Getting range of max product subarray using Kadanes algorithm

The link you have provided seems to assume that all the elements are positive. However in my opinion this is not a safe assumption. I have return code to get sub array for maximum product. I have used the same logic used in Kadane's algorithm. Code seems to work for me for all kinds of input. Please let me know if there are issues.

public static int[] getMaxSubArray(int []arr){

    int maxEndingHere = arr[0], maxSoFar = arr[0], startIndex =0, start =0,end=0;

    for(int i=1;i<arr.length;i++){

        if(maxEndingHere<0){
            maxEndingHere = arr[i];
            startIndex = i;         
        }else{          
            maxEndingHere *= arr[i];
        }           
        if(maxEndingHere>=maxSoFar){
            maxSoFar = maxEndingHere;
            start = startIndex;
            end = i;
        }   
    }       
    if(start<=end)
        return Arrays.copyOfRange(arr, start, end+1);

    return null;
}

1. Sample input = {6, 3, -10, 0, 2} Output = {6,3}
2. Sample input = {-2,1,-3,4,-1,2,1,-5,4} Output = {4}
3. Sample input = {-1,-2,-9,-6} Output = {-1}

def max_subarray(A):
    max_ending_here = max_so_far = 0
    max_start = start = 0
    max_end = end = 0

    # the range is [max_start, max_end)

    for i, x in enumerate(A):
        if max_ending_here + x > 0:
            max_ending_here = max_ending_here + x
            end = i+1
        else:
            max_ending_here = 0
            start = end = i

        if max_ending_here > max_so_far:
            max_so_far = max_ending_here
            max_start = start
            max_end = end

    return (max_start, max_end)

It has basically three cases:

1. current number is +ve
2. current number is -ve
3. current number is 0

You need to have two variable:

• min which hold minimum value till now

• max which holds the maximum value till now.

Now for case 3 min and max will be reset to 1.

For case 1: max will be max * a[i] and min will be minimum of min*a[i] and 1.

For case 2: max will be maximum of a[i] * min and 1, but the min value will be max * a[i].

Below is the code:

private static int getMaxProduct(int[] a){
    int minCurrent = 1, maxCurrent = 1, max = Integer.MIN_VALUE;
    for (int current : a) {
        if (current > 0) {
            maxCurrent = maxCurrent * current;
            minCurrent = Math.min(minCurrent * current, 1);
        } else if (current == 0) {
            maxCurrent = 1;
            minCurrent = 1;
        } else {
            int x = maxCurrent;
            maxCurrent = Math.max(minCurrent * current, 1);
            minCurrent = x * current;
        }
        if (max < maxCurrent) {
            max = maxCurrent;
        }
    }
    //System.out.println(minCurrent);
    return max;
}