C# program to find Largest, Smallest, Second Largest, Second Smallest in a List

Finding the largest, smallest, second largest, and second smallest elements in a list is a common programming task in C#. Using LINQ methods like Max(), Min(), OrderBy(), and Skip() provides an elegant solution for these operations.

Syntax

To find the largest element −

list.Max()

To find the smallest element −

list.Min()

To find the second largest element −

list.OrderByDescending(x => x).Skip(1).First()

To find the second smallest element −

list.OrderBy(x => x).Skip(1).First()

Using LINQ Methods

The most straightforward approach uses LINQ extension methods to find all four values −

using System;
using System.Linq;

public class Program {
   public static void Main() {
      var val = new int[] {
         99,
         35,
         26,
         87
      };

      var maxNum = val.Max();
      Console.WriteLine("Maximum Number: " + maxNum);

      var minNum = val.Min();
      Console.WriteLine("Minimum Number: " + minNum);

      var secondMax = val.OrderByDescending(z => z).Skip(1).First();
      Console.WriteLine("Second Largest Number: " + secondMax);

      var secondMin = val.OrderBy(z => z).Skip(1).First();
      Console.WriteLine("Second Smallest Number: " + secondMin);
   }
}

The output of the above code is −

Maximum Number: 99
Minimum Number: 26
Second Largest Number: 87
Second Smallest Number: 35

Using Distinct Values

When the list contains duplicate values, you may want to find distinct second largest and smallest values −

using System;
using System.Linq;

public class Program {
   public static void Main() {
      var numbers = new int[] { 99, 87, 99, 35, 26, 87, 35 };

      var distinctNumbers = numbers.Distinct().ToList();
      Console.WriteLine("Original: [" + string.Join(", ", numbers) + "]");
      Console.WriteLine("Distinct: [" + string.Join(", ", distinctNumbers.OrderByDescending(x => x)) + "]");

      var maxNum = distinctNumbers.Max();
      var minNum = distinctNumbers.Min();
      var secondMax = distinctNumbers.OrderByDescending(x => x).Skip(1).First();
      var secondMin = distinctNumbers.OrderBy(x => x).Skip(1).First();

      Console.WriteLine("Maximum: " + maxNum);
      Console.WriteLine("Minimum: " + minNum);
      Console.WriteLine("Second Largest: " + secondMax);
      Console.WriteLine("Second Smallest: " + secondMin);
   }
}

The output of the above code is −

Original: [99, 87, 99, 35, 26, 87, 35]
Distinct: [99, 87, 35, 26]
Maximum: 99
Minimum: 26
Second Largest: 87
Second Smallest: 35

Using Manual Iteration

For better performance with large datasets, you can use a single-pass algorithm −

using System;

public class Program {
   public static void Main() {
      int[] numbers = { 99, 35, 26, 87, 45, 12, 78 };

      int max = int.MinValue, secondMax = int.MinValue;
      int min = int.MaxValue, secondMin = int.MaxValue;

      foreach (int num in numbers) {
         if (num > max) {
            secondMax = max;
            max = num;
         } else if (num > secondMax && num != max) {
            secondMax = num;
         }

         if (num 

The output of the above code is −

Array: [99, 35, 26, 87, 45, 12, 78]
Largest: 99
Smallest: 12
Second Largest: 87
Second Smallest: 26


Comparison of Methods

Conclusion

C# provides multiple approaches to find largest, smallest, second largest, and second smallest values in a list. LINQ methods offer clean and readable solutions, while manual iteration provides better performance for large datasets. Choose the approach that best fits your requirements for code readability and performance.