Maximum subarray sum modulo M

Here given code implementation process.

// Include header file
#include <iostream>

#include <set>

using namespace std;
/*
    C++ Program for
    Maximum subarray sum modulo M
*/
class MaximumSubarray
{
	public:
		// Print array elements
		void printArray(int arr[], int n)
		{
			for (int i = 0; i < n; ++i)
			{
				cout << "  " << arr[i];
			}
		}
	int maxValue(int a, int b)
	{
		if (a > b)
		{
			return a;
		}
		return b;
	}
	void maxModuloMSubArray(int arr[], int n, int m)
	{
		int prefix = 0;
		int max = 0;
		set < int > record;
		// Add first element in record
		record.insert(0);
		for (int i = 0; i < n; ++i)
		{
			prefix = (prefix + arr[i]) % m;
			max = this->maxValue(max, prefix);
          	auto v = record.lower_bound(prefix+1);
			if (v != record.end())
			{
				max = this->maxValue(max, prefix - 
                                     (*v) + m);
			}
			// Add prefix into record
			record.insert(prefix);
		}
		// Display calculated result
		cout << "Result : " << max;
	}
};
int main()
{
	MaximumSubarray *task = new MaximumSubarray();
	int arr[] = {
		4 , 3 , 2 , 3 , 7 , 2
	};
	int n = sizeof(arr) / sizeof(arr[0]);
	int m = 7;
	/*
	M = 7
	arr = [4 ,3, 2, 3, 7, 2]
	All subarray
	---------------------------------------------------  
	    [ 4 ] = [4] % 7                           => 4
	    [ 4 +  3 ] = [7] % 7                      => 0
	    [ 4 +  3 +  2 ] = [9] % 7                 => 2
	    [ 4 +  3 +  2 +  3 ] = [12] % 7           => 5
	    [ 4 +  3 +  2 +  3 +  7 ] = [19] % 7      => 5
	    [ 4 +  3 +  2 +  3 +  7 +  2 ] = [21] % 7 => 0
	    [ 3 ] = [3] % 7                           => 3
	    [ 3 +  2 ] = [5] % 7                      => 5
	    [ 3 +  2 +  3 ] = [8] % 7                 => 1
	    [ 3 +  2 +  3 +  7 ] = [15] % 7           => 1
	    [ 3 +  2 +  3 +  7 +  2 ] = [17] % 7      => 3
	    [ 2 ] = [2] % 7                           => 2
	    [ 2 +  3 ] = [5] % 7                      => 5
	    [ 2 +  3 +  7 ] = [12] % 7                => 5
	    [ 2 +  3 +  7 +  2 ] = [14] % 7           => 0
	    [ 3 ] = [3] % 7                           => 3
	    [ 3 +  7 ] = [10] % 7                     => 3
	    [ 3 +  7 +  2 ] = [12] % 7                => 5
	    [ 7 ] = [7] % 7                           => 0
	    [ 7 +  2 ] = [9] % 7                      => 2
	    [ 2 ] = [2] % 7                           => 2
	----------------------------------------------------- 
	    Result : 5
	*/
	task->maxModuloMSubArray(arr, n, m);
	return 0;
}

