C# Program for Largest Sum Contiguous Subarray

Write an efficient program to find the sum of contiguous subarray within a one-dimensional array of numbers that has the largest sum. 

Kadane’s Algorithm:

Initialize:
    max_so_far = INT_MIN
    max_ending_here = 0

Loop for each element of the array
  (a) max_ending_here = max_ending_here + a[i]
  (b) if(max_so_far < max_ending_here)
            max_so_far = max_ending_here
  (c) if(max_ending_here < 0)
            max_ending_here = 0
return max_so_far

Explanation:
The simple idea of Kadane’s algorithm is to look for all positive contiguous segments of the array (max_ending_here is used for this). And keep track of maximum sum contiguous segment among all positive segments (max_so_far is used for this). Each time we get a positive-sum compare it with max_so_far and update max_so_far if it is greater than max_so_far 

    Lets take the example:
    {-2, -3, 4, -1, -2, 1, 5, -3}

    max_so_far = max_ending_here = 0

    for i=0,  a[0] =  -2
    max_ending_here = max_ending_here + (-2)
    Set max_ending_here = 0 because max_ending_here < 0

    for i=1,  a[1] =  -3
    max_ending_here = max_ending_here + (-3)
    Set max_ending_here = 0 because max_ending_here < 0

    for i=2,  a[2] =  4
    max_ending_here = max_ending_here + (4)
    max_ending_here = 4
    max_so_far is updated to 4 because max_ending_here greater 
    than max_so_far which was 0 till now

    for i=3,  a[3] =  -1
    max_ending_here = max_ending_here + (-1)
    max_ending_here = 3

    for i=4,  a[4] =  -2
    max_ending_here = max_ending_here + (-2)
    max_ending_here = 1

    for i=5,  a[5] =  1
    max_ending_here = max_ending_here + (1)
    max_ending_here = 2

    for i=6,  a[6] =  5
    max_ending_here = max_ending_here + (5)
    max_ending_here = 7
    max_so_far is updated to 7 because max_ending_here is 
    greater than max_so_far

    for i=7,  a[7] =  -3
    max_ending_here = max_ending_here + (-3)
    max_ending_here = 4

Program:

C#

// C# program to print largest
// contiguous array sum
using System;

class GFG
{
    static int maxSubArraySum(int []a)
    {
        int size = a.Length;
        int max_so_far = int.MinValue,
            max_ending_here = 0;

        for (int i = 0; i < size; i++)
        {
            max_ending_here = max_ending_here + a[i];

            if (max_so_far < max_ending_here)
                max_so_far = max_ending_here;

            if (max_ending_here < 0)
                max_ending_here = 0;
        }

        return max_so_far;
    }

    // Driver code
    public static void Main ()
    {
        int [] a = {-2, -3, 4, -1, -2, 1, 5, -3};
        Console.Write("Maximum contiguous sum is " +
                                maxSubArraySum(a));
    }

}

// This code is contributed by Sam007_

Output:

Maximum contiguous sum is 7

Time Complexity: O(N), where N represents the size of the given array.
Auxiliary Space: O(1), no extra space is required, so it is a constant.

Another approach:

C#

static int maxSubArraySum(int[] a, int size)
{
    int max_so_far = a[0], max_ending_here = 0;

    for (int i = 0; i < size; i++) {
        max_ending_here = max_ending_here + a[i];
        if (max_ending_here < 0)
            max_ending_here = 0;

        /* Do not compare for all
        elements. Compare only
        when max_ending_here > 0 */
        else if (max_so_far < max_ending_here)
            max_so_far = max_ending_here;
    }
    return max_so_far;
}

// This code is contributed
// by ChitraNayal

Time Complexity: O(n), where n represents the size of the given array.
Auxiliary Space: O(1), no extra space is required, so it is a constant.

Algorithmic Paradigm: Dynamic Programming
Following is another simple implementation suggested by Mohit Kumar. The implementation handles the case when all numbers in the array are negative. 

C#

// C# program to print largest
// contiguous array sum
using System;

class GFG
{
    static int maxSubArraySum(int []a, int size)
    {
    int max_so_far = a[0];
    int curr_max = a[0];

    for (int i = 1; i < size; i++)
    {
        curr_max = Math.Max(a[i], curr_max+a[i]);
        max_so_far = Math.Max(max_so_far, curr_max);
    }

    return max_so_far;
    }

    // Driver code
    public static void Main ()
    {
        int []a = {-2, -3, 4, -1, -2, 1, 5, -3};
        int n = a.Length;
        Console.Write("Maximum contiguous sum is "
                           + maxSubArraySum(a, n));
    }

}

// This code is contributed by Sam007_

Output: 

Maximum contiguous sum is 7

Time Complexity: O(N), where N represents the size of the given array.
Auxiliary Space: O(1), no extra space is required, so it is a constant.

To print the subarray with the maximum sum, we maintain indices whenever we get the maximum sum.  

C#

// C# program to print largest
// contiguous array sum
using System;

class GFG
{
    static void maxSubArraySum(int []a,
                               int size)
    {
        int max_so_far = int.MinValue,
        max_ending_here = 0, start = 0,
        end = 0, s = 0;

        for (int i = 0; i < size; i++)
        {
            max_ending_here += a[i];

            if (max_so_far < max_ending_here)
            {
                max_so_far = max_ending_here;
                start = s;
                end = i;
            }

            if (max_ending_here < 0)
            {
                max_ending_here = 0;
                s = i + 1;
            }
        }
        Console.WriteLine("Maximum contiguous " +
                         "sum is " + max_so_far);
        Console.WriteLine("Starting index " +
                                      start);
        Console.WriteLine("Ending index " +
                                      end);
    }

    // Driver code
    public static void Main()
    {
        int []a = {-2, -3, 4, -1,
                   -2, 1, 5, -3};
        int n = a.Length;
        maxSubArraySum(a, n);
    }
}

// This code is contributed
// by anuj_67.

Output: 

Maximum contiguous sum is 7
Starting index 2
Ending index 6

Kadane’s Algorithm can be viewed both as a greedy and DP. As we can see that we are keeping a running sum of integers and when it becomes less than 0, we reset it to 0 (Greedy Part). This is because continuing with a negative sum is way more worse than restarting with a new range. Now it can also be viewed as a DP, at each stage we have 2 choices: Either take the current element and continue with previous sum OR restart a new range. These both choices are being taken care of in the implementation. 

Time Complexity: O(n)

Auxiliary Space: O(1)

Now try the below question 
Given an array of integers (possibly some elements negative), write a C program to find out the *maximum product* possible by multiplying ‘n’ consecutive integers in the array where n ≤ ARRAY_SIZE. Also, print the starting point of the maximum product subarray.

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.