Java 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: 

import java.io.*;
// Java program to print largest contiguous array sum
import java.util.*;

class Kadane
{
    public static void main (String[] args)
    {
        int [] a = {-2, -3, 4, -1, -2, 1, 5, -3};
        System.out.println("Maximum contiguous sum is " +
                                       maxSubArraySum(a));
    }

    static int maxSubArraySum(int a[])
    {
        int size = a.length;
        int max_so_far = Integer.MIN_VALUE, 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;
    }
}

Output:

Maximum contiguous sum is 7

Time Complexity: O(n) 

Auxiliary Space: O(1)

Another approach:

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 ANKITRAI1

Time Complexity: O(n) 

Auxiliary Space: O(1)

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. 

// Java program to print largest contiguous
// array sum
import java.io.*;

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 program to test maxSubArraySum */
    public static void main(String[] args)
    {
    int a[] = {-2, -3, 4, -1, -2, 1, 5, -3};
    int n = a.length;   
    int max_sum = maxSubArraySum(a, n);
    System.out.println("Maximum contiguous sum is " 
                       + max_sum);
    }
}

// This code is contributed by Prerna Saini

Output: 

Maximum contiguous sum is 7

Time Complexity: O(n) 

Auxiliary Space: O(1)

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

// Java program to print largest 
// contiguous array sum
class GFG {

    static void maxSubArraySum(int a[], int size)
    {
        int max_so_far = Integer.MIN_VALUE,
        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;
            }
        }
        System.out.println("Maximum contiguous sum is " 
                           + max_so_far);
        System.out.println("Starting index " + start);
        System.out.println("Ending index " + end);
    }

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

// This code is contributed by  prerna saini

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.