Adding time and space complexities.

Time complexity: O(n)
Space complexity: O(1)


On 29 March 2016 at 23:44, kumar raja <[email protected]> wrote:

> I think it is straight forward. Correct me if i am wrong or if there is
> better solution.
>
> 1) Do one pass over the list of elements and count the number of 1's. let
> us say it is K
> 2)  count = 0
>       from index 0 to K-1 do
>          if array[index] != 1
>             count ++
>          end
>       end
>
> The variable "count"  indicates the minimum number of steps required to
> obtain a sorted list.
>
>
> On 29 March 2016 at 19:30, Régis Bacra <[email protected]>
> wrote:
>
>> This puzzle comes from a contribution on codingame.com (link to the
>> puzzle <https://www.codingame.com/games/community/?puzzleId=103>). Any
>> idea to solve it efficiently?
>>
>> Given a list of 1 and 0, you must regroup all the 1 at the begin of the
>> list in a minimum number of steps. A step is the interchange of two
>> elements located at different positions.
>> The expected result is the minimum number of steps required to obtain a
>> sorted list.
>>
>> Examples:
>> 1 0 1 0 1 -> 1
>> 0 1 0 1 1 1 0 -> 2
>>
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>
>
>
> --
> Regards
> Kumar Raja
>
>
>


-- 
Regards
Kumar Raja

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