Livspace | Technical Round 2 | SDE 2 | kadane variation + 1D DP
Anonymous User
198

Hi all,
The interview was of 1 hour. I was asked 2 questions and able to answer both. Below are the asked questions:

  • Return the maximum sum subarray as well as the maximum sum in a subarray.
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);
    }

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Keep hustling!

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