Maximum Subarray in C#

Welcome to the Maximum Subarray in C# page! Here, you'll find the source code for this program as well as a description of how the program works.

Current Solution

using System;
using System.Collections.Generic;
using System.Linq;

public static class Program
{
    private static void ShowUsage()
    {
        Console.Error.WriteLine("Usage: Please provide a list of integers in the format: \"1, 2, 3, 4, 5\"");
        Environment.Exit(1);
    }

    private static List<int> ParseIntegerList(string input)
    {
        if (string.IsNullOrWhiteSpace(input))
            ShowUsage();

        var list = input
            .Split(',', StringSplitOptions.RemoveEmptyEntries | StringSplitOptions.TrimEntries)
            .Select(s =>
            {
                if (!int.TryParse(s, out var val))
                    ShowUsage();
                return val;
            })
            .ToList();

        if (list.Count == 0)
            ShowUsage();

        return list;
    }

    private static int MaximumSubarraySum(IReadOnlyList<int> numbers)
    {
        if (numbers.Count == 0)
            return 0;

        int currentSum = numbers[0];
        int maxSum = numbers[0];

        for (int i = 1; i < numbers.Count; i++)
        {
            int number = numbers[i];
            currentSum = Math.Max(number, currentSum + number);
            maxSum = Math.Max(maxSum, currentSum);
        }

        return maxSum;
    }

    public static int Main(string[] args)
    {
        if (args.Length != 1)
            ShowUsage();

        var inputList = ParseIntegerList(args[0]);

        Console.WriteLine(MaximumSubarraySum(inputList));

        return 0;
    }
}

Maximum Subarray in C# was written by:

• Ștefan-Iulian Alecu

If you see anything you'd like to change or update, please consider contributing.

How to Implement the Solution

No 'How to Implement the Solution' section available. Please consider contributing.

How to Run the Solution

No 'How to Run the Solution' section available. Please consider contributing.