Clever. Thanks!

Le mardi 29 mars 2016 20:23:22 UTC+2, kumar raja a écrit :
>
> Adding time and space complexities.
>
> Time complexity: O(n)
> Space complexity: O(1)
>
>
> On 29 March 2016 at 23:44, kumar raja <[email protected] <javascript:>> 
> 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] 
>> <javascript:>> 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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