Amazon Interview Question for Quality Assurance Engineers


Team: Kindle
Country: India
Interview Type: Written Test




Comment hidden because of low score. Click to expand.
3
of 3 vote

The first method is just to simply go through the array and calculate the sums for all previous elements and all next elements, and compare them.
But is is too inefficient.
More efficient method is:
1). To calculate total sum of all elements.
Growing Sum =0
2). Going through the array, subtract elements from the total sum and add them to growing sum. Compare Growing Sum to the Total Sum. If they are equal - we found our element(s).

- sergey.a.kabanov September 18, 2016 | Flag Reply
Comment hidden because of low score. Click to expand.
2
of 2 vote

Example given in question is wrong. Please rectify that to avoid any confusion.

1+0==1
but -11 +1 +12 !=1

- yogeshsharma October 18, 2016 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

var list = new List<int>() {1, 1, -11, 1, 12};
var getListCount = list.Count;
var isListCountOdd = getListCount % 2 == 0;
var getIntegerIncrement = 3;
int takeFromList;

if (isListCountOdd)
{
	takeFromList = getListCount/2;
}
else
{
	if (getListCount <= 5)
	{
		takeFromList = getListCount - getIntegerIncrement;
	}
	else
	{
		getIntegerIncrement++;
		takeFromList = getListCount - getIntegerIncrement;
	}
}

// Get left and right numbers
var leftNumbers = list.Take(takeFromList);
var rightNumbers = list.Skip(takeFromList).Take(getIntegerIncrement);
// Sum them up
var leftNumbersTotal = leftNumbers.Sum();
var rightNumbersTotal = rightNumbers.Sum();

Console.Write(leftNumbersTotal == rightNumbersTotal ? "Left == Right" : "Left != Right");
Console.ReadKey();

- Anonymous September 26, 2016 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

-- This solution is for list contains 4 or more count.
var list = new List<int>() {1, 1, -11, 1, 12};
var getListCount = list.Count;
var isListCountOdd = getListCount % 2 == 0;
var getIntegerIncrement = 3;
int takeFromList;

if (isListCountOdd)
{
	takeFromList = getListCount/2;
}
else
{
	if (getListCount <= 5)
	{
		takeFromList = getListCount - getIntegerIncrement;
	}
	else
	{
		getIntegerIncrement++;
		takeFromList = getListCount - getIntegerIncrement;
	}
}

// Get left and right numbers
var leftNumbers = list.Take(takeFromList);
var rightNumbers = list.Skip(takeFromList).Take(getIntegerIncrement);
// Sum them up
var leftNumbersTotal = leftNumbers.Sum();
var rightNumbersTotal = rightNumbers.Sum();

Console.Write(leftNumbersTotal == rightNumbersTotal ? "Left == Right" : "Left != Right");
Console.ReadKey();

- gunjan.prmr September 26, 2016 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

PYTHON::

>>> def sum(l):
a = 0
d = 0
b = len(l)
c = b/2
for i in range(0, c):
a = a + l[i]
for j in range(c, ((c*2)+1)):
d = d + l[j]
if a == d:
print "both sides of sum is equal"
return a
elif a != d:
a = 0
d = 0
for i in range(0, c+1):
a = a + l[i]
for j in range(c+1, ((c*2)+1)):
d = d + l[j]
if a == d:
print "both sides of sum is equal at 2nd level"
return a
else:
print "both sides of sum is not equal"
return False


>>>
>>> sum([1,1,-11,1,12])
both sides of sum is equal
2
>>> sum([-11,1,12,1,1])
both sides of sum is equal at 2nd level
2
>>> sum([-11,1,12,1,0])
both sides of sum is not equal
False
>>>

- Madhu Mohan May 18, 2017 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

private static boolean recArray ( int[] nums, int index, int sum1, int sum2 ) {
		  if ( index >= nums.length ) {
		    return sum1 == sum2;
		  }
		  int value = nums[index];
		  return (recArray(nums, index + 1, sum1 + value, sum2) || recArray(nums, index + 1, sum1, sum2 + value));
		}

where index =0 at the beginning

- anonymousNg November 12, 2017 | Flag Reply


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