Output

Result : 5
// Include namespace system
using System;
using System.Collections.Generic;
/*
    Csharp Program for
    Maximum subarray sum modulo M
*/
public class MaximumSubarray
{
    // Print array elements
    public void printArray(int[] arr, int n)
    {
        for (int i = 0; i < n; ++i)
        {
            Console.Write("  " + arr[i]);
        }
    }
    public int maxValue(int a, int b)
    {
        if (a > b)
        {
            return a;
        }
        return b;
    }
    public void maxModuloMSubArray(int[] arr, int n, int m)
    {
        int prefix = 0;
        int max = 0;
        SortedSet < int > record = new SortedSet < int > ();
        // Add first element in record
        record.Add(0);
        for (int i = 0; i < n; ++i)
        {
            prefix = (prefix + arr[i]) % m;
            max = this.maxValue(max, prefix);
            if (record.Max > prefix + 1)
            {
                SortedSet < int > ans = 
                record.GetViewBetween(prefix + 1, record.Max);
                if (ans.Count > 0)
                {
                    max = this.maxValue(max, prefix - (ans.Min) + m);
                }
            }
            // Add prefix into record
            record.Add(prefix);
        }
        // Display calculated result
        Console.Write("Result : " + max);
    }
    public static void Main(String[] args)
    {
        MaximumSubarray task = new MaximumSubarray();
        int[] arr = {
            4 , 3 , 2 , 3 , 7 , 2
        };
        int n = arr.Length;
        int m = 7;
        /*
        M = 7
        arr = [4 ,3, 2, 3, 7, 2]
        All subarray
        ---------------------------------------------------  
            [ 4 ] = [4] % 7                           => 4
            [ 4 +  3 ] = [7] % 7                      => 0
            [ 4 +  3 +  2 ] = [9] % 7                 => 2
            [ 4 +  3 +  2 +  3 ] = [12] % 7           => 5
            [ 4 +  3 +  2 +  3 +  7 ] = [19] % 7      => 5
            [ 4 +  3 +  2 +  3 +  7 +  2 ] = [21] % 7 => 0
            [ 3 ] = [3] % 7                           => 3
            [ 3 +  2 ] = [5] % 7                      => 5
            [ 3 +  2 +  3 ] = [8] % 7                 => 1
            [ 3 +  2 +  3 +  7 ] = [15] % 7           => 1
            [ 3 +  2 +  3 +  7 +  2 ] = [17] % 7      => 3
            [ 2 ] = [2] % 7                           => 2
            [ 2 +  3 ] = [5] % 7                      => 5
            [ 2 +  3 +  7 ] = [12] % 7                => 5
            [ 2 +  3 +  7 +  2 ] = [14] % 7           => 0
            [ 3 ] = [3] % 7                           => 3
            [ 3 +  7 ] = [10] % 7                     => 3
            [ 3 +  7 +  2 ] = [12] % 7                => 5
            [ 7 ] = [7] % 7                           => 0
            [ 7 +  2 ] = [9] % 7                      => 2
            [ 2 ] = [2] % 7                           => 2
        ----------------------------------------------------- 
            Result : 5
        */
        task.maxModuloMSubArray(arr, n, m);
    }
}

Output

Result : 5
import java.util.TreeSet;
/*
    Java Program for
    Maximum subarray sum modulo M
*/
public class MaximumSubarray
{
	// Print array elements
	public void printArray(int[] arr, int n)
	{
		for (int i = 0; i < n; ++i)
		{
			System.out.print("  " + arr[i]);
		}
	}
	public int maxValue(int a, int b)
	{
		if (a > b)
		{
			return a;
		}
		return b;
	}
	public void maxModuloMSubArray(int[] arr, int n, int m)
	{
		int prefix = 0;
		int max = 0;
		TreeSet < Integer > record = new TreeSet < Integer > ();
		// Add first element in record
		record.add(0);
		for (int i = 0; i < n; ++i)
		{
			prefix = (prefix + arr[i]) % m;
			max = maxValue(max, prefix);
			if (record.higher(prefix) != null)
			{
				max = maxValue(max, prefix - (record.higher(prefix)) + m);
			}
			// Add prefix into record
			record.add(prefix);
		}
		// Display calculated result
		System.out.print("Result : " + max);
	}
	public static void main(String[] args)
	{
		MaximumSubarray task = new MaximumSubarray();
		int[] arr = {
			4 , 3 , 2 , 3 , 7 , 2
		};
		int n = arr.length;
		int m = 7;
		/*
		M = 7
		arr = [4 ,3, 2, 3, 7, 2]
		All subarray
		---------------------------------------------------  
		    [ 4 ] = [4] % 7                           => 4
		    [ 4 +  3 ] = [7] % 7                      => 0
		    [ 4 +  3 +  2 ] = [9] % 7                 => 2
		    [ 4 +  3 +  2 +  3 ] = [12] % 7           => 5
		    [ 4 +  3 +  2 +  3 +  7 ] = [19] % 7      => 5
		    [ 4 +  3 +  2 +  3 +  7 +  2 ] = [21] % 7 => 0
		    [ 3 ] = [3] % 7                           => 3
		    [ 3 +  2 ] = [5] % 7                      => 5
		    [ 3 +  2 +  3 ] = [8] % 7                 => 1
		    [ 3 +  2 +  3 +  7 ] = [15] % 7           => 1
		    [ 3 +  2 +  3 +  7 +  2 ] = [17] % 7      => 3
		    [ 2 ] = [2] % 7                           => 2
		    [ 2 +  3 ] = [5] % 7                      => 5
		    [ 2 +  3 +  7 ] = [12] % 7                => 5
		    [ 2 +  3 +  7 +  2 ] = [14] % 7           => 0
		    [ 3 ] = [3] % 7                           => 3
		    [ 3 +  7 ] = [10] % 7                     => 3
		    [ 3 +  7 +  2 ] = [12] % 7                => 5
		    [ 7 ] = [7] % 7                           => 0
		    [ 7 +  2 ] = [9] % 7                      => 2
		    [ 2 ] = [2] % 7                           => 2
		----------------------------------------------------- 
		    Result : 5
		*/
		task.maxModuloMSubArray(arr, n, m);
	}
}

Output

Result : 5


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