting the highest and lowest is simpler and can be done using Max() and Min() LINQ methods.

var lowest = numbers.Min();
var highest = numbers.Max();

If you're worried about complexity, you can achieve better result using Selection algorithm. Using it you can perform the operations in O(n) complexity.

Answer 11

You don't specify the complexity requirement: one way is to sort the array in descending order and pick the top, second and third items.

Another is to build a Heap, and then perform remove root 3 times (with the heap being rebuilt after each remove).