We have to consider cases when an element is zero.

On Thu, Aug 16, 2012 at 7:07 PM, shady <sinv...@gmail.com> wrote:

> well we can do with just one array. Overwrite the answer directly on
> left[] array.
>
>
> On Thu, Aug 16, 2012 at 6:47 PM, mohit <mohitsingh1...@gmail.com> wrote:
>
>>
>> here are the steps :
>> 1) Construct a temporary array left[] such that left[i] contains product
>> of all elements on left of A[i] excluding A[i].
>> 2) Construct another temporary array right[] such that right[i] contains
>> product of all elements on on right of A[i] excluding A[i].
>> 3) To get OUT[], multiply left[] and right[].
>>
>> time complexity : O(n)
>>
>>
>> On Thursday, August 16, 2012 2:26:58 PM UTC+5:30, ram wrote:
>>>
>>>
>>> Hi,
>>>
>>>        This is a microsoft question asked in our campus previous year. 
>>> Anyone having idea please share it here...
>>>
>>>        Given an array of n elements A[n]. Write a program to create a new 
>>> array OUT[n],
>>>
>>>
>>>         which has its elements as multiplication of all the elements in the 
>>> input array A[n] except that element (i.e.) OUT[2] = A[0] * A[1] * A[3] * ? 
>>> * A[n-1].
>>>          Constraint is one should not use division operator.
>>>
>>>  --
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