Javascript 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:

Javascript

<script>

// JavaScript program to find maximum
// contiguous subarray

// Function to find the maximum
// contiguous subarray
function maxSubArraySum(a, size)
{
    var maxint = Math.pow(2, 53)
    var max_so_far = -maxint - 1
    var max_ending_here = 0

    for (var 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
var a = [ -2, -3, 4, -1, -2, 1, 5, -3 ]
document.write("Maximum contiguous sum is",
               maxSubArraySum(a, a.length))

// This code is contributed by AnkThon

</script>

Output:

Maximum contiguous sum is 7

Another approach:

Javascript

<script>
        // JavaScript Program to implement
        // the above approach

        function maxSubarraySum(arr, size)
        {
            let max_ending_here = 0, max_so_far = Number.MIN_VALUE;
            for (let i = 0; i < size; i++) {

                // include current element to previous subarray only
                // when it can add to a bigger number than itself.
                if (arr[i] <= max_ending_here + arr[i]) {
                    max_ending_here += arr[i];
                }

                // Else start the max subarray from current element
                else {
                    max_ending_here = arr[i];
                }
                if (max_ending_here > max_so_far) {
                    max_so_far = max_ending_here;
                }
            }
            return max_so_far;
        }

// This code is contributed by Potta Lokesh
    </script>

Time Complexity: O(n) 
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. 

Javascript

<script>
// C# program to print largest
// contiguous array sum

function maxSubArraySum(a,size)
{
  let max_so_far = a[0];
  let curr_max = a[0];

  for (let 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

let a = [-2, -3, 4, -1, -2, 1, 5, -3];
let n = a.length;
document.write("Maximum contiguous sum is ",maxSubArraySum(a, n));

</script>

Output: 

Maximum contiguous sum is 7

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

Javascript

<script>
// javascript program to print largest
// contiguous array sum
function maxSubArraySum(a , size) {
        var max_so_far = Number.MIN_VALUE, max_ending_here = 0, start = 0, end = 0, s = 0;

        for (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;
            }
        }
        document.write("Maximum contiguous sum is " + max_so_far);
        document.write("<br/>Starting index " + start);
        document.write("<br/>Ending index " + end);
    }

    // Driver code

        var a = [ -2, -3, 4, -1, -2, 1, 5, -3 ];
        var n = a.length;
        maxSubArraySum(a, n);

// This code is contributed by Rajput-Ji
</script>

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 refer complete article on Largest Sum Contiguous Subarray for more details!