This will be done in one pass i.e O(n).
On Wed, Apr 14, 2010 at 10:17 PM, gaurav kishan <gauravkis...@gmail.com>wrote:
> Can everyone check this out and let me the issues ?
>
> int[] i=new
> int[]{11,2,3,11,4,11,76,11,11,65,11,44,78,11,13,11,79,11,11,11,56};
> int count=1,element=i[0];
> for(int j=1;j<i.length;j++)
> {
> if(element==i[j])
> count++;
> else
> {
> count--;
> if(count==0)
> {
> element=i[j];
> count=1;
> }
> }
>
> }
> System.out.println("Mode is "+element);
> }
>
> Regards,
> Gaurav Kishan
>
> On Wed, Apr 14, 2010 at 10:01 PM, sharad kumar <aryansmit3...@gmail.com>wrote:
>
>> ya over here its >501 rite?????
>>
>>
>> On Wed, Apr 14, 2010 at 8:24 PM, Prakhar Jain <prakh...@gmail.com> wrote:
>>
>>> If m thinking right,
>>> That works if mode occurs >=n/2 times in the array
>>>
>>> Best,
>>> Prakhar Jain
>>> http://web.iiit.ac.in/~prakharjain/
>>>
>>>
>>> On Wed, Apr 14, 2010 at 8:12 PM, sharad kumar <aryansmit3...@gmail.com
>>> > wrote:
>>>
>>>> can we make use of majority VOTE ALGORITHM?
>>>>
>>>>
>>>> On Wed, Apr 14, 2010 at 4:14 PM, Gauri <gauri...@gmail.com> wrote:
>>>>
>>>>> Say If I have an array of 1,000 32-bit integers .And one of the value
>>>>> is occuring 501 number of times or more in the array. Can someone help
>>>>> me devise an efficient algorithm for the same ?
>>>>>
>>>>> Thanks & Regards
>>>>> Gauri
>>>>>
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