Как найти второе максимальное число в массиве

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.

asked Feb 11, 2013 at 10:38

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

answered Nov 20, 2018 at 16:49

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];
}

answered Aug 13, 2014 at 0:12

 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

answered Mar 14, 2018 at 4:49

/* 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]))

answered May 31, 2019 at 20:43

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

answered Jan 16, 2022 at 18:16

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];
            }
        }

answered Apr 17, 2022 at 11:04

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();

answered Feb 11, 2013 at 10:44

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

answered Nov 14, 2015 at 6:48

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];

       }
   }

}   

answered Nov 4, 2015 at 15:04

Sort the array and take the second to last value?