Find the second maximum number in an array with the smallest complexity

Question

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.

Answer 1

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 2

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];
}

Answer 3

 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

Answer 4

           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);

Answer 5

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

Answer 6

/* 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]))

Answer 7

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");
}

Answer 8

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();

Answer 9

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

Answer 10

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];
            }
        }

Answer 11

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.

Answer 12

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();

Answer 13

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();
        }   
    }
}

Answer 14

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];

       }
   }

}   

Answer 15

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

Answer 16

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];
          }

     }

Answer 17

对数组进行排序并取倒数第二个值？

Answer 18

 var result = (from elements in inputElements
    orderby elements descending
    select elements).Distinct().Skip(1).Take(1);
 return result.FirstOrDefault();

Answer 19

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();
        }
    }
}