GE (General Electric) Interview Question for Software Developers


Country: United States
Interview Type: In-Person




Comment hidden because of low score. Click to expand.
1
of 1 vote

@Stephen, since the question does not give a length for the subarray, applying Kadane's algorithm for Maximum subarray problem would be more optimal.

int getMaxSubarraySum(int[] a) {
	if (a == null || a.length == 0)	throw new IllegalArgumentException();
		
	boolean allNegative = true;
	int max = Integer.MIN_VALUE;
	for (int i : a) {
		max = Math.max(max, i);
		if (i >= 0)	{
			allNegative = false;
			break;
		}
	}

	if (allNegative)	return max;
		
	int tempMax = a[0];
	max = a[0];
	for (int i = 1; i < a.length; i++) {
		tempMax = Math.max(tempMax + a[i], 0);
		max = Math.max(max, tempMax);
	}
	return max;
}

- raitGroup1007 July 25, 2017 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

Just had a quick go at this using a sliding window approach in C assuming an array of integers:

#include <stdio.h>

// Uses a sliding window
int MaxSumOfSubArray(int *intArray, int len, int subLen)
{
	if ( ( ! intArray ) || (subLen > len) )
	{
		return 0;
	}

	int blockSum = 0; // To store each possible sum

	// Calculate the initial sum
	int frontIdx = 0;
	while (frontIdx < subLen)
	{
		blockSum += intArray[frontIdx];
		frontIdx++;
	}

	int maxSum = blockSum; // Initial Result

	// Slide the window through the array
	int backIdx = 0;
	while (frontIdx < len)
	{
		blockSum -= intArray[backIdx];
		blockSum += intArray[frontIdx];

		if (blockSum > maxSum)
		{
			maxSum = blockSum;
		}

		backIdx++;
		frontIdx++;
	}

	return maxSum;
}

#define TEST_ARRAY_LEN		4
#define TEST_LENGTHS_LEN	6

void main(int argc, char * argv[])
{
	int testArray[TEST_ARRAY_LEN] = { 1, 3, 5, 4 };
	int testLengths[TEST_LENGTHS_LEN] = { 0, 1, 2, 3, 4, 5 };

	printf("For testing array: { ");
	for (int i = 0; i < TEST_ARRAY_LEN; i++)
	{
		printf("%d ", testArray[i]);
	}
	printf("}\n");

	for (int i = 0; i < TEST_LENGTHS_LEN; i++)
	{
		printf("Max sum for subarray of length %d: %d\n", testLengths[i], MaxSumOfSubArray(testArray, TEST_ARRAY_LEN, testLengths[i]));
	}
}

Seems to work pretty quickly but doesn't really employ any shortcuts.

- Stephen Vidas July 24, 2017 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

We could use divide and conquer approach in this case.

public class MaxSumOfSubArray {
    public static int findMaxSumOfSubArray(int[] elements) {
        return findMaxSumOfSubArray(elements, 0, elements.length - 1);
    }

    private static int findMaxSumOfSubArray(int[] elements, int start, int end) {
        if(start == end) {
            return elements[start];
        }
        int sum = 0;
        int mid = (start + end)/2;

        int left = findMaxSumOfSubArray(elements, start, mid);

        int right = findMaxSumOfSubArray(elements, mid+1, end);
        int maxSum = 0;

        int leftSum = Integer.MIN_VALUE;
        int rightSum = Integer.MIN_VALUE;
        for(int i = mid + 1; i <= end; i++) {
            sum += elements[i];
            leftSum = Math.max(sum, leftSum);
        }
        sum = 0;
        for(int i = mid; i >= 0; i--) {
            sum += elements[i];
            rightSum = Math.max(rightSum, sum);
        }

        maxSum = Math.max(left, right);
        int leftRightSum = leftSum + rightSum;
        return Math.max(maxSum, leftRightSum);
    }

    public static void main(String[] args) {
        int arr[] = {2, 3, -4, -5, -7};
        int max = MaxSumOfSubArray.findMaxSumOfSubArray(arr);
        System.out.println("Max: " + max);
    }
}

- mani January 13, 2018 | Flag Reply


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