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

Solution 1

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 :)

Solution 2

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();

Solution 3

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.

Solution 4

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).

View more solutions

Comments

[Henk Holterman]

Didn't your teacher give you directions? (If it's homework, please use that tag)

[Noon Silk]

This, IMHO, is a shocking way to do this. Hence the downvote. (providing explanation, not argument).

[Helen]

To format a code block, indent it 4 spaces, or select it and click the code button (101010) on the editor's toolbar.

[marc_s]

@silky: why is this so shocking?? Can you elaborate?? Using LINQ on a small collection like this sounds like a very natural and very obvious choice....

[Chris]

Creating a method that returns the highest value from an array of integers has actually been an interview question before. Max() actually turned out how to be the easiest solution.

[Willy David Jr]

SMax and SMin is only applicable if the array has no duplicate record. SMin still will return 3 as the answer which should be 6.

[Aimal Khan]

does not work in case of int[] myArray = new int[] { 9, 4, 3, 6, 2 };